Free math worksheets

- Focus on the denominator rather than the numerator. - Which input makes the denominator equal to zero?

1. A rational expression is undefined when its denominator is zero. 2. Solve \(x-5=0\), giving the excluded input \(x=5\).

\(x=5\)

- Compare the factors, not the individual terms. - Look for one entire factor that appears in both numerator and denominator.

1. The numerator factors are \(x+4\) and \(x-1\). 2. The denominator factors are \(x+4\) and \(x+2\). 3. The common factor is \(x+4\).

\(x+4\)

- Ask whether the quantity being removed multiplies every term in both numerator and denominator. - Distinguish a factor from a term in a sum.

1. Cancellation is allowed only for a common factor of the entire numerator and denominator. 2. In \(x+6\) and \(x+3\), the \(x\) terms are parts of sums, not common factors. 3. Therefore, the cancellation is not legal.

No. The \(x\) terms are not common factors of the numerator and denominator, so they cannot be canceled.

- Check the denominator in the original expression before simplifying. - Ask whether an algebraic cancellation can change where the original denominator was zero.

1. The original denominator \(x-3\) is zero when \(x=3\), so the original expression is undefined there. 2. Factoring gives \(\frac{(x-3)(x+3)}{x-3}=x+3\) only for \(x\ne3\). 3. Simplification does not restore the excluded input.

No. The original expression is still undefined at \(x=3\), even though the simplified rule \(x+3\) has a value there.

- Factor a common number or variable from a numerator or denominator. - Cancel only common factors of the entire numerator and denominator. - Record the values that make the original denominator zero before canceling. - If the denominator simplifies to \(1\), the fraction bar can be removed.

1. For part a, factor \(5\) from the numerator and \(10a = 5(2a)\): \(\frac{5(a - b)}{5(2a)} = \frac{a - b}{2a}\). The original restriction is \(a \ne 0\). 2. For part b, factor \(x\) from the numerator: \(\frac{x(x + 2)}{x} = x + 2\) for \(x \ne 0\). 3. For part c, factor \(3\) from the denominator: \(\frac{3y}{3(y + 2)} = \frac{y}{y + 2}\). The original restriction is \(y \ne -2\). 4. For part d, factor \(2\) from the numerator: \(\frac{2(z - 2)}{z - 2} = 2\) for \(z \ne 2\).

a) \(\frac{a - b}{2a}\), with \(a \ne 0\). b) \(x + 2\), with \(x \ne 0\). c) \(\frac{y}{y + 2}\), with \(y \ne -2\). d) \(2\), with \(z \ne 2\).

- Factor a common number from a numerator or denominator. - Cancel common variable factors only when they are nonzero. - Cancel factors, not terms in a sum. - Look for identical binomial factors.

1. For part a, divide the numerator and denominator by \(7x\): \(\frac{14x}{21x^2} = \frac{2}{3x}\), with \(x \ne 0\). 2. For part b, factor \(4\) from the numerator: \(\frac{4(2a - 3)}{4} = 2a - 3\). The denominator is the nonzero constant \(4\), so there is no variable restriction. 3. For part c, factor the denominator: \(\frac{x + 2}{2(x + 2)} = \frac{1}{2}\), with \(x \ne -2\). 4. For part d, factor the denominator: \(\frac{5y - 7}{2(5y - 7)} = \frac{1}{2}\), with \(5y - 7 \ne 0\), or \(y \ne \frac{7}{5}\).

a) \(\frac{2}{3x}\), with \(x \ne 0\). b) \(2a - 3\); no variable restriction. c) \(\frac{1}{2}\), with \(x \ne -2\). d) \(\frac{1}{2}\), with \(y \ne \frac{7}{5}\).

- Evaluate each original expression first. - Then evaluate the expression Lisa would obtain. - Compare the two values. - Is the \(5\) a factor or a term in a sum?

1. For part a, the original value is \(\frac{5 \cdot 5}{5 \cdot 10} = \frac{25}{50} = 0.5\). Canceling the common factor gives \(\frac{a}{b} = \frac{5}{10} = 0.5\), so the cancellation is valid. 2. For part b, the original value is \(\frac{5 + 5}{5 + 10} = \frac{10}{15} = \frac{2}{3}\). Crossing out the \(5\) would give \(\frac{a}{b} = \frac{1}{2}\), which is different. The cancellation is invalid. 3. For part c, the original value is \(\frac{5(5 + 1)}{5} = 6\). Canceling the common factor \(5\) gives \(a + 1 = 6\), so the cancellation is valid. 4. Cancellation is valid only for a nonzero factor of the entire numerator and the entire denominator. Terms connected by addition cannot be canceled separately.

a) Valid; both values are \(0.5\). b) Invalid; the values are \(\frac{2}{3}\) and \(\frac{1}{2}\). c) Valid; both values are \(6\). Only common nonzero factors of the entire numerator and denominator may be canceled.

- Factor a common number or variable from the numerator. - Rewrite the numerator as a product before canceling. - The original denominator cannot equal zero. - In part c, compare the factored numerator and denominator.

1. For part a, factor \(6\) from the numerator: \(\frac{6(x + 3)}{12} = \frac{x + 3}{2}\). The denominator is the nonzero constant \(12\), so there is no excluded input. 2. For part b, factor \(x\) from the numerator: \(\frac{x(x - 5)}{x} = x - 5\), with original restriction \(x \ne 0\). 3. For part c, factor: \(\frac{4(x - 2)}{2(x - 2)} = 2\), with original restriction \(x \ne 2\).

a) \(\frac{x + 3}{2}\); no excluded input. b) \(x - 5\), with \(x \ne 0\). c) \(2\), with \(x \ne 2\).

- Distinguish between a product and a sum. - Can each denominator be written with \(x\) as a factor of the entire expression? - Test a value such as \(x=2\) to check the claim.

1. For \(\frac{x}{x^2}\), rewrite the denominator as \(x\cdot x\). For \(x\ne0\), cancel the common factor \(x\): \(\frac{x}{x\cdot x}=\frac{1}{x}\). 2. For \(\frac{x}{x+1}\), the denominator is a sum, and \(x\) is not a factor of the entire denominator. No cancellation is possible. Its restriction is \(x\ne-1\). 3. The claim is false. A repeated variable can be canceled only when it is a nonzero common factor of the entire numerator and denominator, not merely a term that appears in both.

The claim is false. \(\frac{x}{x^2}=\frac{1}{x}\) for \(x\ne0\), but \(\frac{x}{x+1}\) cannot be simplified by cancellation. Cancellation applies to common factors, not terms in a sum.

- Find the greatest common numerical and variable factor in both numerator terms. - Rewrite the numerator as a product. - Cancel only a common nonzero factor of the entire numerator and denominator.

1. Factor the greatest common factor from the numerator: \(14a^2b - 21ab^2 = 7ab(2a - 3b)\). 2. Substitute the factored form: \(\frac{7ab(2a - 3b)}{7ab}\). 3. Cancel the common factor \(7ab\) to obtain \(2a - 3b\). 4. The original denominator requires \(a \ne 0\) and \(b \ne 0\).

\(2a - 3b\), with \(a \ne 0\) and \(b \ne 0\).

- Can you factor a common number from the numerator? - For which value is the original fraction undefined? - Compare the restrictions before and after canceling.

1. Factor the numerator of the fraction: \(4x - 12 = 4(x - 3)\). 2. For \(x \ne 3\), cancel \(x - 3\): \(\frac{4(x - 3)}{x - 3} = 4\). 3. Therefore, the expression simplifies to \(4 + x = x + 4\). 4. The original denominator is zero at \(x = 3\), so \(D_{\text{original}} = \mathbb{R} \setminus \{3\}\). 5. The simplified expression \(x + 4\) is defined for every real number, so \(D_{\text{simplified}} = \mathbb{R}\).

The simplified expression is \(x + 4\). \(D_{\text{original}} = \mathbb{R} \setminus \{3\}\), and \(D_{\text{simplified}} = \mathbb{R}\).

- Find the greatest common factor of the numerator terms. - Rewrite the numerator and denominator as products. - Preserve the zero of the original denominator.

1. Factor the greatest common factor from the numerator: \(12x^2 - 18x = 6x(2x - 3)\). 2. Write the denominator as \(6x^2 = 6x \cdot x\). 3. Cancel the common factor \(6x\): \(\frac{6x(2x - 3)}{6x \cdot x} = \frac{2x - 3}{x}\). 4. The original denominator requires \(x \ne 0\).

\(\frac{2x - 3}{x}\), with \(x \ne 0\).

- When is a rational expression undefined? - Compare the inputs allowed in the original and simplified forms. - What happens when \(x = 0\) is substituted into the original expression?

1. In the original expression, \(x\) is in the denominator. Therefore, \(x \ne 0\), and \(D_{\text{original}} = \mathbb{R} \setminus \{0\}\). 2. Factor the numerator: \(\frac{x(x - 5)}{x}\). For \(x \ne 0\), cancel \(x\) to obtain \(x - 5\). 3. The simplified expression \(x - 5\) is defined for every real number, so \(D_{\text{simplified}} = \mathbb{R}\). 4. Lucas’s statement is incorrect. The domain of an expression must be determined from the form being considered; simplifying can hide restrictions from the original expression.

Lucas is incorrect. \(D_{\text{original}} = \mathbb{R} \setminus \{0\}\), while \(D_{\text{simplified}} = \mathbb{R}\). The original expression remains undefined at \(x = 0\).

- Which operation is undefined in a rational expression? - Set the denominator equal to zero. - Solve the resulting linear equation.

1. Set the denominator equal to zero: \(2x + 14 = 0\). 2. Solve: \(2x = -14\), so \(x = -7\). 3. The expression is undefined at \(x = -7\), so its domain is \(D = \mathbb{R} \setminus \{-7\}\).

The expression is undefined at \(x = -7\). The domain is \(D = \mathbb{R} \setminus \{-7\}\).

