How do you factor x^2 + 4xy - 5y^2x2+4xy5y2?

3 Answers
Aug 19, 2016

(x-1y)(x+5y)(x1y)(x+5y)

Aug 22, 2016

(x + 5y)(x - y)(x+5y)(xy)

Explanation:

All the information you need is in the expression.

Read from right to left.

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 y)(x y) fill in the variables

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

(x + 5y)(x - y)" fill in the signs"(x+5y)(xy) fill in the signs

Aug 22, 2016

x^2+4xy-5y^2 = (x-y)(x+5y)x2+4xy5y2=(xy)(x+5y)

Explanation:

x^2+4xy-5y^2x2+4xy5y2 is a homogeneous expression. We propose that can be formed by the product of two homogeneous expresions.
x^2+4xy-5y^2 = (a x + b y)(c x + d y)x2+4xy5y2=(ax+by)(cx+dy).

So

x^2+4xy-5y^2 = a c x^2 + (bc+ad)xy +bd y^2x2+4xy5y2=acx2+(bc+ad)xy+bdy2

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 = 1/a,d = 5/a

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

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

x^2+4xy-5y^2 = (x-y)(x+5y)