# How do you factor given that f(8)=0 and f(x)=x^3-11x^2+14x+80?

Nov 1, 2016

$f \left(x\right) = \left(x - 8\right) \left(x - 5\right) \left(x + 2\right)$

#### Explanation:

Since $f \left(8\right) = 0$, we know that one of the zeroes is $8$

Divide the function by $x - 8$, then we will have a quadratic function which we can easliy factor.

How to perform polynomial division when the divisor is a binomial:
1. Treat each term of the dividend as a "digit". We will performing division by "digit" (similar to number division)
2. Perform division with the first "digit" of the binomial only
3. When multiplying the resulting quotient back with the divisor, include the remaining digit
4. Subtract the resulting product from the dividend
5. Repeat 1-4 with the resulting difference until division is no longer possible

I can't format the polynomial division properly so I will apologize in advance.

<pre>
x^2 - 3x - 10
-------------------------------
x - 8 / x^3 - 11x^2 + 14x + 80
- (x^3 - 8x^2)
--------------
-3x^2 + 14x + 80
-(-3x^2 + 24x)
------------------
-10x + 80
-10x + 80
-------------
0
</pre>

$f \left(x\right) = {x}^{3} - 11 {x}^{2} + 14 x + 80$

$\implies f \left(x\right) = \left(x - 8\right) \left({x}^{2} - 3 x - 10\right)$

List out all the factors of the last term
$10 = 1 \cdot 10$
$10 = 5 \cdot 2$

We need the product to be negative, the factors need to be of different sign.

$- 10 = - 1 \cdot 10$
$- 10 = 1 \cdot - 10$
$- 10 = - 5 \cdot 2$
$- 10 = 5 \cdot - 2$

Take a look at the sign of the middle term $- 3 x$. Since the sign is negative, we need the larger factor to be negative.

$- 10 = 1 \cdot - 10$
$- 10 = - 5 \cdot 2$

Add the factors, the resulting value should be equal to the middle term

$1 + - 10 = - 9$
$- 5 + 2 = - 3$

$\implies {x}^{2} - 3 x - 10 = \left(x - 5\right) \left(x + 2\right)$

$\implies f \left(x\right) = \left(x - 8\right) \left(x - 5\right) \left(x + 2\right)$