Question

use synthetic division to find the result when $2x^4 + 16x^3 + 11x^2 - 23x - 14$ is divided by $x + 7$.

Explanation:

Step1: Identify the root

For the divisor \(x + 7\), the root \(c=-7\) (since \(x+7 = x-(-7)\)). The coefficients of the dividend \(2x^{4}+16x^{3}+11x^{2}-23x - 14\) are \(2,16,11,-23,-14\). But synthetic division is typically for monic divisors (leading coefficient 1) when the dividend is a polynomial with integer coefficients. Wait, actually, we can adjust. Wait, the dividend is \(2x^{4}+16x^{3}+11x^{2}-23x - 14\) and divisor is \(x + 7\) (which is monic, leading coefficient 1). So we set up synthetic division with \(c=-7\) and coefficients \(2,16,11,-23,-14\).

Wait, no, synthetic division for a divisor \(x - c\) (here \(x+7=x-(-7)\), so \(c = -7\)). The coefficients of the dividend are \(a_4 = 2\), \(a_3=16\), \(a_2 = 11\), \(a_1=-23\), \(a_0=-14\).

Step2: Set up synthetic division

Write down \(c=-7\) and the coefficients:

$$
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$$

of each step:

• Bring down the leading coefficient \(2\).
• Multiply \(2\) by \(-7\) to get \(-14\). Add to the next coefficient \(16\): \(16+(-14)=2\).
• Multiply \(2\) by \(-7\) to get \(-14\). Add to the next coefficient \(11\): \(11+(-14)=-3\).
• Multiply \(-3\) by \(-7\) to get \(21\). Add to the next coefficient \(-23\): \(-23 + 21=-2\).
• Multiply \(-2\) by \(-7\) to get \(14\). Add to the last coefficient \(-14\): \(-14+14 = 0\).

The last number is the remainder (which is \(0\), so \(x + 7\) is a factor). The other numbers are the coefficients of the quotient polynomial, starting from one degree less than the dividend. So the quotient is \(2x^{3}+2x^{2}-3x - 2\) (since the dividend was degree 4, quotient is degree 3, with coefficients \(2,2,-3,-2\)).

Wait, let's verify: \((x + 7)(2x^{3}+2x^{2}-3x - 2)=x(2x^{3}+2x^{2}-3x - 2)+7(2x^{3}+2x^{2}-3x - 2)=2x^{4}+2x^{3}-3x^{2}-2x+14x^{3}+14x^{2}-21x - 14=2x^{4}+(2x^{3}+14x^{3})+(-3x^{2}+14x^{2})+(-2x-21x)-14=2x^{4}+16x^{3}+11x^{2}-23x - 14\), which matches the dividend. So the division gives a quotient of \(2x^{3}+2x^{2}-3x - 2\) and remainder \(0\).

Answer:

The result of the division is \(2x^{3}+2x^{2}-3x - 2\) (since the remainder is \(0\), we can say the quotient is \(2x^{3}+2x^{2}-3x - 2\)).