# How do you use synthetic substitution to evaluate f(2) for f(x)=x^3+2x^2-4x-5?

Nov 4, 2016

$f \left(2\right) = 3$

#### Explanation:

$f \left(x\right) = {x}^{3} + 2 {x}^{2} - 4 x - 5$
When the problem asks to evaluate f(2), simply substitute in 2 for every time you see "x" in the function. So we have:
$f \left(2\right) = {\left(2\right)}^{3} + 2 {\left(2\right)}^{2} - 4 \left(2\right) - 5$
Now we simplify each term:
$f \left(2\right) = 8 + 2 \left(4\right) - 8 - 5$
$f \left(2\right) = 8 + 8 - 8 - 5$
$f \left(2\right) = 3$

Nov 4, 2016

$f \left(2\right) = 3$

#### Explanation:

According to Remainder theorem, if a polynomial $f \left(x\right)$ is divided by a monomial of degree one say $x - a$, the remainder is $f \left(a\right)$.

Hence, to evaluate $f \left(2\right)$ for ${x}^{3} + 2 {x}^{2} - 4 x - 5$, we should divided it by $x - 2$

One Write the coefficients of $x$ in the dividend inside an upside-down division symbol.

$\textcolor{w h i t e}{1} | \textcolor{w h i t e}{X} 1 \text{ "color(white)(X)2color(white)(XX)-4" "" } - 5$
$\textcolor{w h i t e}{1} | \text{ } \textcolor{w h i t e}{X}$
" "stackrel("—————————————)

Two Put the divisor at the left.

$2 | \textcolor{w h i t e}{X} 1 \text{ "color(white)(X)2color(white)(XX)-4" "" } - 5$
$\textcolor{w h i t e}{1} | \text{ } \textcolor{w h i t e}{X}$
" "stackrel("—————————————)

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

$2 | \textcolor{w h i t e}{X} 1 \text{ "color(white)(X)2color(white)(XX)-4" "" } - 5$
$\textcolor{w h i t e}{1} | \text{ } \textcolor{w h i t e}{X}$
" "stackrel("—————————————)
$\textcolor{w h i t e}{1} | \textcolor{w h i t e}{X} \textcolor{red}{1}$

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

$2 | \textcolor{w h i t e}{X} 1 \text{ "color(white)(X)2color(white)(XX)-4" "" } - 5$
$\textcolor{w h i t e}{1} | \text{ } \textcolor{w h i t e}{X X 1} 2$
" "stackrel("—————————————)
$\textcolor{w h i t e}{1} | \textcolor{w h i t e}{X} \textcolor{b l u e}{1}$

$2 | \textcolor{w h i t e}{X} 1 \text{ "color(white)(X)2color(white)(XX)-4" "" } - 5$
$\textcolor{w h i t e}{1} | \text{ } \textcolor{w h i t e}{X X 1} 2$
" "stackrel("—————————————)
$\textcolor{w h i t e}{1} | \textcolor{w h i t e}{X} \textcolor{b l u e}{1} \textcolor{w h i t e}{X 11} \textcolor{red}{4}$

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

$2 | \textcolor{w h i t e}{X} 1 \text{ "color(white)(X)2color(white)(XX)-4" "" } - 5$
$\textcolor{w h i t e}{1} | \text{ } \textcolor{w h i t e}{X X 1} 2 \textcolor{w h i t e}{X X X X} 8 \textcolor{w h i t e}{X X \times} 8$
" "stackrel("—————————————)
$\textcolor{w h i t e}{1} | \textcolor{w h i t e}{X} \textcolor{b l u e}{1} \textcolor{w h i t e}{X 11} \textcolor{red}{4} \textcolor{w h i t e}{X X X x} \textcolor{red}{4} \textcolor{w h i t e}{X X X X} \textcolor{red}{3}$

Remainder is $3$. Hence $f \left(2\right) = 3$