- Compare the divisor with the standard form \(x-c\). - List coefficients in descending powers of \(x\). - Check whether any power is missing and needs a placeholder coefficient.

1. Synthetic division by \(x-c\) uses the number \(c\). Since the divisor is \(x-4\), the outside number is \(4\). 2. The polynomial has no \(x^2\) term, so its coefficient must be recorded as \(0\). 3. The coefficient row is \(1, 0, -5, 2\).

Outside number: \(4\) Coefficient row: \(1, 0, -5, 2\)

- Which term has the highest power in the dividend and in the divisor? - Find a term that would cancel the dividend's highest-power term when multiplied by the divisor.

1. Divide the leading terms: \(\frac{6x^2}{2x}=3x\). 2. Write \(3x\) as the first quotient term. Multiplying \(3x(2x+1)\) will cancel the leading \(6x^2\) term in the dividend.

\(3x\), because \(6x^2\div 2x=3x\).

- Which part is multiplied by the divisor? - What part is left over after that product?

1. Compare the given expression with \(P(x)=d(x)Q(x)+R\). 2. The factor multiplying \(x-3\) is the quotient \(x^2+2\). 3. The number added afterward is the remainder \(5\).

Quotient: \(x^2+2\); remainder: \(5\).

- What value of \(x\) makes the divisor equal zero? - Compare the divisor with the form \(x-c\).

1. Write the divisor as \(x-(-4)\). 2. Synthetic division uses the number that makes the divisor zero, so the number is \(-4\).

\(-4\), because \(x+4=0\) when \(x=-4\).

- Which powers lie between the highest power and the linear term? - How do you record a power that is not written in the expression?

1. List the powers from \(x^4\) down to \(x^0\). 2. The coefficients of \(x^3\) and \(x^2\) are both \(0\). 3. The coefficient row is \(4,\ 0,\ 0,\ -2,\ 7\).

\(4,\ 0,\ 0,\ -2,\ 7\)

- How does the degree of the remainder compare with the divisor's degree? - Which choice contains a squared term?

1. The divisor has degree \(2\). 2. A remainder must have degree less than \(2\), unless it is zero. 3. The linear expression \(3x-2\) is possible, but \(x^2+4\) has degree \(2\) and cannot be the final remainder.

\(3x-2\), because the remainder must have degree less than \(2\).

- Rewrite the divisor mentally in the form \(x-c\) to identify the synthetic value. - Use the coefficients in descending-power order. - The final synthetic entry is the remainder.

1. For divisor \(x+1=x-(-1)\), use synthetic value \(-1\) with coefficients \(1,2,-5,-6\). 2. Bring down \(1\); multiply by \(-1\) and add to get \(1\); multiply by \(-1\) and add to get \(-6\); multiply by \(-1\) and add to get remainder \(0\). 3. The quotient is \(x^2+x-6\) and the remainder is \(0\).

Quotient: \(x^2+x-6\); remainder: \(0\)

- Write a zero coefficient for every missing power before starting synthetic division. - Use the constant from \(x-1\) as the synthetic value. - Match the quotient coefficients to descending powers one degree lower than the dividend.

1. The polynomial has no \(x^3\)-term, so use coefficients \(2,0,-3,5,-4\) with synthetic value \(1\). 2. Bring down \(2\); multiply by \(1\) and add to get \(2\); multiply by \(1\) and add to get \(-1\); multiply by \(1\) and add to get \(4\); multiply by \(1\) and add to get remainder \(0\). 3. The quotient is \(2x^3+2x^2-x+4\).

Quotient: \(2x^3+2x^2-x+4\); remainder: \(0\)

- Include a zero coefficient for the missing \(x^2\)-term. - A nonzero final synthetic entry is the remainder, not another quotient coefficient. - Use the division identity to write the result after completing the synthetic work.

