# How do you factor x^2 + 4xy - 5y^2?

Aug 19, 2016

$\left(x - 1 y\right) \left(x + 5 y\right)$

Aug 22, 2016

$\left(x + 5 y\right) \left(x - y\right)$

#### Explanation:

All the information you need is in the expression.

Find the factors of 5 which subtract to give 4.
The signs will be different (because of the minus), there will be more positives (because of +)

5 is a prime number - the only factors are 1 x 5 and we see 5 -1 = 4.

We need +5 and -1 to give +4

This leads to the two brackets:

(x" "y)(x" "y)" fill in the variables"

(x" "5y)(x" "1y)" fill in the factors"

$\left(x + 5 y\right) \left(x - y\right) \text{ fill in the signs}$

Aug 22, 2016

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

#### Explanation:

${x}^{2} + 4 x y - 5 {y}^{2}$ is a homogeneous expression. We propose that can be formed by the product of two homogeneous expresions.
${x}^{2} + 4 x y - 5 {y}^{2} = \left(a x + b y\right) \left(c x + d y\right)$.

So

${x}^{2} + 4 x y - 5 {y}^{2} = a c {x}^{2} + \left(b c + a d\right) x y + b d {y}^{2}$

so we have

{ (1 = ac),(4=bc+ad),(-5=bd) :}

We have three equations and four incognitas. Solving for $b , c , d$ we obtain

$b = - a , c = \frac{1}{a} , d = \frac{5}{a}$

applying a feasible value for $a$ as $a = 1$ we obtain

$b = - 1 , c = 1 , d = 5$ so

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