- In part a, a product is zero when at least one factor is zero. - In part b, can a real square be negative?

1. For part a, set each denominator factor equal to zero: \(x = 0\) or \(x - 1.5 = 0\), so \(x = 1.5\). 2. Therefore, \(A(x)\) is undefined at \(x = 0\) and \(x = 1.5\), and its domain is \(D = \mathbb{R} \setminus \{0, 1.5\}\). 3. For part b, the denominator would be zero if \(x^2 + 1 = 0\), or \(x^2 = -1\). 4. No real number has a negative square, so the denominator is never zero. The domain is \(D = \mathbb{R}\).

a) Undefined at \(x = 0\) and \(x = 1.5\); \(D = \mathbb{R} \setminus \{0, 1.5\}\). b) The denominator is never zero for real \(x\); \(D = \mathbb{R}\).

- Look for common factors of the entire numerator and denominator. - Record the zeros of the original denominator before any cancellation. - After simplifying, check that the reduced rule is stated together with those original restrictions.

1. Factor the numerator: \(4x-8=4(x-2)\). 2. Factor the denominator: \(2x^2-8=2(x^2-4)=2(x-2)(x+2)\). 3. Cancel the common factor \(x-2\) and reduce \(\frac{4}{2}\): \(\frac{4(x-2)}{2(x-2)(x+2)}=\frac{2}{x+2}\). 4. The original denominator is zero at \(x=2\) and \(x=-2\). Therefore, the claim is correct only with the original restrictions \(x\ne-2,2\).

The claim is correct: \(\frac{4x-8}{2x^2-8}=\frac{2}{x+2}\) for \(x\ne-2,2\).

- Factor the greatest common factor from the denominator. - Rewrite the expression as a quotient of products before canceling. - Preserve all zeros of the original denominator. - Cancel factors, not terms in a sum.

1. Factor the denominator: \(3y^2 - 6y = 3y(y - 2)\). 2. Rewrite the expression as \(\frac{12y}{3y(y - 2)}\). 3. Cancel the common factor \(3y\): \(\frac{12y}{3y(y - 2)} = \frac{4}{y - 2}\). 4. The original denominator is zero at \(y = 0\) and \(y = 2\), so both values remain excluded.

\(\frac{4}{y - 2}\), with \(y \ne 0, 2\).

- Factor the numerator as a difference of squares. - Factor the greatest common factor from the denominator. - Preserve the zero of the original denominator.

1. Factor the numerator as a difference of squares: \(x^2 - 25 = (x - 5)(x + 5)\). 2. Factor the denominator: \(2x + 10 = 2(x + 5)\). 3. Cancel the common factor \(x + 5\): \(\frac{(x - 5)(x + 5)}{2(x + 5)} = \frac{x - 5}{2}\). 4. The original denominator requires \(x \ne -5\).

\(\frac{x - 5}{2}\), with \(x \ne -5\).

- Find the greatest common numerical and variable factor in the numerator terms. - Rewrite the numerator as a product. - Cancel only common factors of the entire numerator and denominator.

1. The denominator confirms the restrictions \(a \ne 0\) and \(b \ne 0\). 2. The greatest common factor of the numerator terms is \(10a^3b^3\). 3. Factor the numerator: \(20a^4b^3 - 30a^3b^4 = 10a^3b^3(2a - 3b)\). 4. Cancel the common factor: \(\frac{10a^3b^3(2a - 3b)}{10a^3b^3} = 2a - 3b\).

\(2a - 3b\), for \(a \ne 0\) and \(b \ne 0\).

- Which value makes the denominator zero? - A rational expression is zero when its numerator is zero and its denominator is not zero. - Substitute the given input and simplify.

1. The denominator is zero when \(x + 4 = 0\), so \(x = -4\). Therefore, \(D = \mathbb{R} \setminus \{-4\}\). 2. Factor the numerator: \(5x - 15 = 5(x - 3)\). The rational expression equals zero when its numerator is zero and its denominator is nonzero. Thus, \(x - 3 = 0\), so the zero is \(x = 3\). 3. Substitute \(x = 1\): \(T(1) = \frac{5(1) - 15}{1 + 4} = \frac{-10}{5} = -2\).

a) \(D = \mathbb{R} \setminus \{-4\}\). b) \(5x - 15 = 5(x - 3)\), so the zero is \(x = 3\). c) \(T(1) = -2\).

- Split a fraction whose numerator is a sum into separate fractions with the same denominator. - Apply the quotient rule for powers with the same base. - Simplify both coefficients and variable factors.

1. Divide each term by \(6a\): \(\frac{12a^2}{6a}+\frac{6}{6a}=2a+\frac{1}{a}\), where \(a\ne0\). 2. Divide each term by \(5x^2\): \(\frac{10x^3}{5x^2}-\frac{5x^2}{5x^2}+\frac{2}{5x^2}=2x-1+\frac{2}{5x^2}\), where \(x\ne0\). 3. Divide each term by \(4m\): \(\frac{8m^2}{4m}-\frac{12m}{4m}+\frac{1}{4m}=2m-3+\frac{1}{4m}\), where \(m\ne0\).

1) \(2a+\frac{1}{a}\), where \(a\ne0\) 2) \(2x-1+\frac{2}{5x^2}\), where \(x\ne0\) 3) \(2m-3+\frac{1}{4m}\), where \(m\ne0\)

- Factor each difference of squares. - Cancel only common factors, not individual terms. - In part 4, compare \(1-b\) with \(b-1\). - Preserve the values excluded by the original denominator.

1. \(x^2-9=(x-3)(x+3)\), so \(\frac{(x-3)(x+3)}{x+3}=x-3\), with \(x\ne-3\). 2. \(16-y^2=(4-y)(4+y)\), so \(\frac{(4-y)(4+y)}{4-y}=4+y\), with \(y\ne4\). 3. \(4a^2-25=(2a-5)(2a+5)\), so the expression simplifies to \(2a-5\), with \(a\ne-\frac{5}{2}\). 4. \(b^2-1=(b-1)(b+1)\) and \(1-b=-(b-1)\). Therefore the expression simplifies to \(-(b+1)=-b-1\), with \(b\ne1\).

1) \(x-3\), where \(x\ne-3\) 2) \(y+4\), where \(y\ne4\) 3) \(2a-5\), where \(a\ne-\frac{5}{2}\) 4) \(-b-1\), where \(b\ne1\)

- Compare each numerator with \(A^2\pm2AB+B^2\). - Cancel a common factor only after factoring completely. - Preserve restrictions from the original denominator.

1. a) \(x^2+14x+49=(x+7)^2\), so the expression simplifies to \(x+7\), where \(x\ne-7\). 2. b) \(4a^2-12ab+9b^2=(2a-3b)^2\), so the expression simplifies to \(2a-3b\), where \(2a-3b\ne0\). 3. c) \(y^4+2y^2+1=(y^2+1)^2\), so the expression simplifies to \(y^2+1\). For real \(y\), the original denominator is always positive.

a) \(x+7\), where \(x\ne-7\) b) \(2a-3b\), where \(2a-3b\ne0\) c) \(y^2+1\)

- Simplify the two rational expressions separately. - Recognize each numerator as a binomial square. - Retain the restrictions from the original denominators.

1. Factor the first numerator: \(k^2-16k+64=(k-8)^2\). Thus the first fraction simplifies to \(k-8\). 2. Factor the second numerator: \(k^2+16k+64=(k+8)^2\). Thus the second fraction simplifies to \(k+8\). 3. Add: \((k-8)+(k+8)=2k\).

\(2k\), with \(k\ne8\) and \(k\ne-8\)

- Division by an expression can be written as a fraction. - A rational expression is undefined when its denominator is zero. - Set each denominator equal to zero and solve. - In part c, the restriction relates two variables.

1. Part a becomes \(\frac{4x + 12}{x - 7}\). The denominator is zero when \(x - 7 = 0\), so the expression is undefined at \(x = 7\). 2. Part b becomes \(\frac{15}{3y + 12} = \frac{15}{3(y + 4)} = \frac{5}{y + 4}\). The original and simplified expressions are undefined at \(y = -4\). 3. Part c becomes \(\frac{a + 3}{a - b}\). The denominator is zero when \(a - b = 0\), so the expression is undefined whenever \(a = b\).

a) \(\frac{4x + 12}{x - 7}\); undefined at \(x = 7\). b) \(\frac{5}{y + 4}\); undefined at \(y = -4\). c) \(\frac{a + 3}{a - b}\); undefined when \(a = b\).

- Cancel common numerical and variable factors. - Apply exponent rules when dividing powers with the same base. - Compare simplified forms only on a domain allowed by every expression being compared.

1. Expression A simplifies by canceling \(5xy\): \(\frac{15x^2y}{20xy^2} = \frac{3x}{4y}\), with \(x \ne 0\) and \(y \ne 0\). 2. Expression B is already simplified as \(\frac{3x}{4y}\), with \(y \ne 0\). 3. Expression C simplifies by canceling \(3x^2\): \(\frac{9x^3}{12x^2y} = \frac{3x}{4y}\), with \(x \ne 0\) and \(y \ne 0\). 4. Expression D simplifies to \(\frac{3x^2}{4y^2}\), with \(y \ne 0\). 5. On the common domain \(x \ne 0\) and \(y \ne 0\), expressions A, B, and C have the same value. Expression D is not equal to them in general.

Expressions A, B, and C are equivalent on the common domain \(x \ne 0\), \(y \ne 0\), and each simplifies to \(\frac{3x}{4y}\). Expression D simplifies to \(\frac{3x^2}{4y^2}\) and is not equivalent to them in general.

- Identify common numerical and variable factors. - Use exponent rules when canceling powers. - Treat an identical binomial factor as one factor. - If all factors in the numerator cancel, a numerator of \(1\) remains.