1. Include the missing \(x^2\)-term and use coefficients \(1,0,-4,7\) with synthetic value \(2\). 2. Bring down \(1\); multiply by \(2\) and add to get \(2\); multiply by \(2\) and add to get \(0\); multiply by \(2\) and add to get remainder \(7\). 3. The quotient is \(x^2+2x\), so \(x^3-4x+7=(x-2)(x^2+2x)+7\).

Quotient: \(x^2+2x\); remainder: \(7\) \(x^3-4x+7=(x-2)(x^2+2x)+7\)

- At each stage, divide the current leading term by the divisor's leading term. - Multiply the entire divisor by the new quotient term before subtracting. - Continue until the remaining degree is less than the divisor's degree.

1. Divide the leading terms: \(2x^3\div x=2x^2\). Multiply \(x+2\) by \(2x^2\) and subtract, leaving \(-x^2-8x-12\). 2. Divide \(-x^2\div x=-x\). Multiply \(x+2\) by \(-x\) and subtract, leaving \(-6x-12\). 3. Divide \(-6x\div x=-6\). Multiply \(x+2\) by \(-6\) and subtract, leaving remainder \(0\). 4. The quotient is \(2x^2-x-6\).

Quotient: \(2x^2-x-6\); remainder: \(0\)

- Look first at the degree and form of each divisor. - Which divisors can be represented by one synthetic value? - Compare the method requirements before choosing.

1. In part a, the divisor \(x-3\) is linear and has the form \(x-c\). Synthetic division is more efficient, although long division would also work. 2. In part b, the divisor \(x^2+1\) has degree \(2\). Polynomial long division is appropriate because standard synthetic division in this course is used for divisors of the form \(x-c\).

a) Synthetic division, because the divisor is of the form \(x-c\). Long division also works but takes more steps. b) Polynomial long division, because the divisor is quadratic. Standard synthetic division applies to divisors of the form \(x-c\).

- Verification reverses polynomial division. - Multiply the divisor by the proposed quotient first. - Add the proposed remainder only after expanding the product.

1. Multiply the divisor and quotient: \((x-2)(x^2+x+4)=x^3-x^2+2x-8\). 2. Add the claimed remainder: \(x^3-x^2+2x-8+5=x^3-x^2+2x-3\). 3. This matches the dividend exactly, so the quotient and remainder are correct.

The claim is correct because \((x-2)(x^2+x+4)+5=x^3-x^2+2x-3\).

- In long division, divide the current leading term by the divisor's leading term at each stage. - After each quotient term, multiply the entire divisor and subtract. - At the end, check that the remainder has lower degree than the divisor and reconstruct the dividend.

1. Divide the leading terms: \(6x^3\div 2x=3x^2\). Multiply back and subtract to obtain \(-4x^2-16x\). 2. Divide again: \(-4x^2\div 2x=-2x\). Multiply back and subtract to obtain \(-10x-6\). 3. Divide once more: \(-10x\div 2x=-5\). Multiply back and subtract to obtain the remainder \(9\). 4. The quotient is \(3x^2-2x-5\), with remainder \(9\). 5. Verification: \((2x+3)(3x^2-2x-5)+9=6x^3+5x^2-16x-6\), which reproduces the dividend.

Quotient: \(3x^2-2x-5\) Remainder: \(9\)

- What should multiply the divisor to cancel the highest power first? - After subtracting, repeat the same step with the term of highest degree that remains. - When does the division process stop?

1. Divide \(x^2\) by \(x\) to get the first quotient term \(x\). 2. Subtract \(x(x+2)=x^2+2x\) to leave \(2x+8\). 3. Divide \(2x\) by \(x\) to get \(2\); subtract \(2(x+2)=2x+4\). 4. The remainder is \(4\), so the quotient is \(x+2\).

Quotient: \(x+2\); remainder: \(4\).

- Which missing-power coefficient should appear before the constant? - In synthetic division, what happens to each new bottom-row entry before you add in the next column? - The final entry represents the remainder.

