# How do you evaluate f(x)=-x^5-4x^3+6x^2-x at x=-2 using direct substitution and synthetic division?

##### 1 Answer
Apr 29, 2017

$f \left(- 2\right) = 90$ for details please see below.

#### Explanation:

To evaluate $f \left(x\right) = - {x}^{5} - 4 {x}^{3} + 6 {x}^{2} - x$ at x=-2 using direct substitution, we just put value of $x$ in the polynomial.

Hence, $f \left(- 2\right) = - {\left(- 2\right)}^{5} - 4 {\left(- 2\right)}^{3} + 6 {\left(- 2\right)}^{2} - \left(- 2\right)$

= $- \left(- 32\right) - 4 \left(- 8\right) + 6 \times 4 + 2$

= $32 + 32 + 24 + 2 = 90$

Before we talk about synthetic division, one thing regarding remainder theorem which states that when a polynomial $f \left(x\right)$ is divided by $x - a$, the remainder is $f \left(a\right)$.

Hence, to find $f \left(- 2\right)$, we should divide $f \left(x\right)$ by $x - \left(- 2\right) = x + 2$, the remainder so obtained is $f \left(- 2\right)$. Hence let us divide $f \left(x\right) = - {x}^{5} - 4 {x}^{3} + 6 {x}^{2} - x$ by $x + 2$ using synthetic division method.

One Write the coefficients of $x$ in the dividend inside an upside-down division symbol. We are using $0$ as coefficient of ${x}^{4}$ and constant term.

$\textcolor{w h i t e}{1} | \textcolor{w h i t e}{X} - 1 \text{ "color(white)(X)0color(white)(XX)-4" "" "6" "" "-1" "" } 0$
$\textcolor{w h i t e}{1} | \text{ } \textcolor{w h i t e}{X}$
" "stackrel("——————————————————------)

Two As $x + 2 = 0$ gives $x = - 2$ put $- 2$ at the left.

$- 2 | \textcolor{w h i t e}{X} - 1 \text{ "color(white)(X)0color(white)(XX)-4" "" "6" "" "-1" "" } 0$
$\textcolor{w h i t e}{\times} | \text{ } \textcolor{w h i t e}{X X}$
" "stackrel("—————————————---------)

Three Drop the first coefficient of the dividend below the division symbol.

$- 2 | \textcolor{w h i t e}{X} - 1 \text{ "color(white)(X)0color(white)(XX)-4" "" "6" "" "-1" "" } 0$
$\textcolor{w h i t e}{\times} | \text{ } \textcolor{w h i t e}{X}$
" "stackrel("—————————————---------)
$\textcolor{w h i t e}{\times} | \textcolor{w h i t e}{x X} \textcolor{red}{- 1}$

Four Multiply the result by the constant, and put the product in the next column.

$- 2 | \textcolor{w h i t e}{X} - 1 \text{ "color(white)(X)0color(white)(XX)-4" "" "6" "" "-1" "" } 0$
$\textcolor{w h i t e}{\times} | \text{ } \textcolor{w h i t e}{\times X \times} 2$
" "stackrel("————————————---------—)
$\textcolor{w h i t e}{\times} | \textcolor{w h i t e}{X} \textcolor{b l u e}{-} 1$

Five Add down the column.

$- 2 | \textcolor{w h i t e}{X} 1 \text{ "color(white)(X)0color(white)(XX)-4" "" "6" "" "-1" "" } 0$
$\textcolor{w h i t e}{\times} | \text{ } \textcolor{w h i t e}{x X x} 2$
" "stackrel("—————————————---------)
$\textcolor{w h i t e}{\times} | \textcolor{w h i t e}{} \textcolor{b l u e}{-} 1 \textcolor{w h i t e}{X 1} \textcolor{red}{2}$

Six Repeat Steps Four and Five until you can go no farther.

$- 2 | \textcolor{w h i t e}{X} 1 \text{ "color(white)(X)0color(white)(X)-4" "" "6" "" "-1" "" } 0$
$\textcolor{w h i t e}{\times} | \text{ } \textcolor{w h i t e}{X \times} 2 \textcolor{w h i t e}{\times} - 4 \textcolor{w h i t e}{\times x} 16 \textcolor{w h i t e}{\times x} - 44 \textcolor{w h i t e}{\times x} 90$
" "stackrel("———--------——————————)
$\textcolor{w h i t e}{\times} | \textcolor{w h i t e}{} \textcolor{b l u e}{-} 1 \textcolor{w h i t e}{\times x} \textcolor{red}{2} \textcolor{w h i t e}{X} \textcolor{red}{-} 8 \textcolor{w h i t e}{X X} \textcolor{red}{22} \textcolor{w h i t e}{X X} \textcolor{red}{-} 45 \textcolor{w h i t e}{\times x} 90$

Hence, Quotient is $- {x}^{4} + 2 {x}^{3} - 8 {x}^{2} + 22 x - 45$ and remainder is $90$ an hence $f \left(- 2\right) = 90$