Question

Please help with this question on factorisation. Please show steps because I want to understand.

Answer 1

Answer:

$(x - 2y)((x - 2y) - 4) + 4$

Step-by-step explanation:

Given

$x^2 - 4xy - 4x + 4y^2 + 8y + 4$

Required

Factorize

$x^2 - 4xy - 4x + 4y^2 + 8y + 4$

Rearrange

$x^2 - 4xy + 4y^2- 4x + 8y + 4$

Group into three

$(x^2 - 4xy + 4y^2)- (4x - 8y) + (4)$

Factorize the first bracket

$(x^2 - 2xy- 2xy + 4y^2) - (4x - 8y) + (4)$

$x(x - 2y) - 2y(x - 2y)- (4x - 8y) + (4)$

$(x - 2y)(x - 2y) - (4x - 8y) + (4)$

Factorize the second bracket

$(x - 2y)(x - 2y) - 4(x - 2y) + 4$

x - 2y is a common factor of $(x - 2y)(x - 2y) - 4(x - 2y)$

So, the expression becomes

$(x - 2y)((x - 2y) - 4) + 4$

The expression cannot be further simplified;

Hence, $x^2 - 4xy - 4x + 4y^2 + 8y + 4$ is equivalent to $(x - 2y)((x - 2y) - 4) + 4$