1. Use the coefficient row \(2,-3,0,5\), including the missing \(x\)-term. 2. Multiply the first bottom-row entry \(2\) by \(2\) and add to \(-3\), giving \(1\). 3. Multiply \(1\) by \(2\) and add to \(0\), giving \(2\). 4. Multiply \(2\) by \(2\) and add to \(5\), giving \(9\).

The next three entries are \(1,\ 2,\ 9\).

- What equation must connect a dividend, divisor, quotient, and remainder? - Multiply the proposed quotient by the divisor first. - Does adding the remainder reproduce the original polynomial?

1. A correct division must satisfy \(P(x)=(x-1)Q(x)+5\). 2. Multiply: \((x-1)(x^2+x+2)=x^3+x-2\). 3. Add \(5\) to get \(x^3+x+3=P(x)\). The proposed division is correct.

Correct: \((x-1)(x^2+x+2)+5=x^3+x+3\).

- What is the actual divisor when the number in synthetic division is \(2\)? - Compare the degrees of that divisor and the one in the question. - Would using only one factor of a product be equivalent to dividing by the whole product?

1. Synthetic division with \(2\) divides by the linear expression \(x-2\). 2. The requested divisor \(x^2-4=(x-2)(x+2)\) has two linear factors, so one synthetic division by \(x-2\) is not the complete requested division. 3. The student's claim is false. Long division by \(x^2-4\) would divide by the whole quadratic divisor.

No. Synthetic division with \(2\) divides by \(x-2\), not by \(x^2-4\).

- A known zero \(x=a\) gives a factor \(x-a\) and a synthetic value \(a\). - Keep the polynomial coefficients in descending-power order during synthetic division. - Factor the quadratic quotient after the division is complete. - In complete factored form, the exponent on each factor gives the multiplicity.

1. Since \(x=-1\) is a zero, divide by \(x+1\) using synthetic value \(-1\). Use coefficients \(1,-2,-13,-10\): bring down \(1\); multiply by \(-1\) and add to get \(-3\); multiply by \(-1\) and add to get \(-10\); multiply by \(-1\) and add to get remainder \(0\). The quotient is \(x^2-3x-10\). Factor the quotient: \(x^2-3x-10=(x-5)(x+2)\). Therefore, the zeros are \(x=-2\), \(x=-1\), and \(x=5\). 2. The factored form is \(f(x)=(x+1)(x-5)(x+2)\). 3. Each linear factor occurs once, so each zero has multiplicity \(1\).

1. \(x=-2\), \(x=-1\), and \(x=5\) 2. \(f(x)=(x+1)(x-5)(x+2)\) 3. Each zero has multiplicity \(1\).

- A known zero identifies the synthetic value and corresponding linear factor. - Keep the coefficients in descending-power order. - A zero remainder confirms that the known zero's factor divided evenly. - Factor the quadratic quotient to find the remaining zeros.

1. The zero \(x=-1\) corresponds to the factor \(x+1\), so use synthetic value \(-1\) with coefficients \(1,-6,3,10\). 2. Bring down \(1\). Multiply by \(-1\) and add to get \(-7\). Multiply by \(-1\) and add to get \(10\). Multiply by \(-1\) and add to get remainder \(0\). The quotient is \(x^2-7x+10\). 3. Factor the quotient: \(x^2-7x+10=(x-2)(x-5)\). 4. Therefore, the remaining zeros are \(x=2\) and \(x=5\).

\(x=2\) and \(x=5\)

- Substituting a value that produces \(0\) confirms a zero and its corresponding linear factor. - Use the confirmed zero as the synthetic value. - Keep track of the leading coefficient when factoring the quadratic quotient. - Check that the final product has the same degree and leading coefficient as the original polynomial.