1. For part a, cancel \(5x^2y^2\): \(\frac{15x^3y^2}{20x^2y^4} = \frac{3x}{4y^2}\), with \(x \ne 0\) and \(y \ne 0\). 2. For part b, cancel \(3(a + b)\): \(\frac{6(a + b)^2}{9(a + b)} = \frac{2(a + b)}{3}\), with \(a + b \ne 0\). 3. For part c, cancel \(4u(v - 3)\): \(\frac{4u(v - 3)}{8u^2(v - 3)} = \frac{1}{2u}\), with \(u \ne 0\) and \(v \ne 3\).

a) \(\frac{3x}{4y^2}\), with \(x \ne 0\) and \(y \ne 0\). b) \(\frac{2(a + b)}{3}\), with \(a + b \ne 0\). c) \(\frac{1}{2u}\), with \(u \ne 0\) and \(v \ne 3\).

- Simplify the numerical coefficients and each variable separately. - Subtract exponents when dividing powers with the same base. - If a larger power remains in the denominator, keep that power below the fraction bar. - Preserve every original denominator restriction.

1. For part a, cancel \(5x^2y^2\): \(\frac{15x^3y^2}{10x^2y^4} = \frac{3x}{2y^2}\), with \(x \ne 0\) and \(y \ne 0\). 2. For part b, cancel \(4a^3b^3\): \(\frac{4a^5b^3}{12a^3b^6} = \frac{a^2}{3b^3}\), with \(a \ne 0\) and \(b \ne 0\). 3. For part c, cancel \(9pq^2\): \(\frac{18p^2q^4}{27pq^2} = \frac{2pq^2}{3}\), with \(p \ne 0\) and \(q \ne 0\). 4. For part d, apply exponent rules: \(\frac{z^7w^3}{z^4w^5} = \frac{z^3}{w^2}\), with \(z \ne 0\) and \(w \ne 0\).

a) \(\frac{3x}{2y^2}\), with \(x \ne 0\), \(y \ne 0\). b) \(\frac{a^2}{3b^3}\), with \(a \ne 0\), \(b \ne 0\). c) \(\frac{2pq^2}{3}\), with \(p \ne 0\), \(q \ne 0\). d) \(\frac{z^3}{w^2}\), with \(z \ne 0\), \(w \ne 0\).

- Use \(\frac{-A}{B} = \frac{A}{-B} = -\frac{A}{B}\). - Reversing the order in a difference introduces a factor of \(-1\). - Factor \(-1\) from either the numerator or denominator.

1. For part a, factor \(-1\) from the numerator: \(7 - x = -(x - 7)\). Thus, \(\frac{7 - x}{z} = -\frac{x - 7}{z}\), with \(z \ne 0\). 2. For part b, factor \(-1\) from the denominator: \(5 - b = -(b - 5)\). Thus, \(\frac{a}{5 - b} = -\frac{a}{b - 5}\), with \(b \ne 5\). 3. For part c, factor \(-1\) from the numerator: \(x - y = -(y - x)\). Thus, \(\frac{x - y}{x + y} = -\frac{y - x}{x + y}\), with \(x + y \ne 0\). 4. For part d, factor \(-1\) from the numerator: \(-3 - m = -(3 + m)\). Thus, \(\frac{-3 - m}{n} = -\frac{3 + m}{n}\), with \(n \ne 0\).

a) \(-\frac{x - 7}{z}\), with \(z \ne 0\). b) \(-\frac{a}{b - 5}\), with \(b \ne 5\). c) \(-\frac{y - x}{x + y}\), with \(x + y \ne 0\). d) \(-\frac{3 + m}{n}\), with \(n \ne 0\).

- Factor common numerical factors first. - Reversing a difference introduces a factor of \(-1\). - Look for factors that are identical except for sign.

1. For part a, \(8 - x = -(x - 8)\), so \(\frac{x - 8}{8 - x} = \frac{x - 8}{-(x - 8)} = -1\), with \(x \ne 8\). 2. For part b, factor \(4a - 12 = 4(a - 3)\) and \(3 - a = -(a - 3)\). Then \(\frac{4(a - 3)}{-(a - 3)} = -4\), with \(a \ne 3\). 3. For part c, factor \(15 - 3y = 3(5 - y) = -3(y - 5)\). Then \(\frac{y - 5}{-3(y - 5)} = -\frac{1}{3}\), with \(y \ne 5\).

a) \(-1\), with \(x \ne 8\). b) \(-4\), with \(a \ne 3\). c) \(-\frac{1}{3}\), with \(y \ne 5\).

- Find a common factor in every term of the numerator. - Factor the denominator when that reveals a common numerical factor. - Cancel only factors joined by multiplication. - Preserve the conditions that make each original denominator nonzero.

1. Factor \(8\) from the numerator: \(\frac{8(a - b)}{16a} = \frac{a - b}{2a}\), with \(a \ne 0\). 2. Factor \(3\) from the numerator and \(12\) from the denominator: \(\frac{3(x + y)}{12(z + w)} = \frac{x + y}{4(z + w)}\), with \(z + w \ne 0\). 3. Factor \(5\) from the numerator: \(\frac{5(k - 3m)}{20k} = \frac{k - 3m}{4k}\), with \(k \ne 0\).

1. \(\frac{a - b}{2a}\), with \(a \ne 0\). 2. \(\frac{x + y}{4(z + w)}\), with \(z + w \ne 0\). 3. \(\frac{k - 3m}{4k}\), with \(k \ne 0\).

- Factor common numerical and variable factors. - Look for a difference of squares and a perfect-square trinomial. - Cancel only identical factors.

1. Factor the numerator and denominator: \(7x^2 - 7xy = 7x(x - y)\) and \(x^2 - y^2 = (x - y)(x + y)\). Cancel \(x - y\) to obtain \(\frac{7x}{x + y}\), with \(x \ne y\) and \(x \ne -y\). 2. Factor the numerator as \((a + 2b)^2\) and the denominator as \(2(a + 2b)\). Cancel \(a + 2b\) to obtain \(\frac{a + 2b}{2}\), with \(a \ne -2b\).

1) \(\frac{7x}{x + y}\), with \(x \ne y\) and \(x \ne -y\). 2) \(\frac{a + 2b}{2}\), with \(a \ne -2b\).

- Factor the greatest common factor from each numerator and denominator. - Rewrite sums or differences as products before canceling. - In part 3, compare the two binomials after factoring out \(-1\). - Cancel only factors, not individual terms.

1. Factor the numerator: \(12ab - 18ac = 6a(2b - 3c)\). Cancel \(6a\) to obtain \(2b - 3c\), with \(a \ne 0\). 2. Factor the numerator and denominator: \(x^2 + xy = x(x + y)\) and \(x^2 - xy = x(x - y)\). Cancel \(x\) to obtain \(\frac{x + y}{x - y}\), with \(x \ne 0\) and \(x \ne y\). 3. Factor the numerator as \(5(u - v)\). Since \(v - u = -(u - v)\), the expression simplifies to \(-5\), with \(u \ne v\).

1) \(2b - 3c\), with \(a \ne 0\). 2) \(\frac{x + y}{x - y}\), with \(x \ne 0\) and \(x \ne y\). 3) \(-5\), with \(u \ne v\).

- Look for a difference of squares and a perfect-square trinomial. - Factor any common constant first. - Cancel only common factors, not terms. - Preserve the zeros of the original denominators.

1. Factor the numerator and denominator: \(x^2 - 16 = (x - 4)(x + 4)\) and \(x^2 + 8x + 16 = (x + 4)^2\). Cancel one factor of \(x + 4\) to obtain \(\frac{x - 4}{x + 4}\), with \(x \ne -4\). 2. Factor: \(2a - 6 = 2(a - 3)\) and \(a^2 - 6a + 9 = (a - 3)^2\). Cancel one factor of \(a - 3\) to obtain \(\frac{2}{a - 3}\), with \(a \ne 3\).

1) \(\frac{x - 4}{x + 4}\), with \(x \ne -4\). 2) \(\frac{2}{a - 3}\), with \(a \ne 3\).

- Division by zero is undefined. - Focus on each denominator. - A product is zero when at least one factor is zero. - A quadratic equation such as \(x^2 = 16\) can have two real solutions.

1. A rational expression is undefined when its denominator equals zero. 2. For part a, solve \(x + 4 = 0\), giving \(x = -4\). 3. For part b, solve \(3x - 15 = 0\), giving \(x = 5\). 4. For part c, factor \(x^2 - 16 = (x - 4)(x + 4)\). The denominator is zero at \(x = 4\) and \(x = -4\). 5. For part d, use the zero-product property: \((x - 3)(x + 5) = 0\) when \(x = 3\) or \(x = -5\).

a) \(x = -4\). b) \(x = 5\). c) \(x = -4\) and \(x = 4\). d) \(x = -5\) and \(x = 3\).

- Factor the numerator before comparing the two algebraic rules. - A canceled factor can simplify a formula without restoring an input excluded by the original denominator. - Use the original denominator, not only the simplified rule, when stating the domain.

1. Factor the numerator: \(x^2-9=(x-3)(x+3)\). 2. For \(x\ne3\), \(r(x)=\frac{(x-3)(x+3)}{x-3}=x+3=s(x)\), so the two graphs coincide wherever \(r\) is defined. 3. In the original expression for \(r\), the factor \(x-3\) is in the denominator. At \(x=3\), that denominator is \(0\), so \(3\) is excluded even though the simplified rule is \(x+3\). 4. The corresponding line value is \(s(3)=6\), which explains the open circle at \((3,6)\) on \(r\). The function \(s\) has no such restriction. 5. Therefore, the domain of \(r\) is \(\mathbb{R}\setminus\{3\}\), while the domain of \(s\) is \(\mathbb{R}\).

a) \(r(x)=x+3\) for \(x\ne3\). b) The graphs coincide for every \(x\ne3\). The original denominator factor \(x-3\) makes \(r\) undefined at \(x=3\), which explains the open circle at \((3,6)\); \(s\) is defined there. c) \(\operatorname{dom}(r)=\mathbb{R}\setminus\{3\}\) and \(\operatorname{dom}(s)=\mathbb{R}\).

