# How do you use the remainder theorem to evaluate f(x)=x^3+x^2-5x-6 at x=2?

Oct 1, 2017

$f \left(2\right) = - 4$

#### Explanation:

We use the remainder theorem to establish what the remainder is when we divide a polynomial function by a linear factor.

Here we are given a function and asked to evaluate the value of the function at a given value, so we can just evaluate it:

We have:

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

And so, when $x = 2$ we have:

$f \left(2\right) = 8 + 4 - 10 - 6 = - 4$

We can also use the remainder theorem to establish a value of $f \left(a\right)$. as the remainder theorem tells us that is we divide $f \left(x\right)$ by a linear factor $\left(x - a\right)$ the remainder is $f \left(a\right)$.

So, let us divide the given polynomial, $f \left(x\right)$ by $x - 2$ using algebraic long division, yielding a remainder of $f \left(2\right)$

 {: ( , , , , x^2 , + , 3x , +, 1 , ), ( , ,"----", "----", "----", "----", "----", "----", "----" , ), (x-2, ")" , x^3, +,x^2, -,5x,-,6, ), ( , , x^3, -, 2x^2, , , , , - ), ( , , , ,"----", "----", "----", "----", "----", ), ( , , , , 3x^2, -, 5x, -, 6, ), ( , , , , 3x^2, -, 6x, , , - ), ( , , , , , ,"----", "----", "----" , ), ( , , , , , , x, -, 6, ), ( , , , , , , x, -, 2, -), ( , , , , , , , "----", "----" , ), ( , , , , , , , -, 4, ) :}

And we have remainder $- 4$ indicating $f \left(2\right) = - 4$ , as above.