1. Evaluate the polynomial: \(p(1)=2-4-22+24=0\). Therefore, \(x=1\) is a zero and \(x-1\) is a factor. 2. Use synthetic value \(1\) with coefficients \(2,-4,-22,24\). Bring down \(2\); multiply by \(1\) and add to get \(-2\); multiply by \(1\) and add to get \(-24\); multiply by \(1\) and add to get remainder \(0\). The quotient is \(2x^2-2x-24\). 3. Factor the quotient: \(2x^2-2x-24=2(x^2-x-12)=2(x-4)(x+3)\). 4. Therefore, the complete factorization is \(p(x)=2(x-1)(x-4)(x+3)\).

\(p(x)=2(x-1)(x-4)(x+3)\)

- A known zero determines both the divisor factor and its synthetic-division value. - Keep coefficients in descending-power order during synthetic division. - Use the remainder to check the factor before finding the other zeros. - Factor the quadratic quotient to find the remaining zeros.

1. Since \(x=3\) is a zero, \(x-3\) is a factor and the synthetic-division value is \(3\). 2. Use coefficients \(1,-6,-1,30\). Bring down \(1\), then multiply by \(3\) and add successively to obtain \(-3\), \(-10\), and remainder \(0\). The quotient is \(x^2-3x-10\). 3. Factor the quotient: \(x^2-3x-10=(x-5)(x+2)\). The remaining zeros are \(5\) and \(-2\), and \(f(x)=(x-3)(x-5)(x+2)\).

a) Since \(f(3)=0\), \(x-3\) is a factor and the synthetic-division value is \(3\). b) Remaining zeros: \(x=-2\) and \(x=5\); \(f(x)=(x-3)(x-5)(x+2)\)

- Substitute each proposed value and check whether the output is \(0\). - Use one confirmed zero as the synthetic value to reduce the cubic to a quadratic. - Keep the coefficients in descending-power order. - Factor the quadratic quotient to finish finding the zeros.

1. Evaluate the proposed values: \(f(1)=4\), \(f(-1)=0\), \(f(2)=0\), and \(f(-2)=-20\). Therefore, \(x=-1\) and \(x=2\) are zeros among the tested values. 2. Use synthetic value \(-1\) with coefficients \(1,-4,1,6\). Bring down \(1\); multiply by \(-1\) and add to get \(-5\); multiply by \(-1\) and add to get \(6\); multiply by \(-1\) and add to get remainder \(0\). The quotient is \(x^2-5x+6\). 3. Factor the quotient: \(x^2-5x+6=(x-2)(x-3)\). 4. Thus, all zeros are \(x=-1\), \(x=2\), and \(x=3\).

Tested zeros: \(x=-1\) and \(x=2\) All zeros: \(x=-1\), \(x=2\), and \(x=3\)

- Use each known zero as a synthetic value, one after the other. - Keep the quotient coefficients in descending-power order before the second division. - A zero remainder at each stage confirms that the corresponding linear factor divides evenly. - Factor the final quadratic quotient.

1. Use synthetic value \(-2\) with coefficients \(1,-4,-7,22,24\). Bring down \(1\); multiply by \(-2\) and add to get \(-6\); multiply by \(-2\) and add to get \(5\); multiply by \(-2\) and add to get \(12\); multiply by \(-2\) and add to get remainder \(0\). The quotient is \(x^3-6x^2+5x+12\). 2. Use synthetic value \(3\) with quotient coefficients \(1,-6,5,12\). Bring down \(1\); multiply by \(3\) and add to get \(-3\); multiply by \(3\) and add to get \(-4\); multiply by \(3\) and add to get remainder \(0\). The new quotient is \(x^2-3x-4\). 3. Factor the quadratic: \(x^2-3x-4=(x-4)(x+1)\). 4. Therefore, the remaining zeros are \(x=4\) and \(x=-1\).

\(x=-1\) and \(x=4\)

- Use each known zero as a synthetic value, one at a time. - Each zero remainder lowers the degree by \(1\) and confirms the corresponding linear factor. - Factor the remaining quadratic after the two divisions. - Check the signs in the factors: a zero \(r\) corresponds to \(x-r\).