- The numerator \(15\) never changes sign. - A quotient is negative when its numerator and denominator have opposite signs. - Translate the required sign of \(x-3\) into an inequality for \(x\).

1. At \(x=3\), the denominator is \(3-3=0\). Division by zero is undefined, so \(x=3\) is excluded from the domain. 2. \(T(0)=\frac{15}{-3}=-5\), \(T(4)=\frac{15}{1}=15\), and \(T(8)=\frac{15}{5}=3\). 3. The numerator \(15\) is always positive. Therefore, \(T(x)<0\) exactly when the denominator is negative. 4. Solve \(x-3<0\), which gives \(x<3\). Thus, \(T(x)<0\) for every \(x\in(-\infty,3)\).

a) \(x=3\) is excluded because it makes the denominator \(0\). b) \(T(0)=-5\), \(T(4)=15\), and \(T(8)=3\). c) \(T(x)<0\) exactly for \(x\in(-\infty,3)\).

- You may cancel common factors, not separate terms in a sum. - Factor the entire numerator and denominator before canceling. - Substitute a simple value to test whether an equation is plausible. - Check the original denominator for restrictions.

1. Part a is incorrect. Terms in a sum cannot be canceled separately. Instead, \(\frac{x+5}{5}=\frac{x}{5}+1\). 2. Part b is correct because \(3\) is a factor of the entire numerator and denominator: \(\frac{3(x-1)}{3}=x-1\). 3. Part c is correct for \(a\ne0\): \(\frac{a(a-1)}{a}=a-1\). 4. Part d is incorrect because Tereza canceled terms rather than factors. Factor \(2\) from the numerator and denominator: \(\frac{2(x+2)}{2(x+4)}=\frac{x+2}{x+4}\), with \(x\ne-4\).

a) Incorrect; \(\frac{x+5}{5}=\frac{x}{5}+1\). b) Correct. c) Correct for \(a\ne0\). d) Incorrect; \(\frac{2x+4}{2x+8}=\frac{x+2}{x+4}\), with \(x\ne-4\).

- Look for a difference of squares or a perfect-square trinomial. - Factor before canceling. - Cancel only identical factors. - Record all zeros of the original denominator before simplifying.

1. For part a, factor the numerator as a difference of squares: \(\frac{(x - 2)(x + 2)}{x + 2} = x - 2\), with \(x \ne -2\). 2. For part b, factor the numerator as a perfect-square trinomial: \(\frac{(x - 3)^2}{x - 3} = x - 3\), with \(x \ne 3\). 3. For part c, factor the numerator and denominator: \(\frac{2(x + 1)}{(x - 1)(x + 1)} = \frac{2}{x - 1}\). The original restrictions are \(x \ne -1\) and \(x \ne 1\).

a) \(x - 2\), with \(x \ne -2\). b) \(x - 3\), with \(x \ne 3\). c) \(\frac{2}{x - 1}\), with \(x \ne -1, 1\).

- Look for a difference of squares or a perfect-square trinomial. - Factor a numerical common factor from the denominator when needed. - Use \(y - 5 = -(5 - y)\) to match factors. - Preserve every zero of the original denominator.

1. For part a, factor the numerator as a difference of squares: \(\frac{(x - 7)(x + 7)}{x - 7} = x + 7\), with \(x \ne 7\). 2. For part b, factor the numerator and denominator: \(\frac{(a + 3)^2}{3(a + 3)} = \frac{a + 3}{3}\), with \(a \ne -3\). 3. For part c, factor the denominator: \(25 - y^2 = (5 - y)(5 + y)\). Because \(y - 5 = -(5 - y)\), \(\frac{y - 5}{(5 - y)(5 + y)} = -\frac{1}{y + 5}\). The original restrictions are \(y \ne -5\) and \(y \ne 5\).

a) \(x + 7\), with \(x \ne 7\). b) \(\frac{a + 3}{3}\), with \(a \ne -3\). c) \(-\frac{1}{y + 5}\), with \(y \ne -5, 5\).

- Factor common constants before using special-product patterns. - Squaring a factored binomial also squares its numerical factor. - Use \(2 - z = -(z - 2)\) to track the sign. - Record the original denominator restrictions before canceling.

1. For part a, factor the numerator and denominator: \(\frac{2(x^2 - 9)}{(x + 3)^2} = \frac{2(x - 3)(x + 3)}{(x + 3)^2} = \frac{2(x - 3)}{x + 3}\), with \(x \ne -3\). 2. For part b, factor: \(10z - 5z^2 = 5z(2 - z) = -5z(z - 2)\) and \(z^2 - 4 = (z - 2)(z + 2)\). Cancel \(z - 2\) to obtain \(-\frac{5z}{z + 2}\). The original restrictions are \(z \ne -2\) and \(z \ne 2\). 3. For part c, rewrite \((4a - 8)^2 = [4(a - 2)]^2 = 16(a - 2)^2\) and \(8a - 16 = 8(a - 2)\). Cancel to obtain \(2(a - 2) = 2a - 4\), with \(a \ne 2\).

a) \(\frac{2(x - 3)}{x + 3}\), with \(x \ne -3\). b) \(-\frac{5z}{z + 2}\), with \(z \ne -2, 2\). c) \(2a - 4\), with \(a \ne 2\).

- Substitute the same value into both expressions. - What would the numerator have to be if it were \((x+3)^2\)? - Compare \(x^2+9\) with the relevant special-product expansion.

1. Substitute \(x=1\) into the original expression: \(\frac{1^2+9}{1+3}=\frac{10}{4}=2.5\). 2. Substitute \(x=1\) into the claimed result: \(1+3=4\). 3. Because \(2.5\ne4\), the claim is false. 4. The numerator \(x^2+9\) is a sum of squares and is not equal to \((x+3)^2\). In fact, \((x+3)^2=x^2+6x+9\). Since \(x+3\) is not a factor of the numerator, it cannot be canceled.

Yasmine is incorrect. At \(x=1\), the original expression equals \(2.5\), while \(x+3=4\). Also, \(x^2+9\ne(x+3)^2\), so \(x+3\) is not a common factor.

- Which values make an original denominator equal to zero? - Look for a common factor in each numerator and denominator. - Compare the excluded values before and after simplifying.

1. For part a, factor the numerator as a difference of squares: \(\frac{(x - 5)(x + 5)}{x - 5}\). For \(x \ne 5\), cancel \(x - 5\) to obtain \(x + 5\). 2. The original expression is undefined at \(x = 5\), so \(D_{\text{original}} = \mathbb{R} \setminus \{5\}\). The simplified expression \(x + 5\) is defined for all real numbers, so \(D_{\text{simplified}} = \mathbb{R}\). The domains are not the same. 3. For part b, factor and simplify: \(\frac{3(x + 2)}{x} \cdot \frac{x^2}{x + 2} = 3x\), with the original restrictions \(x \ne 0\) and \(x \ne -2\). 4. Thus, \(D_{\text{original}} = \mathbb{R} \setminus \{-2, 0\}\). The simplified expression \(3x\) is defined for all real numbers, so \(D_{\text{simplified}} = \mathbb{R}\). The domains are not the same.

a) Simplified expression: \(x + 5\); \(D_{\text{original}} = \mathbb{R} \setminus \{5\}\); \(D_{\text{simplified}} = \mathbb{R}\). The domains are not the same. b) Simplified expression: \(3x\); \(D_{\text{original}} = \mathbb{R} \setminus \{-2, 0\}\); \(D_{\text{simplified}} = \mathbb{R}\). The domains are not the same.

- Which value makes the denominator of \(T_1\) equal to zero? - Factor the numerator of \(T_1\). - Compare what happens to the two expressions at \(x = -5\).

1. The denominator of \(T_1\) is zero when \(x = -5\), so \(D_1 = \mathbb{R} \setminus \{-5\}\). The expression \(T_2(x) = x\) is defined for every real number, so \(D_2 = \mathbb{R}\). 2. Factor the numerator: \(T_1(x) = \frac{x(x + 5)}{x + 5}\). For \(x \ne -5\), cancel \(x + 5\) to obtain \(T_1(x) = x = T_2(x)\). 3. At \(x = -5\), \(T_1\) is undefined because its denominator is zero, while \(T_2(-5) = -5\). Therefore, the expressions do not have the same domain and cannot be treated as identical for all real numbers.

a) \(D_1 = \mathbb{R} \setminus \{-5\}\); \(D_2 = \mathbb{R}\). b) \(T_1(x) = \frac{x(x + 5)}{x + 5} = x\) for \(x \ne -5\). c) \(T_1\) is undefined at \(x = -5\), but \(T_2\) is defined there.

- When is a rational expression undefined? - Factor the denominator and find its zeros. - What condition makes a rational expression equal to zero? - Which part of the expression determines its domain restrictions?

1. Factor the denominator: \(x^2 + 3x = x(x + 3)\). 2. The denominator is zero at \(x = 0\) and \(x = -3\), so the domain is \(D = \mathbb{R} \setminus \{-3, 0\}\). 3. These values are excluded because division by zero is undefined. 4. Keep the same denominator to preserve the domain, but choose a numerator with a different zero. For example, \(S(x) = \frac{x + 1}{x^2 + 3x}\) has the same domain and has a zero at \(x = -1\).

a) \(D = \mathbb{R} \setminus \{-3, 0\}\). b) At \(x = -3\) and \(x = 0\), the denominator is zero, so the expression is undefined. c) One possible answer is \(S(x) = \frac{x + 1}{x^2 + 3x}\), which has a zero at \(x = -1\).

- Find every value that makes each original denominator zero. - Check whether \(x=3\) belongs to both domains. - Simplifying an expression does not erase restrictions from its original form. - Must equivalent expressions have the same domain?

