# How do you factor y= x^2-x-20 ?

Mar 14, 2016

y = (x + 4)( x- 5)

#### Explanation:

Find 2 numbers knowing sum (-1) and product (-20). It is the factor pairs (4, - 5). The 2 numbers are 4 and -5
y = (x + 4)(x - 5)

Mar 14, 2016

Find its solutions and write it as:

$a {x}^{2} + b x + c = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

${x}^{2} - x - 20 = \left(x - 5\right) \left(x + 4\right)$

#### Explanation:

If you find all it's solutions $\left({x}_{1} , {x}_{2} , {x}_{3.} . .\right)$ you can write it as a product of its solutions. If the polynomial has two solutions $\left({x}_{1} , {x}_{2}\right)$ you can solve it like this:

$a {x}^{2} + b x + c = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

For $y = {x}^{2} - x - 20$ the two solutions:

$y = 0$

${x}^{2} - x - 20 = 0$

$a = 1$
$b = - 1$
$c = - 20$

Δ=(-1)^2-4*1*(-20)=81

x_(1,2)=(-b+-sqrt(Δ))/(2a)=(-(-1)+-sqrt(81))/(2*1)=(1+-9)/2

${x}_{1} = 5$

${x}_{2} = - 4$

Therefore the equation can be written:

${x}^{2} - x - 20 = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) = \left(x - 5\right) \left(x + 4\right)$