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

3 Answers
Aug 19, 2016

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

Aug 22, 2016

Answer:

#(x + 5y)(x - y)#

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

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

Aug 22, 2016

Answer:

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

Explanation:

#x^2+4xy-5y^2# 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)#.

So

#x^2+4xy-5y^2 = a c x^2 + (bc+ad)xy +bd 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 = 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)#