Answer
To divide $4{{x}^{3}}-8x-2$ by $x+5$ using synthetic division, the first step is to write $\left. {\underline {\,
-5 \,}}\! \right| $ , $4$ , $0$ , $-8$ , $-2$.
Work Step by Step
Let us consider,
$x+5\overline{\left){4{{x}^{3}}-8x-2}\right.}$.
And the coefficient of the dividend is:
$\underbrace{4}_{\text{coe}\text{. of }{{x}^{3}}}\text{ }\underbrace{0}_{\text{coe}\text{. of }{{x}^{2}}}\text{ }\underbrace{-8}_{\text{coe}\text{. of }{{x}^{1}}}\text{ }\underbrace{-2}_{\text{coe}\text{. of }{{x}^{0}}}$.
And the constant for the divisor is:
$x-\text{c}=x+5\Rightarrow \text{c}=-5$.
We know that according to the synthetic division procedure, the first step is:
$\left. {\underline {\,
-5 \,}}\! \right| \text{ }4\text{ }0\text{ }-8\text{ }-2$.
Thus, the first step is to write $\left. {\underline {\,
-5 \,}}\! \right| $ $4$ $0$ $-8$ $-2$.