1. For \(A(x)\), the denominator is zero at \(x=-4\), so \(D_A=\mathbb{R}\setminus\{-4\}\). 2. For \(B(x)\), the denominator is zero at \(x=-4\) and \(x=3\), so \(D_B=\mathbb{R}\setminus\{-4,3\}\). 3. Evaluate \(A\) at \(x=3\): \(A(3)=\frac{2}{3+4}=\frac{2}{7}\). 4. The value \(x=3\) is not in \(D_B\), so \(B(3)\) is undefined. 5. Although \(2x-6=2(x-3)\) and \(B(x)\) simplifies to \(\frac{2}{x+4}\) for \(x\ne3\), the original expressions have different domains. Ren's claim is false.

a) \(D_A=\mathbb{R}\setminus\{-4\}\); \(D_B=\mathbb{R}\setminus\{-4,3\}\). b) \(A(3)=\frac{2}{7}\); \(B(3)\) is undefined. c) The claim is false because the expressions have different domains. They agree only where both are defined.

- Factor out the common numerical factor first. - Factor the difference of squares in the denominator. - Use \(3 - y = -(y - 3)\). - Preserve every zero of the original denominator.

1. Factor \(2\) from the numerator and denominator: \(\frac{2(3 - y)}{2(y^2 - 9)} = \frac{3 - y}{y^2 - 9}\). 2. Factor the denominator as a difference of squares: \(y^2 - 9 = (y - 3)(y + 3)\). 3. Rewrite the numerator as \(3 - y = -(y - 3)\). 4. Cancel \(y - 3\): \(\frac{-(y - 3)}{(y - 3)(y + 3)} = -\frac{1}{y + 3}\). 5. The original denominator is zero at \(y = -3\) and \(y = 3\), so both values remain excluded.

\(-\frac{1}{y + 3}\), with \(y \ne -3, 3\).

- Determine the zeros of the original denominator before simplifying. - Keep track of every excluded value from the original expression. - Compare the original expression and its simplified rule specifically at \(x=3\).

1. Set the original denominator equal to zero: \(x^2-9=0\). 2. Factor: \((x-3)(x+3)=0\), so \(x=3\) or \(x=-3\). 3. Therefore, the domain of the original expression is \(D=\mathbb{R}\setminus\{-3,3\}\). 4. For \(x\ne3\), the expression simplifies as \(\frac{x-3}{(x-3)(x+3)}=\frac{1}{x+3}\). 5. Canceling \(x-3\) does not add \(x=3\) to the original domain. At that input, the original denominator is zero, so the original expression remains undefined.

a) The expression is undefined at \(x=-3\) and \(x=3\). b) \(D=\mathbb{R}\setminus\{-3,3\}\). c) The cancellation is valid only when \(x\ne3\). The original expression still has a zero denominator at \(x=3\).

- Factor out the \(5\) in the numerator. - Look for a difference of squares and a perfect-square trinomial. - Cancel only common factors and preserve the original restriction.

1. Factor the numerator: \(5x^2 - 20 = 5(x^2 - 4) = 5(x - 2)(x + 2)\). 2. Factor the denominator as a perfect-square trinomial: \(x^2 - 4x + 4 = (x - 2)^2\). 3. Cancel one factor of \(x - 2\): \(\frac{5(x - 2)(x + 2)}{(x - 2)^2} = \frac{5(x + 2)}{x - 2}\). 4. The original denominator requires \(x \ne 2\).

\(\frac{5(x + 2)}{x - 2}\), with \(x \ne 2\).

- Factor the numerator as a perfect square. - Factor the denominator as a difference of squares. - Preserve both factors that can make the original denominator zero.

1. Factor the numerator as a perfect-square trinomial: \(a^2 + 2ab + b^2 = (a + b)^2\). 2. Factor the denominator as a difference of squares: \(a^2 - b^2 = (a + b)(a - b)\). 3. Cancel one factor of \(a + b\): \(\frac{(a + b)^2}{(a + b)(a - b)} = \frac{a + b}{a - b}\). 4. The original denominator requires \(a + b \ne 0\) and \(a - b \ne 0\), equivalently \(a \ne -b\) and \(a \ne b\).

\(\frac{a + b}{a - b}\), with \(a \ne -b\) and \(a \ne b\).

- Does the numerator match a perfect-square trinomial? - Factor out a common number from the denominator before using the difference of squares. - Which values make the original denominator zero? - Which factors appear in both the numerator and denominator?

1. Factor the numerator as a perfect square: \(x^2-6x+9=(x-3)^2\). 2. Factor the denominator: \(4x^2-36=4(x^2-9)=4(x-3)(x+3)\). 3. The original denominator is zero at \(x=-3\) and \(x=3\), so the domain is \(D_f=\mathbb{R}\setminus\{-3, 3\}\). 4. Cancel one common factor of \(x-3\): \(f(x)=\frac{x-3}{4(x+3)}\), with the original domain restrictions still in effect.

a) \(f(x)=\frac{(x-3)^2}{4(x-3)(x+3)}\) b) \(D_f=\mathbb{R}\setminus\{-3, 3\}\), and the simplified expression is \(f(x)=\frac{x-3}{4(x+3)}\).

- Use the zeros of a quadratic to write it as a product of linear factors. - First factor out the leading coefficient from the numerator. - Only factors, not terms in a sum, may be canceled. - Which values make the original denominator zero?

1. Factor out \(2\) from the numerator: \(2x^2-4x-6=2(x^2-2x-3)=2(x-3)(x+1)\). 2. Factor the denominator: \(x^2+5x+4=(x+4)(x+1)\). 3. The original denominator is zero at \(x=-4\) and \(x=-1\), so \(D_g=\mathbb{R}\setminus\{-4, -1\}\). 4. Cancel the common factor \(x+1\): \(g(x)=\frac{2(x-3)}{x+4}\), with the original domain restrictions still in effect.

Factored form: \(g(x)=\frac{2(x-3)(x+1)}{(x+4)(x+1)}\) Domain: \(D_g=\mathbb{R}\setminus\{-4, -1\}\) Simplified expression: \(g(x)=\frac{2(x-3)}{x+4}\)

- Compare the values excluded by the two denominators. - Factor the numerator of each rational expression. - Can the separate \(1\) in \(f(x)\) be combined with the fraction? - Two functions are identical only when both their domains and outputs agree.

1. Both denominators are zero at \(x=2\), so both functions have the domain \(\mathbb{R}\setminus\{2\}\). 2. Factor and simplify \(f\): \(f(x)=\frac{(x-2)(x+3)}{x-2}+1=x+3+1=x+4\) for \(x\neq 2\). 3. Simplify \(g\): \(g(x)=\frac{2(x^2+2x-8)}{2(x-2)}=\frac{(x-2)(x+4)}{x-2}=x+4\) for \(x\neq 2\). 4. The two functions have the same domain and the same output \(x+4\) for every input in that domain. Therefore, they are identical.

Both functions have domain \(\mathbb{R}\setminus\{2\}\), and both simplify to \(x+4\) on that domain. Therefore, \(f=g\).

- Distribute \(x^2\) through the numerator first. - Can you factor out a common power of \(x\) from the numerator and denominator? - Does the remaining quadratic match a perfect-square trinomial? - Which complete factors appear in both the numerator and denominator?

1. The original expression requires \(x\neq 0\) because of \(x^{-1}\). The denominator also requires \(x\neq 3\). 2. Distribute \(x^2\) in the numerator: \(x^2(x-6+9x^{-1})=x^3-6x^2+9x\). 3. Factor the numerator: \(x^3-6x^2+9x=x(x^2-6x+9)=x(x-3)^2\). 4. Factor the denominator: \(x^2-3x=x(x-3)\). 5. Cancel the common factors to obtain \(x-3\), while retaining the original restrictions \(x\neq 0\) and \(x\neq 3\).

\(x-3\), for \(x\neq 0,3\)

- Distribute \(x^4\) through the terms in parentheses. - Factor the resulting numerator and denominator completely. - Look for a difference of squares and a perfect-square trinomial. - Which values make any part of the original expression undefined?

1. The original expression requires \(x\neq 0\). The denominator also requires \(x\neq -1\). 2. Simplify the numerator: \(x^4\left(\frac{1}{x}-\frac{1}{x^3}\right)=x^3-x\). 3. Factor the numerator: \(x^3-x=x(x^2-1)=x(x-1)(x+1)\). 4. Factor the denominator: \(x^2+2x+1=(x+1)^2\). 5. Cancel one common factor of \(x+1\): \(\frac{x(x-1)}{x+1}\), with the original restrictions \(x\neq 0\) and \(x\neq -1\).

\(\frac{x(x-1)}{x+1}\), for \(x\neq 0,-1\)

- Before canceling powers of the variable, what value would make the original denominator zero? - Simplify each variable quotient before doing the remaining equation steps. - After solving, compare the result with the excluded value from the original expression.

1. In (a), the original denominator requires \(x \ne 0\). Simplify \((12x^2) \div (4x)\) to \(3x\), so \(3x=18\) and \(x=6\). 2. In (b), the original denominator requires \(y \ne 0\). Simplify \((40y^3) \div (8y^2)\) to \(5y\), so \(5y+13=48\) and \(y=7\). 3. In (c), the original denominator requires \(z \ne 0\). Simplify \((18z^2) \div (6z)\) to \(3z\), so \(22z-3z=38\), giving \(z=2\). 4. Each solution is allowed by its original nonzero restriction.

a) \(x = 6\), with excluded value \(x=0\) b) \(y = 7\), with excluded value \(y=0\) c) \(z = 2\), with excluded value \(z=0\)

- What value is excluded by the original denominator \(9y\)? - Simplify the quotient term by term before solving the equation. - Check the final value against the original restriction.

1. The original denominator requires \(y \ne 0\). 2. Divide each numerator term by \(9y\): \((18y^2-9y)\div(9y)=2y-1\). 3. Solve \(14+2y-1=20\), so \(13+2y=20\), \(2y=7\), and \(y=\frac{7}{2}\). 4. Since \(\frac{7}{2}\ne0\), the solution is allowed.

