Free Math Worksheets

- Move all terms to one side. - Recognize a difference of squares. - Equal function outputs correspond to graph intersections.

1. Rewrite the equation with all terms on one side: \(x^2 - 4 = 0\). 2. Factor the difference of squares: \((x - 2)(x + 2) = 0\). 3. Therefore, \(x = -2\) or \(x = 2\). 4. Geometrically, these values are the x-coordinates of the intersection points of the graphs of \(f\) and \(g\).

a) \(x^2 - 4 = 0\), so \((x - 2)(x + 2) = 0\) b) \(\{-2, 2\}\) c) They are the x-coordinates of the graphs’ intersection points.

- A product equals zero when at least one factor equals zero. - Find the greatest common factor of the terms in each equation. - Check the factorization by multiplying.

1. For part a, factor out \(7x\): \(7x(2x - 3) = 0\). 2. Apply the zero-product property: \(7x = 0\) or \(2x - 3 = 0\). Thus, \(x = 0\) or \(x = \frac{3}{2}\). 3. For part b, factor out \(0.5x\): \(0.5x(x + 8) = 0\). 4. Apply the zero-product property: \(0.5x = 0\) or \(x + 8 = 0\). Thus, \(x = 0\) or \(x = -8\).

a) \(x \in \left\{0, \frac{3}{2}\right\}\) b) \(x \in \{-8, 0\}\)

- Rewrite the constant as a perfect square. - Use the difference-of-squares pattern. - Apply the zero-product property after factoring.

1. Recognize a difference of squares: \(x^2 - 16 = x^2 - 4^2\). 2. Factor: \((x - 4)(x + 4) = 0\). 3. Apply the zero-product property: \(x = 4\) or \(x = -4\).

\(x \in \{-4, 4\}\)

- Choose an efficient method for each expression, such as factoring or reasoning about squares. - Simplify by factoring out a common factor first. - Use the zero-product property after factoring. - Consider the possible values of \(x^2\) for real \(x\).

1. For \(p\), factor out \(2\): \(2(x^2-6x+5)=2(x-1)(x-5)\). By the zero-product property, the zeros are \(x=1\) and \(x=5\). 2. For \(q\), \(x^2+4=0\) would require \(x^2=-4\). No real number has a negative square, so there are no real zeros. 3. For \(r\), factor: \(5x^2-10x=5x(x-2)\). By the zero-product property, the zeros are \(x=0\) and \(x=2\).

a) \(x=1\) and \(x=5\) b) No real zeros, because \(x^2=-4\) has no real solution. c) \(x=0\) and \(x=2\)

- Identify the form of each quadratic before choosing a method. - Use the zero-product property when an expression is factored. - Look for a greatest common factor. - Factor a trinomial when two integers have the required sum and product. - If the equation contains only \(x^2\) and a constant, isolate \(x^2\).

1. For a), use the zero-product property: \(x - 6 = 0\) or \(x + 4 = 0\). The zeros are \(x = 6\) and \(x = -4\). 2. For b), factor out \(x\): \(x(x - 5) = 0\). The zeros are \(x = 0\) and \(x = 5\). 3. For c), factor the trinomial: \(x^2 - 4x - 12 = (x - 6)(x + 2)\). The zeros are \(x = 6\) and \(x = -2\). 4. For d), factor or isolate \(x^2\): \(4x^2 - 16 = 4(x - 2)(x + 2)\). The zeros are \(x = 2\) and \(x = -2\).

a) \(x = -4\), \(x = 6\) b) \(x = 0\), \(x = 5\) c) \(x = -2\), \(x = 6\) d) \(x = -2\), \(x = 2\)

- Move all terms to one side so the other side is \(0\). - Factor out the greatest common factor. - Apply the zero-product property.

1. For part a, move all terms to the left: \(4x^2 - 12x = 0\). 2. Factor: \(4x(x - 3) = 0\). Thus, \(x = 0\) or \(x = 3\). 3. For part b, move all terms to one side: \(-x^2 + 5x = 0\). 4. Factor: \(-x(x - 5) = 0\). Thus, \(x = 0\) or \(x = 5\).

a) \(x \in \{0, 3\}\) b) \(x \in \{0, 5\}\)

- Find a factor pair of \(-21\) whose sum is \(-4\). - Set each linear factor equal to zero. - Compare a zero \(r\) with its factor \(x - r\).

1. Find two numbers whose product is \(-21\) and whose sum is \(-4\): \(-7\) and \(3\). 2. Therefore, \(x^2 - 4x - 21 = (x - 7)(x + 3)\). 3. The zeros are \(7\) and \(-3\). A zero \(r\) corresponds to the factor \(x - r\), so the zero \(-3\) produces the factor \(x - (-3) = x + 3\).