1. Use synthetic value \(1\) with coefficients \(1,2,-7,-8,12\). Bring down \(1\); multiply by \(1\) and add to get \(3\); multiply by \(1\) and add to get \(-4\); multiply by \(1\) and add to get \(-12\); multiply by \(1\) and add to get remainder \(0\). The quotient is \(x^3+3x^2-4x-12\). 2. Use synthetic value \(-2\) with coefficients \(1,3,-4,-12\). Bring down \(1\); multiply by \(-2\) and add to get \(1\); multiply by \(-2\) and add to get \(-6\); multiply by \(-2\) and add to get remainder \(0\). The quotient is \(x^2+x-6\). 3. Factor the remaining quadratic: \(x^2+x-6=(x-2)(x+3)\). 4. Therefore, \(p(x)=(x-1)(x+2)(x-2)(x+3)\). The zeros are \(x=-3\), \(x=-2\), \(x=1\), and \(x=2\).

The complete factorization is \(p(x)=(x-1)(x+2)(x-2)(x+3)\). The zeros are \(x=-3\), \(x=-2\), \(x=1\), and \(x=2\).

- Use the zero \(2\) as the synthetic value each time you test another copy of the factor \(x-2\). - Count consecutive zero remainders to determine multiplicity. - When a quotient no longer gives zero at \(x=2\), stop dividing by \(x-2\). - Factor the final quotient completely.

1. Verify the zero: \(f(2)=16-32+12+8-4=0\). 2. Use synthetic value \(2\) with coefficients \(1,-4,3,4,-4\). Bring down \(1\); multiply by \(2\) and add to get \(-2\); multiply by \(2\) and add to get \(-1\); multiply by \(2\) and add to get \(2\); multiply by \(2\) and add to get remainder \(0\). The quotient is \(x^3-2x^2-x+2\). 3. Divide again using synthetic value \(2\) with coefficients \(1,-2,-1,2\). Bring down \(1\); multiply by \(2\) and add to get \(0\); multiply by \(2\) and add to get \(-1\); multiply by \(2\) and add to get remainder \(0\). The quotient is \(x^2-1\). 4. The remaining quotient is not divisible by \(x-2\) because \(2^2-1=3\ne0\). Therefore, \(x=2\) has multiplicity \(2\). 5. Factor \(x^2-1=(x-1)(x+1)\). Thus, \(f(x)=(x-2)^2(x-1)(x+1)\).

The zero \(x=2\) has multiplicity \(2\). The complete factorization is \(f(x)=(x-2)^2(x-1)(x+1)\).

- Use \(2\) as the synthetic value for each successive division. - Each zero remainder confirms one more copy of the factor \(x-2\). - After the third successful division, test the remaining factor at \(x=2\).

1. Use synthetic value \(2\) with coefficients \(1,-3,-6,28,-24\). Bring down \(1\); multiply by \(2\) and add to get \(-1\); multiply by \(2\) and add to get \(-8\); multiply by \(2\) and add to get \(12\); multiply by \(2\) and add to get remainder \(0\). The quotient is \(x^3-x^2-8x+12\). 2. Divide again with coefficients \(1,-1,-8,12\). Bring down \(1\); multiply by \(2\) and add to get \(1\); multiply by \(2\) and add to get \(-6\); multiply by \(2\) and add to get remainder \(0\). The quotient is \(x^2+x-6\). 3. Divide a third time with coefficients \(1,1,-6\). Bring down \(1\); multiply by \(2\) and add to get \(3\); multiply by \(2\) and add to get remainder \(0\). The remaining factor is \(g(x)=x+3\). 4. Since \(g(2)=5\ne0\), a fourth division by \(x-2\) would not have zero remainder. Therefore, \(x=2\) has multiplicity \(3\), not \(4\).

After three divisions, \(g(x)=x+3\). Because \(g(2)=5\ne0\), \(x=2\) has multiplicity \(3\) and not multiplicity \(4\).

