- Line up terms that have the same power of \(x\). - A missing power in one polynomial has coefficient \(0\). - Combine coefficients without changing the exponents.

1. Add coefficients of like powers of \(x\). 2. The cubic terms give \(3x^3-x^3=2x^3\), the quadratic term is \(4x^2\), the linear terms give \(-2x+7x=5x\), and the constants give \(5-1=4\). 3. Therefore, \(A(x)+B(x)=2x^3+4x^2+5x+4\).

\(2x^3+4x^2+5x+4\)

- Multiply the monomial by each term inside the parentheses. - Multiply coefficients and add exponents on like bases. - Check the sign of each product.

1. Distribute \(-3x^2\) to each term. 2. The products are \(-6x^5\), \(3x^3\), and \(-12x^2\). 3. Therefore, the simplified polynomial is \(-6x^5+3x^3-12x^2\).

\(-6x^5+3x^3-12x^2\)

- Focus only on the highest-degree term in each factor. - Degrees add when powers of \(x\) are multiplied. - The leading coefficient comes from multiplying the leading coefficients of the factors.

1. The highest-degree terms are \(-3x^2\) and \(2x^4\). 2. Their product is \(-6x^6\), and no other pair of terms can produce degree \(6\). 3. Therefore, \(F\) has degree \(6\) and leading coefficient \(-6\).

Degree \(6\); leading coefficient \(-6\).

- Compare the variable part of each term, not just its coefficient. - Which terms contain exactly the same power of \(x\)? - Add coefficients only after identifying like terms.

1. The terms \(4x^2\) and \(-7x^2\) have the same power of \(x\), so they are like terms. 2. Their sum is \((4-7)x^2=-3x^2\). The terms \(3x\) and \(x^3\) have different powers.

\(4x^2\) and \(-7x^2\) combine to \(-3x^2\). The other terms have different powers of \(x\).

- Make a place for each power between the highest power and the constant term. - Find the term with the greatest exponent. - What number multiplies that highest-power term?

1. There is no \(x^3\) term, so its coefficient is \(0\). 2. The highest exponent is \(4\), so the degree is \(4\). The coefficient of \(x^4\) is \(6\).

Coefficient of \(x^3\): \(0\); degree: \(4\); leading coefficient: \(6\)

- The term outside the parentheses multiplies every term inside. - Multiply the numerical coefficients and add exponents when multiplying powers of \(x\). - Watch the sign of the constant term in the parentheses.

1. Distribute \(5x\) to both terms: \(5x\cdot2x=10x^2\) and \(5x\cdot(-3)=-15x\). 2. The product is \(10x^2-15x\).

\(10x^2-15x\)

- The first and last terms look like squares. - What middle term comes from squaring \(x+a\)? - Check the sign that gives a positive middle term.

1. The first and last terms are \(x^2\) and \(5^2\). 2. The middle term is \(2\cdot x\cdot5=10x\), so the polynomial is \((x+5)^2\).

\((x+5)^2\)

- Test a zero of \(f\) against the other two graphs. - What equation relates the function values at an intersection of \(f\) and \(g\)? - Substitute the relation from part c into the definition of \(h\).

1. Statement a is false. The graph gives \(f(-2)=0\) and \(g(-2)=4\), so \(h(-2)=4\ne0\). 2. Statement b is true. At an intersection, \(f(x)=g(x)\), so \(h(x)=f(x)+g(x)=2f(x)\). 3. Statement c is true. If \(f(x)=-g(x)\), then \(h(x)=f(x)+g(x)=0\).

a) False. At \(x=-2\), \(f(-2)=0\) but \(h(-2)=4\). b) True. Where \(f(x)=g(x)\), their sum is \(h(x)=2f(x)\). c) True. If \(f(x)=-g(x)\), their sum is \(0\).

- Choose a simple input such as \(x=1\) and evaluate both combined functions. - Compare those outputs with the candidate graphs. - Alternatively, factor each combined polynomial and compare its zeros with the graphs.

1. Test \(x=1\): \(s(1)=1^2+2\cdot 1=3\), while \(d(1)=1^2-2\cdot 1=-1\). 2. Graph p passes through \((1, 3)\), so p represents \(s(x)=x^2+2x\). 3. Graph q passes through \((1, -1)\), so q represents \(d(x)=x^2-2x\). 4. The zeros confirm the matches: \(s(x)=x(x+2)\) has zeros \(-2\) and \(0\), while \(d(x)=x(x-2)\) has zeros \(0\) and \(2\).

p: \(s(x)=x^2+2x\); q: \(d(x)=x^2-2x\)

- Compare \(x^2\) and \(x^4\) when \(|x|<1\). - Determine the sign of each difference on that interval. - Notice that the two difference functions are opposites.