1. \((x - 7)(x + 3)\) 2. Each zero \(r\) gives a factor \(x - r\), so the sign of the number in the factor is opposite the sign of the zero.

- Move all terms to one side when necessary. - Factor out the greatest common factor containing \(x\). - Apply the zero-product property.

1. For part a, factor: \(7x(x + 3) = 0\). Thus, \(x = 0\) or \(x = -3\). 2. For part b, move all terms to one side: \(0.5x^2 - 2x = 0\). 3. Factor: \(0.5x(x - 4) = 0\). Thus, \(x = 0\) or \(x = 4\).

a) \(x \in \{-3, 0\}\) b) \(x \in \{0, 4\}\)

- First factor out the greatest common factor. - Then factor the remaining quadratic trinomial. - Check your result by multiplying the factors.

1. Factor out the greatest common factor: \(45n^2 - 60n + 15 = 15(3n^2 - 4n + 1)\). 2. Factor the trinomial: \(3n^2 - 4n + 1 = (3n - 1)(n - 1)\). 3. Therefore, \(45n^2 - 60n + 15 = 15(3n - 1)(n - 1)\).

\(15(3n - 1)(n - 1)\)

- Look for a common numerical factor first. - Consider factoring out a negative common factor so the leading term inside the parentheses is positive. - Check your result by multiplying the factors.

1. Factor out \(-4\): \(-8b^2 + 20b + 12 = -4(2b^2 - 5b - 3)\). 2. Factor the trinomial: \(2b^2 - 5b - 3 = (2b + 1)(b - 3)\). 3. Therefore, \(-8b^2 + 20b + 12 = -4(2b + 1)(b - 3)\).

\(-4(2b + 1)(b - 3)\)

- First factor out the greatest common factor. - Then factor the remaining quadratic trinomial. - Multiply the factors to check your work.

1. Factor out the greatest common factor: \(12z^2 - 30z + 12 = 6(2z^2 - 5z + 2)\). 2. Factor the trinomial: \(2z^2 - 5z + 2 = (2z - 1)(z - 2)\). 3. Therefore, \(12z^2 - 30z + 12 = 6(2z - 1)(z - 2)\).

\(6(2z - 1)(z - 2)\)

- Begin by factoring out the greatest common factor. - Use the signs of the constant and middle terms to choose the signs in the binomial factors. - Check by multiplying the factors.

1. Factor out the greatest common factor: \(20y^2 + 10y - 30 = 10(2y^2 + y - 3)\). 2. Factor the trinomial: \(2y^2 + y - 3 = (2y + 3)(y - 1)\). 3. Therefore, \(20y^2 + 10y - 30 = 10(2y + 3)(y - 1)\).

\(10(2y + 3)(y - 1)\)

- Factor out the greatest common factor first. - Then find two binomial factors whose product gives the remaining trinomial. - Make sure the expression is fully factored, and check by multiplying.

1. Factor out the greatest common factor: \(18x^2 - 12x - 6 = 6(3x^2 - 2x - 1)\). 2. Factor the trinomial: \(3x^2 - 2x - 1 = (3x + 1)(x - 1)\). 3. Therefore, \(18x^2 - 12x - 6 = 6(3x + 1)(x - 1)\).

\(6(3x + 1)(x - 1)\)

- Test the value that makes \(x - 4\) equal to zero in the original equation. - Ask when division by \(x - 4\) is not allowed. - Move all terms to one side and look for a common binomial factor.

1. For part a, solving \(x - 4 = 3\) gives \(x = 7\). 2. Dividing by \(x - 4\) assumes that \(x - 4 \ne 0\). This excludes \(x = 4\), even though \(x = 4\) makes both sides of the original equation equal to zero. 3. Move all terms to one side: \((x - 4)^2 - 3(x - 4) = 0\). 4. Factor out \(x - 4\): \((x - 4)((x - 4) - 3) = 0\), so \((x - 4)(x - 7) = 0\). 5. Apply the zero-product property: \(x = 4\) or \(x = 7\).

a) \(x = 7\) b) Dividing by \(x - 4\) discards the case \(x - 4 = 0\). Division by zero is undefined, and \(x = 4\) is a solution of the original equation. c) \(x \in \{4, 7\}\)

- Look for a greatest common factor or a difference of squares. - Clear a decimal coefficient if that makes the trinomial easier to factor. - After factoring, use the zero-product property.