- Possible integer zeros are related to factors of the constant term. - Once an integer zero is confirmed, use it as a synthetic value to lower the degree. - Repeat the process with another confirmed zero. - Solve the final quadratic equation exactly.

1. The possible integer zeros are among \(\pm1\), \(\pm2\), \(\pm5\), and \(\pm10\). Testing shows that \(f(1)=0\) and \(f(-2)=0\). 2. Use synthetic value \(1\) with coefficients \(1,1,-7,-5,10\). Bring down \(1\); multiply by \(1\) and add to get \(2\); multiply by \(1\) and add to get \(-5\); multiply by \(1\) and add to get \(-10\); multiply by \(1\) and add to get remainder \(0\). The quotient is \(x^3+2x^2-5x-10\). 3. Use synthetic value \(-2\) with coefficients \(1,2,-5,-10\). Bring down \(1\); multiply by \(-2\) and add to get \(0\); multiply by \(-2\) and add to get \(-5\); multiply by \(-2\) and add to get remainder \(0\). The quotient is \(x^2-5\). 4. Solve \(x^2-5=0\): \(x=-\sqrt{5}\) or \(x=\sqrt{5}\). 5. Therefore, the zeros are \(x=-\sqrt{5}\), \(x=-2\), \(x=1\), and \(x=\sqrt{5}\).

The zeros are \(x=-\sqrt{5}\), \(x=-2\), \(x=1\), and \(x=\sqrt{5}\).

- Start by testing integer factors of the constant term. - Use each confirmed zero as a synthetic value. - After each zero remainder, continue with the quotient coefficients. - Stop when the quotient is quadratic, then solve that equation exactly.

1. Possible integer zeros are among \(\pm1\), \(\pm2\), \(\pm3\), and \(\pm6\). Testing identifies \(x=1\), \(x=-1\), and \(x=2\) as zeros. 2. Use synthetic value \(1\) with coefficients \(1,-2,-4,8,3,-6\). Bring down \(1\); multiply and add successively to obtain \(-1,-5,3,6\), then remainder \(0\). The quotient is \(x^4-x^3-5x^2+3x+6\). 3. Use synthetic value \(-1\) with coefficients \(1,-1,-5,3,6\). Bring down \(1\); multiply and add successively to obtain \(-2,-3,6\), then remainder \(0\). The quotient is \(x^3-2x^2-3x+6\). 4. Use synthetic value \(2\) with coefficients \(1,-2,-3,6\). Bring down \(1\); multiply and add successively to obtain \(0,-3\), then remainder \(0\). The quotient is \(x^2-3\). 5. Solve \(x^2-3=0\): \(x=-\sqrt{3}\) or \(x=\sqrt{3}\). 6. Therefore, the zeros are \(x=-\sqrt{3}\), \(x=-1\), \(x=1\), \(x=\sqrt{3}\), and \(x=2\).

The zeros are \(x=-\sqrt{3}\), \(x=-1\), \(x=1\), \(x=\sqrt{3}\), and \(x=2\).

- Set the two polynomial expressions equal and move all terms to one side. - After finding an integer solution, use it as the synthetic value. - Remember to include a zero coefficient for any missing power in the cubic. - Factor the quadratic quotient and then use either original function for the y-coordinates.

1. Set the functions equal: \(-x^3+3x^2=4\), so \(x^3-3x^2+4=0\). 2. Test small integer values. Since \((-1)^3-3(-1)^2+4=0\), \(x=-1\) is a solution and \(x+1\) is a factor. 3. The cubic has coefficients \(1,-3,0,4\); include the zero coefficient for the missing \(x\)-term. Use synthetic value \(-1\): bring down \(1\); multiply by \(-1\) and add to get \(-4\); multiply by \(-1\) and add to get \(4\); multiply by \(-1\) and add to get remainder \(0\). The quotient is \(x^2-4x+4\). 4. Factor the quotient: \(x^2-4x+4=(x-2)^2\). Therefore, the x-coordinates are \(x=-1\) and \(x=2\). 5. Because \(g(x)=4\), both corresponding y-coordinates are \(4\). 6. The intersection points are \((-1, 4)\) and \((2, 4)\).