\(y = \frac{7}{2}\), with excluded value \(y=0\)

- Identify the value that would make either original denominator zero. - Simplify the two quotients separately before combining them. - Be careful with the subtraction in front of the second simplified expression.

1. The original denominators require \(x \ne 0\). 2. Simplify the first quotient: \((15x^2+10x)\div(5x)=3x+2\). 3. Simplify the second quotient: \((12x^2-18x)\div(6x)=2x-3\). 4. Solve \((3x+2)-(2x-3)=1\). This gives \(x+5=1\), so \(x=-4\). 5. Since \(-4\ne0\), the solution is allowed.

\(x = -4\), with excluded value \(x=0\)

- Divide every term of the numerator by \(2x\). - The polynomial part is the part with no variable in a denominator. - Use the rewritten expression for parts b and c.

1. Divide each term in the numerator by \(2x\): \(T(x)=\frac{4x^2}{2x}-\frac{2x}{2x}+\frac{7}{2x}=2x-1+\frac{7}{2x}\), where \(x\ne0\). 2. For \(x=0.5\), \(T(0.5)=2\cdot0.5-1+\frac{7}{2\cdot0.5}=7\). 3. The polynomial part is \(2x-1\). Solve \(2x-1=10\): \(2x=11\), so \(x=5.5\).

a) \(2x-1+\frac{7}{2x}\), where \(x\ne0\) b) \(7\) c) \(x=5.5\)

- Look for algebraic structures that reveal common factors in each expression. - Pay close attention to the sign relationship between \(A-B\) and \(B-A\). - Preserve every input excluded by an original denominator.

1. Factor \(25-z^2=(5-z)(5+z)\). Dividing by \(D=5+z\) gives \(5-z\), provided \(z\ne-5\). 2. Factor \(k^2-16=(k-4)(k+4)\), while \(4-k=-(k-4)\). Thus \(\frac{(k-4)(k+4)}{-(k-4)}=-(k+4)=-k-4\), for \(k\ne4\). The claim is false. 3. Factor \(9y^2-64=(3y-8)(3y+8)\). The expression simplifies to \(3y-8\), where \(y\ne-\frac{8}{3}\).

a) \(D=5+z\), with \(z\ne-5\) b) The claim is false; the result is \(-k-4\), with \(k\ne4\). c) \(3y-8\), with \(y\ne-\frac{8}{3}\)

- Translate “quotient,” “sum,” and “difference” one part at a time. - Factor \(k^2 - 25\) as a difference of squares. - Which values make the original denominator zero?

1. The numerator is \(k + 5\), and the denominator is \(k^2 - 25\). The rational expression is \(\frac{k + 5}{k^2 - 25}\). 2. Factor the denominator as a difference of squares: \(k^2 - 25 = (k - 5)(k + 5)\). 3. The original denominator is zero at \(k = 5\) and \(k = -5\), so the original domain is \(D = \mathbb{R} \setminus \{-5, 5\}\). 4. For values in the original domain, cancel \(k + 5\) to obtain \(\frac{1}{k - 5}\). The original restrictions \(k \ne -5\) and \(k \ne 5\) must still be retained when this simplified form represents the original expression.

a) \(\frac{k + 5}{k^2 - 25}\). b) The expression simplifies to \(\frac{1}{k - 5}\), with original domain \(D = \mathbb{R} \setminus \{-5, 5\}\). The excluded values make the original denominator zero.

- Which input makes the denominator zero? - Factor \(x\) from the numerator before canceling. - Division by zero is undefined. - Does a zero numerator guarantee a value of zero when the denominator is also zero?

1. The denominator is \(x\), so \(x \ne 0\). The original domain is \(D = \mathbb{R} \setminus \{0\}\). 2. Factor the numerator: \(x^2 - 4x = x(x - 4)\). For \(x \ne 0\), \(T(x) = \frac{x(x - 4)}{x} = x - 4\). 3. Evaluate at \(x = 4\): \(T(4) = 4 - 4 = 0\). 4. At \(x = 0\), the original denominator is also zero. A rational expression with denominator zero is undefined, even when the numerator is zero. Therefore, \(T(0)\) has no value.

a) \(D = \mathbb{R} \setminus \{0\}\). b) \(T(x) = x - 4\) for \(x \ne 0\), and \(T(4) = 0\). c) \(T(0)\) is undefined because the original denominator equals zero.

- Substitute \(3a\) and \(3b\) using parentheses. - Simplify the transformed expression and compare it with the original. - Factor common constants from the entire numerator and denominator. - Apply exponent rules carefully to expressions such as \((3a)^2\).

1. For part a, \(\frac{5(3a)}{3b} = \frac{15a}{3b} = \frac{5a}{b}\), so the value is unchanged. 2. For part b, \(\frac{3a - 3b}{3(3a)} = \frac{3(a - b)}{9a} = \frac{a - b}{3a}\), so the value is unchanged. 3. For part c, \(\frac{(3a)^2}{(3b)^2 + 3a} = \frac{9a^2}{9b^2 + 3a} = \frac{3a^2}{3b^2 + a}\). This is not generally equal to \(\frac{a^2}{b^2 + a}\). For example, when \(a = 1\) and \(b = 1\), the values are \(\frac{1}{2}\) and \(\frac{3}{4}\). 4. For part d, \(\frac{(3a)^2 + (3b)^2}{(3a)(3b)} = \frac{9(a^2 + b^2)}{9ab} = \frac{a^2 + b^2}{ab}\), so the value is unchanged.

a) Unchanged. b) Unchanged. c) Not unchanged in general. d) Unchanged. Thus, a), b), and d) are invariant under the simultaneous scaling.

- Factor before canceling; do not cancel terms in a sum. - Find factors common to every term in a numerator or denominator. - In part c, use \(n - m = -(m - n)\). - Preserve all zeros of each original denominator.

1. For part a, factor \(4\) from the numerator: \(\frac{4(a + 2b)}{12} = \frac{a + 2b}{3}\). The denominator is a nonzero constant, so there is no variable restriction. 2. For part b, factor \(x\): \(\frac{x(x - 5)}{x} = x - 5\), with \(x \ne 0\). 3. For part c, factor \(7\) from the numerator and \(2\) from the denominator: \(\frac{7(m - n)}{2(n - m)}\). Since \(n - m = -(m - n)\), the result is \(-\frac{7}{2}\), with \(m \ne n\). 4. For part d, factor: \(\frac{3k(k + 2)}{k(k + 2)} = 3\), with \(k \ne 0\) and \(k \ne -2\).

a) \(\frac{a + 2b}{3}\); no variable restriction. b) \(x - 5\), with \(x \ne 0\). c) \(-\frac{7}{2}\), with \(m \ne n\). d) \(3\), with \(k \ne 0, -2\).

- Find the input that makes the denominator zero. - Factor the numerator as a difference of squares. - Factor a common number from the denominator. - Use order of operations when substituting.

1. The denominator is zero when \(3x + 12 = 0\), so \(x = -4\). Thus, the original domain is \(D = \mathbb{R} \setminus \{-4\}\). 2. Factor the numerator and denominator: \(x^2 - 16 = (x - 4)(x + 4)\) and \(3x + 12 = 3(x + 4)\). 3. Cancel \(x + 4\): \(\frac{(x - 4)(x + 4)}{3(x + 4)} = \frac{x - 4}{3}\), with \(x \ne -4\). 4. Using the original form, \(T(2) = \frac{2^2 - 16}{3 \cdot 2 + 12} = \frac{-12}{18} = -\frac{2}{3}\). 5. Using the simplified form, \(T(2) = \frac{2 - 4}{3} = -\frac{2}{3}\). The values agree.

a) \(D = \mathbb{R} \setminus \{-4\}\). b) \(\frac{x - 4}{3}\), with \(x \ne -4\). c) Both forms give \(-\frac{2}{3}\).

- Factor before canceling; do not cancel terms in a sum. - Look for common numerical or variable factors. - Use a difference of squares when it appears. - Cancel only identical factors. - Preserve the zeros of the original denominators.

1. For part a, factor: \(\frac{3(x + 2)}{x(x + 2)} = \frac{3}{x}\), with \(x \ne 0\) and \(x \ne -2\). 2. For part b, factor: \(\frac{a(a - 1)}{(a - 1)(a + 1)} = \frac{a}{a + 1}\), with \(a \ne -1\) and \(a \ne 1\). 3. For part c, factor: \(\frac{5y(y + 2)}{15y} = \frac{y + 2}{3}\), with \(y \ne 0\). 4. For part d, factor: \(\frac{2(b - 2)}{(b - 2)(b + 2)} = \frac{2}{b + 2}\), with \(b \ne -2\) and \(b \ne 2\).

a) \(\frac{3}{x}\), with \(x \ne 0, -2\). b) \(\frac{a}{a + 1}\), with \(a \ne -1, 1\). c) \(\frac{y + 2}{3}\), with \(y \ne 0\). d) \(\frac{2}{b + 2}\), with \(b \ne -2, 2\).

- Factor common constants and variables. - Reversing the order of a difference introduces a factor of \(-1\). - Reduce numerical coefficients separately from variable factors. - Look for binomial factors that match except for sign.

1. Factor the numerator: \(7a - 7b = 7(a - b)\). Since \(b - a = -(a - b)\), the expression simplifies to \(-7\), with \(a \ne b\). 2. Factor: \(x^2 - 3x = x(x - 3)\) and \(6 - 2x = 2(3 - x) = -2(x - 3)\). Cancel \(x - 3\) to obtain \(-\frac{x}{2}\), with \(x \ne 3\). 3. Use \(s - 2r = -(2r - s)\). Then cancel \(2r - s\), reduce \(\frac{12}{18}\), and cancel one factor each of \(u\) and \(v\): the result is \(-\frac{2u}{3v}\). The original restrictions are \(u \ne 0\), \(v \ne 0\), and \(s \ne 2r\).