1. For part a, factor out \(2x\): \(2x(x - 9) = 0\). Thus, \(x = 0\) or \(x = 9\). 2. For part b, factor out \(3\) and use the difference of squares: \(3((x - 1)^2 - 9) = 3(x - 4)(x + 2) = 0\). Thus, \(x = 4\) or \(x = -2\). 3. For part c, multiply by \(2\): \(x^2 - 2x - 8 = 0\). Factor: \((x - 4)(x + 2) = 0\). Thus, \(x = 4\) or \(x = -2\). 4. For part d, apply the zero-product property: \(2x - 4 = 0\) or \(x + 5 = 0\). Thus, \(x = 2\) or \(x = -5\).

a) \(x \in \{0, 9\}\) b) \(x \in \{-2, 4\}\) c) \(x \in \{-2, 4\}\) d) \(x \in \{-5, 2\}\)

- Check whether the constant is a perfect square and the middle term is missing. - Check whether every term has a common factor. - For a trinomial, look for two numbers whose product is the constant term and whose sum is the coefficient of \(x\).

1. The first equation is a difference of squares: \(x^2 - 144 = (x - 12)(x + 12)\). Therefore, \(x = -12\) or \(x = 12\). 2. The second equation has a greatest common factor of \(2x\): \(2x(x + 5) = 0\). Therefore, \(x = -5\) or \(x = 0\). 3. The third equation is a factorable trinomial: \(x^2 - 2x - 15 = (x - 5)(x + 3)\). Therefore, \(x = -3\) or \(x = 5\).

1. Difference of squares; \(x \in \{-12, 12\}\) 2. Greatest common factor; \(x \in \{-5, 0\}\) 3. Trinomial factoring; \(x \in \{-3, 5\}\)

- Represent the longer leg in terms of the shorter leg. - Substitute both expressions into the Pythagorean theorem. - Factor the resulting quadratic and reject any negative length.

1. Let the shorter leg be \(a\). Then the longer leg is \(a+7\). 2. Apply the Pythagorean theorem: \(a^2+(a+7)^2=13^2\). 3. Simplify: \(2a^2+14a-120=0\), so \(a^2+7a-60=0\). 4. Factor: \((a+12)(a-5)=0\). The positive solution is \(a=5\,\text{cm}\). 5. The other leg is \(5+7=12\,\text{cm}\).

The legs are \(5\,\text{cm}\) and \(12\,\text{cm}\).

- Move all terms to one side and simplify. - Look for a difference of squares. - If the equation becomes \(x^2 + k = 0\) with \(k > 0\), consider whether real factors exist.

1. For part a, simplify to \(2x^2 - 18 = 0\), then divide by \(2\): \(x^2 - 9 = 0\). Factor: \((x - 3)(x + 3) = 0\). Thus, \(x = -3\) or \(x = 3\). 2. For part b, simplify to \(\frac{1}{3}x^2 + 9 = 0\), then multiply by \(3\): \(x^2 + 27 = 0\). This quadratic does not factor over the real numbers and has no real solution. 3. For part c, expand the difference of squares: \(x^2 - 25 = 11\), so \(x^2 - 36 = 0\). Factor: \((x - 6)(x + 6) = 0\). Thus, \(x = -6\) or \(x = 6\). 4. For part d, write \(0.5x^2 - 0.005 = 0\), then divide by \(0.5\): \(x^2 - 0.01 = 0\). Factor: \((x - 0.1)(x + 0.1) = 0\). Thus, \(x = -0.1\) or \(x = 0.1\).

a) \(x \in \{-3, 3\}\) b) \(\varnothing\) c) \(x \in \{-6, 6\}\) d) \(x \in \{-0.1, 0.1\}\)

- Clear the denominator by multiplying by \(p\). - Recognize the resulting difference of squares. - Use the condition \(p \ne 0\) to justify that the two roots are different.

1. Multiply the equation by \(p\): \(x^2 - p^2 = 0\). 2. Factor the difference of squares: \((x - p)(x + p) = 0\). 3. Therefore, \(x = p\) or \(x = -p\). Since \(p \ne 0\), these solutions are distinct. 4. For \(p = 5\), the solutions are \(x = 5\) and \(x = -5\).

a) \(x \in \{-5, 5\}\) b) \((x - p)(x + p) = 0\), so the two distinct solutions are \(x = p\) and \(x = -p\). c) \(x \in \{-p, p\}\)

- Move every term to one side and combine like terms. - Clear fractions if that makes the equation easier to read. - When there is no constant term, factor out \(x\).