The intersection points are \((-1, 4)\) and \((2, 4)\).

- Synthetic division is not appropriate for a quadratic divisor, so use long division. - Divide leading terms to determine each new quotient term. - Subtract the full product at every stage and keep like powers aligned.

1. Divide \(2x^4\) by \(x^2\) to get \(2x^2\). Multiply the divisor by \(2x^2\) and subtract, leaving \(-x^3-4x^2-x+6\). 2. Divide \(-x^3\) by \(x^2\) to get \(-x\). Multiply the divisor by \(-x\) and subtract, leaving \(-3x^2-3x+6\). 3. Divide \(-3x^2\) by \(x^2\) to get \(-3\). Multiply the divisor by \(-3\) and subtract, leaving remainder \(0\). 4. The quotient is \(2x^2-x-3\).

Quotient: \(2x^2-x-3\); remainder: \(0\)

- Insert zero-coefficient terms so like powers stay aligned. - Long division stops when the remainder's degree is less than the divisor's degree. - Subtract the entire product at each stage, including signs.

1. Write the dividend with missing terms: \(x^4+0x^3+2x^2+0x-3\). 2. Divide \(x^4\) by \(x^2\) to get \(x^2\). Subtract \(x^2(x^2-x+1)\), leaving \(x^3+x^2-3\). 3. Divide \(x^3\) by \(x^2\) to get \(x\). Subtract \(x(x^2-x+1)\), leaving \(2x^2-x-3\). 4. Divide \(2x^2\) by \(x^2\) to get \(2\). Subtract \(2(x^2-x+1)\), leaving \(x-5\). 5. Because the remainder has degree less than \(2\), the quotient is \(x^2+x+2\) and the remainder is \(x-5\).

Quotient: \(x^2+x+2\); remainder: \(x-5\)

- Reverse the division identity to reconstruct the dividend. - Expand the divisor-quotient product before adding the remainder. - The check value comes from the zero of the divisor \(x+3\).

1. Use the division identity: \(P(x)=(x+3)(2x^2-x+4)-5\). 2. Expand \((x+3)(2x^2-x+4)=2x^3+5x^2+x+12\), so \(P(x)=2x^3+5x^2+x+7\). 3. Evaluate \(P(-3)=-54+45-3+7=-5\), which agrees with the stated remainder.

a) \(P(x)=2x^3+5x^2+x+7\) b) \(P(-3)=-5\)

- Relate the synthetic value to the zero of the linear divisor. - Check the sign carefully before starting the synthetic-division arithmetic. - Use the final remainder to check the factor conclusion.

1. The divisor \(x+2\) is \(x-(-2)\), so the correct synthetic value is \(-2\), not \(2\). 2. Use coefficients \(1,3,-1,-6\). Bring down \(1\); multiply by \(-2\) and add to get \(1\); multiply by \(-2\) and add to get \(-3\); multiply by \(-2\) and add to get remainder \(0\). 3. The quotient is \(x^2+x-3\), the remainder is \(0\), and \(x+2\) is a factor.

Elif used the wrong synthetic value. The correct value is \(-2\), giving quotient \(x^2+x-3\) and remainder \(0\).

- Use the same synthetic value repeatedly while the remainder stays \(0\). - Each successful division removes one copy of the corresponding linear factor. - Stop when the next quotient no longer evaluates to \(0\) at the same value.

1. Use synthetic value \(1\) with coefficients \(1,-1,-3,5,-2\). The quotient coefficients are \(1,0,-3,2\) with remainder \(0\), so one factor \(x-1\) is removed. 2. Divide the quotient again by \(x-1\): using coefficients \(1,0,-3,2\) gives quotient \(x^2+x-2\) with remainder \(0\). 3. Divide once more by \(x-1\): using coefficients \(1,1,-2\) gives quotient \(x+2\) with remainder \(0\). 4. The remaining quotient has value \(3\) at \(x=1\), so it is not divisible by \(x-1\) again. Therefore, \(x=1\) has multiplicity \(3\). 5. The complete factorization is \(f(x)=(x-1)^3(x+2)\).