1. For \(-1<x<1\), \(x^4<x^2\) except at \(x=0\), where they are equal. 2. Therefore, \(h(x)=x^2-x^4\) is positive on \((-1, 0)\cup(0, 1)\) and equals zero at \(x=0\). Graph q has this behavior. 3. Also, \(k(x)=x^4-x^2=-h(x)\), so it is negative on \((-1, 0)\cup(0, 1)\). Graph p has this behavior.

p: \(k(x)=x^4-x^2\); q: \(h(x)=x^2-x^4\)

- Translate the relative positions of the graphs into an inequality. - Subtract the same quantity from both sides. - Use the definition of \(h\).

1. If the graph of \(f\) is above the graph of \(g\) at an input \(x\), then \(f(x)>g(x)\). 2. Subtract \(g(x)\) from both sides to obtain \(f(x)-g(x)>0\). 3. Since \(h(x)=f(x)-g(x)\), it follows that \(h(x)>0\). Therefore, the statement is always true.

True. The condition \(f(x)>g(x)\) is equivalent to \(f(x)-g(x)>0\), so \(h(x)>0\).

- Treat subtraction as adding the opposite of the entire second polynomial. - Be especially careful with the signs of every term in \(Q(x)\). - Combine only terms with the same exponent.

1. Distribute the subtraction sign to every term of \(Q\): \(P(x)-Q(x)=5x^4-3x^2+2x-7-2x^4-x^3+5x-4\). 2. Combine like terms to get \(3x^4-x^3-3x^2+7x-11\).

\(3x^4-x^3-3x^2+7x-11\)

- Each term in one binomial must multiply each term in the other. - Keep the signs attached to their terms while distributing. - Combine the two linear terms after multiplying.

1. Distribute each term of the first binomial: \((2x-3)(x+5)=2x^2+10x-3x-15\). 2. Combine the linear terms: \(10x-3x=7x\). 3. The product is \(2x^2+7x-15\).

\(2x^2+7x-15\)

- Notice that the quadratic factor has no \(x\)-term. - Distribute each term of \(x-2\) across both terms of the quadratic. - Put the final terms in descending powers of \(x\).

1. Distribute \(x\): \(x(3x^2+5)=3x^3+5x\). 2. Distribute \(-2\): \(-2(3x^2+5)=-6x^2-10\). 3. Combine the results: \(3x^3-6x^2+5x-10\).

\(3x^3-6x^2+5x-10\)

- Track how the subtraction sign affects every term in the second polynomial. - Compare coefficients of like powers after the parentheses are removed. - Check whether the constant term also has the correct sign.

1. The subtraction sign must change the sign of every term in the second polynomial. 2. Rewrite the expression as \(4x^3-2x+5-x^3-6x+7\). 3. Combine like terms: \(3x^3-8x+12\). 4. Miguel's error was failing to distribute the negative sign correctly to all terms of the second polynomial.

The subtraction sign was not distributed to every term. The correct result is \(3x^3-8x+12\).

- Write the area of each square before subtracting. - The difference of two squared expressions factors into a product of their sum and difference. - Keep the final area unit squared.

1. Subtract the two square areas: \(A=(x+3)^2-(x-3)^2\). 2. Use the difference of squares: \(A=[(x+3)-(x-3)][(x+3)+(x-3)]=(6)(2x)=12x\). 3. The outside area is \(12x\,\text{ft}^2\).

\(12x\,\text{ft}^2\)

- Replace the square with two identical factors. - Does multiplying the first term by the last term produce a cross product? - Check all products, not just the first and last squares.

1. The square means \((2x-1)(2x-1)\), which contains two cross products. 2. Together those cross products give \(-4x\), so the correct expansion is \(4x^2-4x+1\).

She left out the middle term \(-4x\). The correct expansion is \(4x^2-4x+1\).

- Think of the difference as a known result with a missing subtrahend. - Isolate \(Q(x)\) before combining like terms. - Check by subtracting the expression you found.

1. Rearrange to \(Q(x)=P(x)-(P(x)-Q(x))\). 2. Subtract: \(Q(x)=(3x^2+5x-1)-(2x^2-x+4)=x^2+6x-5\). 3. Checking: \(P(x)-Q(x)=2x^2-x+4\).

\(Q(x)=x^2+6x-5\)

- Read both graph heights at the same x-value. - Add the two y-values before moving to the other input. - Compare the two resulting sums.

1. At \(x=0\), the graph gives \(f(0)=-2\) and \(g(0)=2\), so the sum is \(0\). 2. At \(x=2\), the graph gives \(f(2)=2\) and \(g(2)=0\), so the sum is \(2\). 3. The sums are not equal.

\(f(0)+g(0)=0\), while \(f(2)+g(2)=2\). They are not equal.