1. For part a, move all terms to the left: \(\frac{1}{2}x^2 + 3x = 0\). 2. Factor: \(x\left(\frac{1}{2}x + 3\right) = 0\). Thus, \(x = 0\) or \(x = -6\). 3. For part b, expand and simplify: \(2x^2 + 8x = x^2 + 3x\), so \(x^2 + 5x = 0\). 4. Factor: \(x(x + 5) = 0\). Thus, \(x = 0\) or \(x = -5\).

a) \(x \in \{-6, 0\}\) b) \(x \in \{-5, 0\}\)

- Expand or use the difference-of-squares pattern to simplify each side. - Move all terms to one side. - Look for a difference of squares before applying the zero-product property.

1. For equation 1, expand: \(x^2 - 12x = 64 - 12x\). Add \(12x\) to both sides: \(x^2 - 64 = 0\). 2. Factor the difference of squares: \((x - 8)(x + 8) = 0\). Thus, \(x = -8\) or \(x = 8\). 3. For equation 2, use the difference-of-squares pattern: \(x^2 - 9 + x^2 - 16 = 7\). 4. Simplify: \(2x^2 - 32 = 0\), then divide by \(2\): \(x^2 - 16 = 0\). 5. Factor: \((x - 4)(x + 4) = 0\). Thus, \(x = -4\) or \(x = 4\).

1) \(x \in \{-8, 8\}\) 2) \(x \in \{-4, 4\}\)

- Expand each product or square carefully. - Move all terms to one side and combine like terms. - Factor the resulting quadratic and use the zero-product property.

1. For equation 1, expand and simplify: \(3x^2 - 12x = x^2 - 2x\), so \(2x^2 - 10x = 0\). 2. Factor: \(2x(x - 5) = 0\). Thus, \(x = 0\) or \(x = 5\). 3. For equation 2, expand and simplify: \(x^2 - 4x + 4 = 3x - 2\), so \(x^2 - 7x + 6 = 0\). 4. Factor: \((x - 1)(x - 6) = 0\). Thus, \(x = 1\) or \(x = 6\).

1) \(x \in \{0, 5\}\) 2) \(x \in \{1, 6\}\)

- Expand the squared binomials and the difference of squares. - Distribute the negative sign before combining like terms. - After simplifying, look for a difference of squares.

1. Expand the expressions: \(9x^2 - 6x + 1 - (x^2 + 4x + 4) + 4x^2 - 25 = 4x^2 - 10x + 4\). 2. Combine like terms: \(12x^2 - 10x - 28 = 4x^2 - 10x + 4\). 3. Move all terms to one side: \(8x^2 - 32 = 0\). 4. Divide by \(8\) and factor: \(x^2 - 4 = (x - 2)(x + 2) = 0\). 5. Therefore, \(x = -2\) or \(x = 2\).

\(x \in \{-2, 2\}\)

- Expand each square and product carefully. - Move all terms to one side and combine like terms. - Factor the resulting trinomial and use the zero-product property.

1. Expand both sides: \(x^2 - 6x + 9 + x^2 + 4x + 4 = x^2 - 16 + 2x + 34\). 2. Combine like terms: \(2x^2 - 2x + 13 = x^2 + 2x + 18\). 3. Move all terms to one side: \(x^2 - 4x - 5 = 0\). 4. Factor: \((x - 5)(x + 1) = 0\). 5. Therefore, \(x = 5\) or \(x = -1\).

\(x \in \{-1, 5\}\)

- Factor out \(x\). - Check when the two solution expressions coincide. - For part b, substitute the given value of \(x\) into the original equation.

1. Factor: \(x(3x + p) = 0\). 2. Therefore, \(x = 0\) or \(x = -\frac{p}{3}\). 3. When \(p = 0\), these expressions give the same solution, \(x = 0\). When \(p \ne 0\), they are distinct. 4. For part b, substitute \(x = -4\): \(3(-4)^2 + p(-4) = 0\). 5. Simplify: \(48 - 4p = 0\), so \(p = 12\).

a) For \(p \ne 0\), \(x \in \left\{0, -\frac{p}{3}\right\}\). For \(p = 0\), \(x = 0\). b) \(p = 12\)

- Find the side length of the larger square from its area. - Write one equation using \(a+b\). - Write a second equation using the Pythagorean theorem. - Substitute and factor the resulting quadratic.

1. The larger square has side length \(\sqrt{49}=7\,\text{cm}\), so \(a+b=7\). 2. In each right triangle, the inner square’s side is the hypotenuse, so \(a^2+b^2=5^2=25\). 3. Substitute \(b=7-a\): \(a^2+(7-a)^2=25\). 4. Simplify: \(2a^2-14a+24=0\), so \(a^2-7a+12=0\). 5. Factor: \((a-3)(a-4)=0\). Thus the leg lengths are \(3\,\text{cm}\) and \(4\,\text{cm}\).

The legs are \(3\,\text{cm}\) and \(4\,\text{cm}\), in either order.