The zero \(x=1\) has multiplicity \(3\), and \(f(x)=(x-1)^3(x+2)\).

- Reverse the division identity to construct the polynomial. - For the second division, pay attention to the sign in \(x+1\). - Check each remainder by substituting the zero of its divisor.

1. Use the division identity: \(P(x)=(x-2)(x^3-3x+1)+5=x^4-2x^3-3x^2+7x+3\). 2. For \(x+1\), use synthetic value \(-1\) with coefficients \(1,-2,-3,7,3\). The quotient is \(x^3-3x^2+7\) and the remainder is \(-4\). 3. Evaluate \(P(2)=5\) and \(P(-1)=-4\); both agree with the division remainders.

a) \(P(x)=x^4-2x^3-3x^2+7x+3\) b) Quotient \(x^3-3x^2+7\); remainder \(-4\) c) \(P(2)=5\) and \(P(-1)=-4\)

- Start with the degree restriction on a remainder when the divisor is quadratic. - Use \(P(x)=d(x)q(x)+r(x)\) at each of the two given x-values. - The two evaluations give two conditions on the two coefficients of the remainder.

1. Because the divisor has degree \(2\), the remainder must have degree less than \(2\). Write \(r(x)=ax+b\). 2. The division identity is \(P(x)=(x^2+1)(2x^2-x+3)+ax+b\). 3. At \(x=1\), the divisor-quotient product equals \(8\), so \(a+b=11-8=3\). 4. At \(x=-1\), the divisor-quotient product equals \(12\), so \(-a+b=5-12=-7\). 5. Solving the two equations gives \(a=5\) and \(b=-2\). Therefore, \(r(x)=5x-2\).

\(r(x)=5x-2\)

- What product of the first quotient term and the width cancels the \(x^2\) term? - Subtract that product from the total area, then continue dividing. - How does the division identity separate the rectangular area from the leftover fixed area?

1. Divide the leading term: \(2x^2\div x=2x\). Subtract \((x+2)(2x)=2x^2+4x\) from \(A(x)\) to leave \(3x+11\). 2. Divide \(3x\div x=3\). Subtract \(3(x+2)=3x+6\) to leave remainder \(5\). 3. Thus \(A(x)=(x+2)(2x+3)+5\). The quotient \(2x+3\) gives the platform's length in feet, and the remainder \(5\) is the ramp's area in square feet.

Quotient: \(2x+3\), so the platform's length is \((2x+3)\,\text{ft}\). Remainder: \(5\), so the ramp's area is \(5\,\text{ft}^2\).

- First express the area of the rectangular base as one polynomial. - Which term should the quotient start with when you divide the leading terms? - Multiply the resulting height by the base area to check the volume.

1. The base area is \((x+2)(x+3)=x^2+5x+6\) square inches. 2. Divide \(V(x)\) by \(x^2+5x+6\). The first quotient term is \(2x\); subtracting \(2x^3+10x^2+12x\) leaves \(3x^2+15x+18\). 3. The next quotient term is \(3\). Subtracting \(3x^2+15x+18\) leaves remainder \(0\). 4. The height is \((2x+3)\,\text{in}\). Checking: \((x^2+5x+6)(2x+3)=2x^3+13x^2+27x+18\).

Divide by the base area \(x^2+5x+6\). Subtract \(2x(x^2+5x+6)\) to leave \(3x^2+15x+18\), then subtract \(3(x^2+5x+6)\) to leave \(0\). Quotient \(2x+3\); height \((2x+3)\,\text{in}\). Check: \((x^2+5x+6)(2x+3)=2x^3+13x^2+27x+18\).