- Use the distributive property to multiply two binomials first. - Multiply the resulting quadratic by the remaining binomial before applying the outside coefficient. - Identify how the highest-degree term is formed from the factors. - Track what the outside coefficient does to the leading term.

1. Multiply the first two factors: \((x - 1)(x + 2) = x^2 + x - 2\). 2. Multiply by the third factor: \((x^2 + x - 2)(x - 4) = x^3 - 3x^2 - 6x + 8\). 3. Multiply by the outside coefficient: \(f(x) = 3x^3 - 9x^2 - 18x + 24\). 4. Thus, \(a_3 = 3\), which equals the coefficient outside the product. 5. In a product of \(n\) monic linear factors, the term \(x^n\) is produced only by multiplying the \(x\)-term from every factor. Its coefficient is \(1\). Multiplying the entire product by \(a_n\) makes the leading term \(a_nx^n\), so the leading coefficient is \(a_n\).

1. \(f(x) = 3x^3 - 9x^2 - 18x + 24\) 2. \(a_3 = 3\). In general, multiplying the \(x\)-term from each monic linear factor produces \(x^n\), and the outside coefficient \(a_n\) makes the leading term \(a_nx^n\).

- Use the square of a binomial to expand \((x + 1)^2\). - The leading coefficient is the coefficient of the highest-degree term. - The constant term contains no variable. - To find only the leading term, multiply only the highest-degree term from each factor.

1. Expand the squared factor: \((x + 1)^2 = x^2 + 2x + 1\). 2. Multiply by the remaining factor: \((x^2 + 2x + 1)(x - 3) = x^3 - x^2 - 5x - 3\). 3. Multiply by \(-2\): \(p(x) = -2x^3 + 2x^2 + 10x + 6\). 4. Therefore, the leading coefficient is \(a_3 = -2\), and the constant term is \(a_0 = 6\). 5. Each linear factor is monic, so multiplying the leading terms gives \(x^2 \cdot x = x^3\). The outside coefficient \(-2\) therefore becomes the coefficient of \(x^3\).

1. \(p(x) = -2x^3 + 2x^2 + 10x + 6\) 2. \(a_3 = -2\) and \(a_0 = 6\) 3. Because the linear factors are monic, their leading terms multiply to \(x^3\); the outside coefficient \(-2\) is therefore the leading coefficient.

- Expand the squared binomial first. - Use the distributive property on the remaining factor. - Standard form lists the terms in descending powers of \(x\). - Read the degree and leading coefficient from the highest-power term.

1. Expand the square: \((x-1)^2=x^2-2x+1\). 2. Multiply by \(x+2\): \((x+2)(x^2-2x+1)=x^3-2x^2+x+2x^2-4x+2\). 3. Combine like terms: \(f(x)=x^3-3x+2\). 4. The highest power is \(3\), so the degree is \(3\) and the leading coefficient is \(1\).

a) \(f(x)=x^3-3x+2\) b) Degree \(3\); leading coefficient \(1\)

- Add the two polynomial expressions by combining like terms. - If \(f(x)+g(x)=f(x)\), what does that tell you about \(g(x)\)? - Can the shop make money overall even if one kind of sale loses money?

1. Add like terms: \(h(x)=(-0.1x^2+0.8x+1)+(0.5x-2)=-0.1x^2+1.3x-1\). 2. Evaluate \(h(2)=-0.4+2.6-1=1.2\), or \(\$1200\), and \(h(8)=-6.4+10.4-1=3\), or \(\$3000\). 3. If \(h(x)=f(x)\), then \(g(x)=0\). Solving \(0.5x-2=0\) gives \(x=4\). 4. No: in month \(2\), \(h(2)=1.2>0\), but \(g(2)=-1<0\). Parts sales lose money even though the total profit is positive.

a) \(h(x)=-0.1x^2+1.3x-1\); month \(2\): \(\$1200\); month \(8\): \(\$3000\) b) Month \(4\), because \(g(4)=0\) c) No. In month \(2\), \(h(2)>0\) but \(g(2)<0\).

- Think of the equation as a missing-addend problem with polynomials. - Isolate \(B(x)\) before combining terms. - Distribute the subtraction sign to the entire known polynomial.

1. Rearrange the equation: \(B(x)=S(x)-A(x)\). 2. Substitute the given polynomials: \(B(x)=5x^3+3x^2-2x+9-(2x^3-x^2+4)\). 3. Combine like terms to obtain \(B(x)=3x^3+4x^2-2x+5\).

\(B(x)=3x^3+4x^2-2x+5\)

- The cost polynomial is subtracted as a group. What happens to the signs of its terms? - Simplify the polynomial before substituting the order size. - What do the sign and dollar unit tell you about the value at \(x=20\)?