1) \(-7\), with \(a \ne b\). 2) \(-\frac{x}{2}\), with \(x \ne 3\). 3) \(-\frac{2u}{3v}\), with \(u \ne 0\), \(v \ne 0\), and \(s \ne 2r\).

- Identify common factors of the entire numerator and denominator before canceling. - Ask whether \(x\) divides every term in the numerator. - Check the original denominator for restrictions and use either equivalent form to verify the numerical value.

1. Factor the numerator: \(4x-12=4(x-3)\). Then \(\frac{4(x-3)}{8x}=\frac{x-3}{2x}\), with \(x\ne0\). 2. Ximena's method is incorrect because \(x\) is not a factor of the entire numerator \(4x-12\). Terms in a difference cannot be canceled separately. 3. Substitute \(x=5\) into the simplified expression: \(T(5)=\frac{5-3}{2\cdot5}=\frac{2}{10}=\frac{1}{5}=0.2\).

a) \(\frac{x-3}{2x}\), with \(x\ne0\). b) Cancellation applies to common factors of the entire numerator and denominator, not to individual terms in a difference. c) \(T(5)=\frac{1}{5}=0.2\).

- Factor common constants or variables first. - Look for differences of squares and perfect-square trinomials. - Cancel only identical factors. - Preserve all zeros of the original denominators.

1. Factor: \(\frac{7(x - 2)}{(x - 2)(x + 2)} = \frac{7}{x + 2}\), with \(x \ne -2\) and \(x \ne 2\). 2. Factor: \(b^2 + 10b + 25 = (b + 5)^2\) and \(b^2 - 25 = (b - 5)(b + 5)\). Cancel one factor of \(b + 5\) to obtain \(\frac{b + 5}{b - 5}\), with \(b \ne -5\) and \(b \ne 5\). 3. Factor: \(\frac{z(z - 1)}{(z - 1)^2} = \frac{z}{z - 1}\), with \(z \ne 1\).

1) \(\frac{7}{x + 2}\), with \(x \ne -2, 2\). 2) \(\frac{b + 5}{b - 5}\), with \(b \ne -5, 5\). 3) \(\frac{z}{z - 1}\), with \(z \ne 1\).

- Factor out a common constant before using a special-product pattern. - Look for a difference of squares. - In part 3, factor in more than one stage. - Continue checking for common factors after each step.

1. Factor: \(2a^2 - 32 = 2(a^2 - 16) = 2(a - 4)(a + 4)\) and \(a^2 - 8a + 16 = (a - 4)^2\). Cancel one factor of \(a - 4\) to obtain \(\frac{2(a + 4)}{a - 4}\), with \(a \ne 4\). 2. Factor: \(x^2 - y^2 = (x - y)(x + y)\) and \(2x + 2y = 2(x + y)\). Cancel \(x + y\) to obtain \(\frac{x - y}{2}\), with \(x + y \ne 0\). 3. Factor: \(m^3 - m = m(m^2 - 1) = m(m - 1)(m + 1)\) and \(m^2 - m = m(m - 1)\). Cancel \(m(m - 1)\) to obtain \(m + 1\), with \(m \ne 0\) and \(m \ne 1\).

1) \(\frac{2(a + 4)}{a - 4}\), with \(a \ne 4\). 2) \(\frac{x - y}{2}\), with \(x + y \ne 0\). 3) \(m + 1\), with \(m \ne 0, 1\).

- Factor common constants first. - Look for differences of squares and perfect-square trinomials. - Cancel only factors, not terms in a sum. - Remember that squaring a binomial removes the effect of reversing its sign.

1. Factor: \(\frac{5(x - 3)}{(x - 3)(x + 3)} = \frac{5}{x + 3}\), with \(x \ne -3\) and \(x \ne 3\). 2. Factor: \(\frac{(a + 2)^2}{3(a + 2)} = \frac{a + 2}{3}\), with \(a \ne -2\). 3. Factor the numerator: \(y^2 - x^2 = (y - x)(y + x)\). Because \((x - y)^2 = (y - x)^2\), cancel one factor of \(y - x\) to obtain \(\frac{x + y}{y - x}\), with \(x \ne y\).

1) \(\frac{5}{x + 3}\), with \(x \ne -3, 3\). 2) \(\frac{a + 2}{3}\), with \(a \ne -2\). 3) \(\frac{x + y}{y - x}\), with \(x \ne y\).

- Look for common factors while keeping track of sign changes. - Record all zeros of the original denominator before simplifying. - Distinguish an indeterminate-looking symbol such as \(\frac00\) from an actual defined value.

1. Factor the numerator as \(x(x-1)\). Factor the denominator as \(1-x^2=(1-x)(1+x)=-(x-1)(x+1)\). 2. Cancel \(x-1\): \(T(x)=-\frac{x}{x+1}\), with the original restrictions \(x\ne-1\) and \(x\ne1\). 3. At \(x=1\), the original denominator is zero. Division by zero is undefined, and \(\frac{0}{0}\) does not equal zero. 4. Substitute \(x=-2\) into the simplified form: \(T(-2)=-\frac{-2}{-2+1}=\frac{2}{-1}=-2\).

1. \(-\frac{x}{x+1}\), with \(x\ne-1,1\). 2. \(\frac{0}{0}\) is undefined, not zero. 3. \(T(-2)=-2\).

- Factor differences of squares. - Factor out \(-1\) when two binomials have reversed terms. - Remove common numerical and variable factors before canceling binomials.

1. Factor the numerator as \((x - 4y)(x + 4y)\). Factor the denominator as \(12y - 3x = -3(x - 4y)\). Cancel \(x - 4y\) to obtain \(-\frac{x + 4y}{3}\), with \(x \ne 4y\). 2. Factor the numerator: \(3a^3b - 3ab^3 = 3ab(a^2 - b^2) = 3ab(a - b)(a + b)\). Factor the denominator: \(6a^2b + 6ab^2 = 6ab(a + b)\). Cancel \(3ab(a + b)\) to obtain \(\frac{a - b}{2}\), with \(a \ne 0\), \(b \ne 0\), and \(a \ne -b\).

1) \(-\frac{x + 4y}{3}\), with \(x \ne 4y\). 2) \(\frac{a - b}{2}\), with \(a \ne 0\), \(b \ne 0\), and \(a \ne -b\).

- Use a difference of squares when possible. - Factor common variables before applying a special-product pattern. - Factor the numerator and denominator separately. - Cancel only identical factors.

1. Factor: \(\frac{(z - 3)(z + 3)}{z(z + 3)} = \frac{z - 3}{z}\), with \(z \ne 0\) and \(z \ne -3\). 2. Factor: \(\frac{ab(a - b)}{(a - b)(a + b)} = \frac{ab}{a + b}\), with \(a \ne b\) and \(a \ne -b\). 3. Factor: \(2x^3 - 2xy^2 = 2x(x^2 - y^2) = 2x(x - y)(x + y)\) and \(x^2 + xy = x(x + y)\). Cancel \(x(x + y)\) to obtain \(2(x - y)\), with \(x \ne 0\) and \(x \ne -y\).

1) \(\frac{z - 3}{z}\), with \(z \ne 0, -3\). 2) \(\frac{ab}{a + b}\), with \(a \ne b\) and \(a \ne -b\). 3) \(2(x - y)\), with \(x \ne 0\) and \(x \ne -y\).

- View \(x^4\) as \((x^2)^2\). - Apply the difference-of-squares pattern. - Look for a factor common to the numerator and denominator. - Check whether any remaining denominator can be zero over the real numbers.

1. Factor the denominator as a difference of squares: \(x^4 - 81 = (x^2 - 9)(x^2 + 9)\). Cancel \(x^2 - 9\) to obtain \(\frac{1}{x^2 + 9}\). The original denominator is zero at \(x = -3\) and \(x = 3\). 2. Factor the numerator as a difference of squares: \(16a^4 - 1 = (4a^2 - 1)(4a^2 + 1)\). Cancel \(4a^2 + 1\) to obtain \(4a^2 - 1\). Because \(4a^2 + 1 > 0\) for every real \(a\), there is no real restriction.

1) \(\frac{1}{x^2 + 9}\), with \(x \ne -3, 3\). 2) \(4a^2 - 1\); no real restriction.

- Look for perfect-square trinomials and differences of squares. - Factor out common numerical factors first. - Cancel only factors common to the entire numerator and denominator. - Preserve every zero of the original denominator.

1. For part a, factor the numerator and denominator: \(x^2 - 10x + 25 = (x - 5)^2\) and \(x^2 - 25 = (x - 5)(x + 5)\). Cancel one factor of \(x - 5\) to obtain \(\frac{x - 5}{x + 5}\), with \(x \ne -5\) and \(x \ne 5\). 2. For part b, factor: \(2a^2 + 4ab + 2b^2 = 2(a + b)^2\) and \(6a^2 - 6b^2 = 6(a - b)(a + b)\). Reduce \(\frac{2}{6}\) and cancel one factor of \(a + b\) to obtain \(\frac{a + b}{3(a - b)}\), with \(a \ne b\) and \(a \ne -b\).

a) \(\frac{x - 5}{x + 5}\), with \(x \ne -5, 5\). b) \(\frac{a + b}{3(a - b)}\), with \(a \ne b\) and \(a \ne -b\).

- First factor out any common numerical factors. - Look for a perfect-square trinomial and a difference of squares. - Cancel only factors common to the entire numerator and denominator. - Preserve the zeros of the original denominator.

1. Factor out the greatest common factors: \(\frac{4(a^2 - 4ab + 4b^2)}{6(a^2 - 4b^2)}\). 2. Factor the perfect-square trinomial and the difference of squares: \(a^2 - 4ab + 4b^2 = (a - 2b)^2\) and \(a^2 - 4b^2 = (a - 2b)(a + 2b)\). 3. Substitute the factors: \(\frac{4(a - 2b)^2}{6(a - 2b)(a + 2b)}\). 4. Reduce \(\frac{4}{6}\) and cancel one factor of \(a - 2b\) to obtain \(\frac{2(a - 2b)}{3(a + 2b)}\). 5. Preserve the restrictions from the original denominator: \(a \ne 2b\) and \(a \ne -2b\).

