Question

A) Use synthetic division to divide  f(x) x - k

B) Rewrite f(x) as the divisor times the quotient plus the remainder, 
i.e., f(x) = (x - k) ( quotient )  +  remainder

C) Find f(k) using the Remainder Theorem.

f(x) = 5x4 + 7x3 + 6x + 3, k = -2

Answer 1

part A)

$\bf \begin{array}{llll} f(x)=5x^4+7x^3+6x+3\\ f(x)=5x^4+7x^3\boxed{+0x^2}+6x+3 \end{array} \qquad \begin{cases} x=-2\\ x+2=0\\ (x+2)\leftarrow factor \end{cases}\\\\ -----------------------------\\\\ \begin{array}{l|rrrrrr} -2&5&7&0&6&3\\ &&-10&6&-12&12\\ --&--&--&--&--&--\\ &5&-3&6&-6&\boxed{15}&\leftarrow remainder \end{array}$

now, the part B), what factors does that division give then? well

$\bf 5x^3-3x^2+6x-6+\boxed{\frac{15}{x+2}}\impliedby \begin{array}{llll} \textit{that'd be the quotient}\\ \textit{with the remainder added} \end{array} \\\\\\ thus \\\\\\ f(x)=5x^4+7x^3+6x+3\implies (x+2)\left( 5x^3-3x^2+6x-6+\frac{15}{x+2} \right)$