1. Subtract the cost polynomial: \(P(x)=(-2x^2+90x)-(20x+300)=-2x^2+70x-300\). 2. Evaluate: \(P(20)=-2(20)^2+70(20)-300=300\). 3. The model predicts a profit of \(\$300\) on an order of \(20\) shirts.

a) \(P(x)=-2x^2+70x-300\) b) \(P(20)=300\) c) The predicted profit for \(20\) shirts is \(\$300\).

- Read the y-value of \(f\) directly from the graph at each requested x-value. - Evaluate \(g\) separately before adding outputs. - For part b, simplify the equation \(f(x)+g(x)=g(x)\) before using the graph.

1. From the graph, \(f(-2)=3\), \(f(0)=-1\), and \(f(2)=3\). 2. Also, \(g(-2)=-1\), \(g(0)=3\), and \(g(2)=7\). Therefore, \(h(-2)=2\), \(h(0)=2\), and \(h(2)=10\). 3. Since \(h(x)=f(x)+g(x)\), the equation \(h(x)=g(x)\) holds exactly when \(f(x)=0\). The graph shows zeros at \(x=-1\) and \(x=1\).

a) \(f(-2)=3\), \(f(0)=-1\), \(f(2)=3\); \(h(-2)=2\), \(h(0)=2\), \(h(2)=10\) b) \(x=-1\) and \(x=1\), because \(h(x)=g(x)\) exactly when \(f(x)=0\).

- Compare the highest powers before doing unnecessary expansion. - For a product, look at how the leading terms combine. - Check whether any operation produces negative or fractional powers of \(x\).

1. The sum \(P+Q\) is a polynomial. Its highest-degree term is \(2x^3\), so its degree is \(3\). 2. The difference \(P-Q\) is a polynomial. Its highest-degree term is \(-2x^3\), so its degree is \(3\). 3. The product \(PQ\) is a polynomial. The leading terms multiply to \(x^2\cdot2x^3=2x^5\), so its degree is \(5\). 4. More generally, combining polynomial terms by addition or subtraction keeps only whole-number, nonnegative powers. Multiplication adds such exponents and produces finitely many terms. All three operations therefore give a polynomial.

1. All three results are polynomials. 2. \(P+Q\): degree \(3\); \(P-Q\): degree \(3\); \(PQ\): degree \(5\). Yes. Adding or subtracting keeps whole-number, nonnegative powers; multiplying adds exponents of those powers. Each result is a polynomial.

- Multiply the stated width and unknown linear length first. - Like powers must have equal coefficients in two identical polynomials. - Use the middle coefficient to check the two values.

1. Multiply the dimensions: \((x+2)(mx+n)=mx^2+(2m+n)x+2n\). 2. Compare with \(3x^2+7x+2\). The \(x^2\) coefficient gives \(m=3\), and the constant term gives \(2n=2\), so \(n=1\). 3. The middle coefficient checks: \(2m+n=6+1=7\). The length is \((3x+1)\,\text{ft}\).

From \((x+2)(mx+n)=mx^2+(2m+n)x+2n\), match coefficients: \(m=3\), \(2n=2\), and \(2m+n=7\). Thus \(n=1\), and the length is \((3x+1)\,\text{ft}\).

- For the product and difference, track only the leading terms at first. - What happens to the degree-\(5\) terms when the polynomials are added? - Could all the lower-degree terms also cancel? - For part d, choose lower-degree terms so the first nonzero term of each sum has the desired power.

1. The leading term of \(PQ\) is \((4x^5)(-4x^5)=-16x^{10}\), so its degree is \(10\) and leading coefficient is \(-16\). 2. The leading term of \(P-Q\) is \(4x^5-(-4x^5)=8x^5\), so its degree is \(5\) and leading coefficient is \(8\). 3. The fifth-degree terms cancel in \(P+Q\). If the result is nonzero, its degree is at most \(4\). If \(Q=-P\), then \(P+Q\) is the zero polynomial, whose degree is not defined under the usual high-school convention. 4. For degree \(4\), choose \(P_1(x)=4x^5+x^4\) and \(Q_1(x)=-4x^5+2x^4\). Their sum is \(3x^4\). 5. For degree \(2\), choose \(P_2(x)=4x^5+x^4+x^2\) and \(Q_2(x)=-4x^5-x^4+2x^2\). Their sum is \(3x^2\).

a) Degree \(10\); leading coefficient \(-16\) b) Degree \(5\); leading coefficient \(8\) c) \(P+Q\) is either the zero polynomial (no defined degree under the usual school convention) or a nonzero polynomial of degree at most \(4\). Its exact degree is undetermined. d) For degree \(4\): \(P_1(x)=4x^5+x^4\), \(Q_1(x)=-4x^5+2x^4\). For degree \(2\): \(P_2(x)=4x^5+x^4+x^2\), \(Q_2(x)=-4x^5-x^4+2x^2\).