Question

use synthetic division to find the quotient and remainder when (-x^4 + 6x^3 + 3x - 9) is divided by (x - 6) by completing the parts below. (a) complete this synthetic division table. (\begin{array}{r|rrrrr}6&-1&6&0&3&-9\\&&&&&\hline&&&&&end{array}) (b) write your answer in the following form: quotient (+\frac{\text{remainder}}{x - 6}). (\frac{-x^4 + 6x^3 + 3x - 9}{x - 6}=square+\frac{square}{x - 6})

Explanation:

Step1: Recall Synthetic Division Rules

For dividing a polynomial \( f(x) = a_nx^n + \dots + a_1x + a_0 \) by \( x - c \), we use the coefficients of \( f(x) \) (including zeros for missing terms) and bring down the leading coefficient, then multiply by \( c \), add to the next coefficient, and repeat. Here, \( f(x)=-x^4 + 6x^3+0x^2 + 3x - 9 \), \( c = 6 \), coefficients are \(-1, 6, 0, 3, -9\).

Step2: Perform Synthetic Division

• Bring down the first coefficient: \(-1\).
• Multiply by \( 6 \): \(-1\times6=-6\). Add to next coefficient: \( 6 + (-6)=0 \).
• Multiply \( 0 \) by \( 6 \): \( 0\times6 = 0 \). Add to next coefficient: \( 0+0 = 0 \).
• Multiply \( 0 \) by \( 6 \): \( 0\times6 = 0 \). Add to next coefficient: \( 3+0 = 3 \).
• Multiply \( 3 \) by \( 6 \): \( 3\times6 = 18 \). Add to last coefficient: \( -9 + 18 = 9 \).

So the synthetic division table (the second row of products: \(-6, 0, 0, 18\); the third row of results: \(-1, 0, 0, 3, 9\)).

Step3: Determine Quotient and Remainder

The quotient is a polynomial of degree one less than the dividend. The coefficients of the quotient are \(-1, 0, 0, 3\), so quotient is \( -x^3+0x^2 + 0x + 3=-x^3 + 3 \). The remainder is the last number, \( 9 \).

Answer:

(a) The synthetic division table (second row: \(-6, 0, 0, 18\); third row: \(-1, 0, 0, 3, 9\))
(b) \(\frac{-x^4 + 6x^3 + 3x - 9}{x - 6}=-x^3 + 3+\frac{9}{x - 6}\)
So the quotient is \(-x^3 + 3\) and the remainder is \(9\). For part (b), the first box is \(-x^3 + 3\) and the second box is \(9\).