Question

Use synthetic division and the Remainder Theorem to evaluate P(c), where p(x)=x^4+7x^3+2x^2+20x+50 c=-7

Answer 1

p(c) = 8

Explanation:

Given polynomial:

`p(x)=x^4+7x^3+2x^2+20x+50`

Remainder Theorem

When we divide a polynomial f(x) by (x − c) the remainder is f(c).

Therefore, if c = -7, then the divisor is:

`\implies (x - (-7)) = (x + 7)`

Perform Synthetic Division

Place "c" in the division box.

Write the coefficients of the dividend in descending order.

(Note: As no terms are missing, we do not need to use any zeros to fill in missing terms).

`\begin{array}ccccc-7 & 1 & 7 & 2 & 20 & 50\\\cline{1-1}\end{array}`

Bring the leading coefficient straight down:

`\begin{array}ccccc-7 & 1 & 7 & 2 & 20 & 50\\\cline{1-1}&\downarrow &&&&\\\cline{2-6}&1\end{array}`

Multiply the number you brought down with the number in the division box and put the result in the next column (under the 7):

`\begin{array}ccccc-7 & 1 & 7 & 2 & 20 & 50\\\cline{1-1}&\downarrow &-7&&&\\\cline{2-6}&1\end{array}`

Add the two numbers together and put the result in the bottom row:

`\begin{array}ccccc-7 & 1 & 7 & 2 & 20 & 50\\\cline{1-1}&\downarrow &-7&&&\\\cline{2-6}&1&0\end{array}`

Repeat:

`\begin{array}c-7 & 1 & 7 & 2 & 20 & 50\\\cline{1-1}&\downarrow &-7&0&&\\\cline{2-6}&1&0&2\end{array}`

`\begin{array}ccccc-7 & 1 & 7 & 2 & 20 & 50\\\cline{1-1}&\downarrow &-7&0&-14&\\\cline{2-6}&1&0&2&6\end{array}`

`\begin{array}c-7 & 1 & 7 & 2 & 20 & 50\\\cline{1-1}&\downarrow &-7&0&-14&-42\\\cline{2-6}&1&0&2&6&8\end{array}`

The bottom row (except the last number) gives the coefficients of the quotient. The degree of the quotient is one less than that of the dividend.

The last number in the bottom row is the remainder.

`\textsf{Quotient}: \quad x^3+2x+6`

`\textsf{Remainder}: \quad 8`

Therefore, as the remainder is p(c) then:

`p(c)=8`

Check by evaluating p(-7):

`\begin{aligned}\implies p(-7)&=(-7)^4+7(-7)^3+2(-7)^2+20(-7)+50\\&=2401-2401+98-140+50\\&=0+98-140+50\\&=98-140+50\\&=-42+50\\&=8 \end{aligned}`