\(\frac{2(a - 2b)}{3(a + 2b)}\), with \(a \ne 2b\) and \(a \ne -2b\).

- Look for perfect-square trinomials inside the numerator and denominator. - Rewrite each expression as a difference of squares. - Factor completely before canceling. - Substitute the values only after simplifying.

1. Rewrite the numerator as a difference of squares: \(x^2 + 2xy + y^2 - z^2 = (x + y)^2 - z^2 = (x + y - z)(x + y + z)\). 2. Rewrite the denominator: \(x^2 - y^2 + 2yz - z^2 = x^2 - (y - z)^2 = (x - y + z)(x + y - z)\). 3. Cancel the common factor \(x + y - z\) to obtain \(\frac{x + y + z}{x - y + z}\). 4. Substitute the given values: \(\frac{5.6 + 2.4 + 3.2}{5.6 - 2.4 + 3.2} = \frac{11.2}{6.4} = 1.75\). 5. The original denominator requires \(x + y - z \ne 0\) and \(x - y + z \ne 0\).

\(\frac{x + y + z}{x - y + z} = 1.75\), with \(x + y - z \ne 0\) and \(x - y + z \ne 0\).

- Combine the last three terms in the numerator into a perfect-square trinomial. - Use the difference-of-squares pattern. - A rational expression is undefined when its denominator equals zero.

1. Group the last three terms in the numerator: \(a^2 - (b^2 + 2bc + c^2)\). 2. Recognize a perfect-square trinomial: \(a^2 - (b + c)^2\). 3. Factor the difference of squares: \((a - b - c)(a + b + c)\). 4. Cancel \(a + b + c\) to obtain \(T = a - b - c\), provided \(a + b + c \ne 0\). 5. Substitute the values: \(15.5 - 2.75 - 1.25 = 11.5\). 6. The original expression is undefined when \(a + b + c = 0\).

a) \(a - b - c\), with \(a + b + c \ne 0\). b) \(11.5\) c) Undefined when \(a + b + c = 0\).

- Group the four terms into two pairs. - Factor a common factor from each pair. - Look for the divisor as a common binomial factor.

1. Group the numerator: \((12a^2-4ab)+(15a-5b)\). 2. Factor each group: \(4a(3a-b)+5(3a-b)\). 3. Factor out the common binomial: \((4a+5)(3a-b)\). 4. Divide by \(3a-b\) to get \(4a+5\). 5. The original quotient requires \(3a-b\ne0\).

\(4a+5\), where \(3a-b\ne0\)

- Factor each difference of squares. - Addition is commutative, but reversing a subtraction changes its sign. - Keep restrictions from the original divisors.

1. a) \(169p^2-144q^2=(13p-12q)(13p+12q)\). Since \(12q+13p=13p+12q\), the result is \(13p-12q\), where \(13p+12q\ne0\). 2. b) \(k^4-16=(k^2-4)(k^2+4)\), so the result is \(k^2+4\), where \(k^2-4\ne0\). 3. c) \(25r^2-36s^2=(5r-6s)(5r+6s)\), while \(6s-5r=-(5r-6s)\). Thus the result is \(-5r-6s\), where \(6s-5r\ne0\).

a) \(13p-12q\), where \(13p+12q\ne0\) b) \(k^2+4\), where \(k\ne2\) and \(k\ne-2\) c) \(-5r-6s\), where \(6s-5r\ne0\)

- Can a real square plus \(1\) ever equal zero? - In a sum of rational expressions, every denominator must be nonzero. - Which part of a rational expression determines its domain restrictions?

1. For part a, factor the denominator: \(x^2 - 9 = (x - 3)(x + 3)\). It is zero at \(x = 3\) and \(x = -3\), so the domain is \(\mathbb{R} \setminus \{-3, 3\}\). 2. For part b, \(x^2 \ge 0\) for every real number, so \(x^2 + 1 \ge 1\). The denominator is never zero, and the domain is \(\mathbb{R}\). 3. For part c, both fractions must be defined. The first denominator excludes \(x = 0\), and the second excludes \(x = 7\). The domain is \(\mathbb{R} \setminus \{0, 7\}\).

a) \(\mathbb{R} \setminus \{-3, 3\}\). b) \(\mathbb{R}\), because \(x^2 + 1\) is always positive. c) \(\mathbb{R} \setminus \{0, 7\}\).

- A rational expression is undefined when its denominator is zero. - Compare the real solutions of the two denominator equations. - One counterexample is enough to disprove a universal claim.

1. For \(A(x)\), the denominator is zero when \(x^2-16=0\). Factoring gives \((x-4)(x+4)=0\), so \(x=-4\) or \(x=4\). Thus, \(D_A=\mathbb{R}\setminus\{-4,4\}\). 2. For \(B(x)\), the denominator would be zero if \(x^2=-16\). No real number has a negative square, so \(D_B=\mathbb{R}\). 3. The claim is false. The expression \(B(x)\) is a counterexample: its denominator contains \(x^2\), but no real input is excluded.

\(D_A=\mathbb{R}\setminus\{-4,4\}\). \(D_B=\mathbb{R}\). Noor's claim is false because \(B(x)\) has no excluded real inputs.

- What denominator is zero at exactly \(x = 2\)? - Substitute the given input and output into the formula containing \(k\). - Can \(x^2 + 1\) equal zero for any real number?

1. To exclude only \(x = 2\), choose a denominator that is zero at \(x = 2\). One possible formula is \(h(x) = \frac{1}{x - 2}\). 2. Substitute \(x = 3\) and \(h(3) = 10\) into \(h(x) = \frac{k}{x - 2}\): \(10 = \frac{k}{3 - 2} = k\). Thus, \(k = 10\), so \(h(x) = \frac{10}{x - 2}\). 3. After the replacement, \(h(x) = \frac{10}{x^2 + 1}\). For every real number \(x\), \(x^2 \ge 0\), so \(x^2 + 1 \ge 1\). The denominator is never zero, so the new domain is \(\mathbb{R}\).

a) One possible formula is \(h(x) = \frac{1}{x - 2}\). b) \(k = 10\), so \(h(x) = \frac{10}{x - 2}\). c) The new domain is \(\mathbb{R}\) because \(x^2 + 1\) is never zero for real \(x\).

- Factor the numerator and denominator before canceling. - Determine the domain from each original expression. - A function is determined by both its rule and its domain. - Check whether one function is defined at an input where the other is not.

1. Factor the denominator of \(f\): \(x^2 - 9 = (x + 3)(x - 3)\). Therefore, \(f\) is undefined at \(x = -3\) and \(x = 3\), so its domain is \(\mathbb{R} \setminus \{-3, 3\}\). 2. The function \(g\) is undefined only at \(x = 3\), so its domain is \(\mathbb{R} \setminus \{3\}\). 3. Factor the numerator: \(2x + 6 = 2(x + 3)\). For inputs in the domain of \(f\), \(f(x) = \frac{2(x + 3)}{(x + 3)(x - 3)} = \frac{2}{x - 3}\). 4. Although the simplified expressions agree wherever \(f\) is defined, the domains are different. In particular, \(g(-3)\) is defined but \(f(-3)\) is not. 5. Therefore, \(f\) and \(g\) are not the same function.

a) \(D_f = \mathbb{R} \setminus \{-3, 3\}\); \(D_g = \mathbb{R} \setminus \{3\}\) b) For \(x \in D_f\), \(f(x) = \frac{2}{x - 3}\); the hole at \(x = -3\) remains. c) No. The expressions agree on \(D_f\), but the functions have different domains.

- How can the excluded inputs be represented as factors in a denominator? - What factor in the numerator creates a zero at \(x = 0\)? - Use an unknown constant in the numerator. - Substitute \(x = 1\) to determine that constant.

1. To exclude \(x = -2\) and \(x = 5\), use the denominator \((x + 2)(x - 5)\). 2. To create a zero at \(x = 0\), include a factor of \(x\) in the numerator. Let \(f(x) = \frac{cx}{(x + 2)(x - 5)}\). 3. Apply the condition \(f(1) = -\frac{1}{12}\): \(-\frac{1}{12} = \frac{c}{(1 + 2)(1 - 5)} = \frac{c}{-12}\). 4. Therefore, \(c = 1\), and one possible function is \(f(x) = \frac{x}{(x + 2)(x - 5)}\). 5. The numerator shares no factor with the denominator, so both excluded values remain outside the domain.

One possible function is \(f(x) = \frac{x}{(x + 2)(x - 5)}\).

- Treat a complicated numerator as a difference of two squares. - Recognize perfect-square trinomials inside the factors. - Factor repeatedly until denominator factors appear. - Apply the difference-of-squares pattern more than once.

1. Treat the numerator as a difference of squares: \((x^2 + y^2 - 2xy)(x^2 + y^2 + 2xy)\). 2. Recognize perfect squares: \(x^2 + y^2 - 2xy = (x - y)^2\) and \(x^2 + y^2 + 2xy = (x + y)^2\). Thus, the numerator is \((x - y)^2(x + y)^2 = (x^2 - y^2)^2\). 3. Cancel one factor of \(x^2 - y^2\) to obtain \(x^2 - y^2\), with \(x \ne y\) and \(x \ne -y\). 4. For part 2, factor repeatedly: \(b^8 - 1 = (b^4 - 1)(b^4 + 1) = (b^2 - 1)(b^2 + 1)(b^4 + 1)\). 5. Cancel \(b^4 + 1\) and \(b^2 + 1\) to obtain \(b^2 - 1\). Both original denominator factors are positive for every real \(b\), so there is no real restriction.

1) \(x^2 - y^2\), with \(x \ne y\) and \(x \ne -y\). 2) \(b^2 - 1\); no real restriction.