Question

Factorisation of \[32{(x + y)^2} - 2x - 2y \] results in
A. \[(x + y)(x + y + 3) \]
B. \[2(x + y)(16x + 16y - 1) \]
C. \[(x + y)(16x + 16y - 1) \]
D. \[2(x + y)(x + y - 3) \] .

Answer 1

Hint: Factorization is a method of writing numbers as their factors or divisors. When the factors of a number are multiplied together, they give the original equation. In this question we can solve by taking common 2 in \[2x \] and \[2y \] , or expanding the term \[{(x + y)^2} \] . We simplify the above equation until we get the answer that is given in the options. How to do this is explained below.

Complete step-by-step answer:
Given, \[32{(x + y)^2} - 2x - 2y \]
Taking common 2 we get,
 \[ = 32{(x + y)^2} - 2(x + y) \]
As we can see \[(x + y) \] present in both the terms,
Taking common \[(x + y) \] we get,
 \[ = (x + y)[32(x + y) - 2] \]
Taking 2 as common, we get
 \[ = 2(x + y)[16(x + y) - 1] \]
Removing brackets we get
 \[ = 2(x + y)[16x + 16y - 1] \]
We stop here, because we can see that \[2(x + y)[16x + 16y - 1] \] is option (b).
If we do not get the answer we need to solve it into a simplified form, so that it matches the given option. If we multiply and solve the given answer we get the original equation (original equation must be in simplified form).
So, the correct answer is “Option B”.

Note: In factorization we solve quadratic equations. The method of solving quadratic equations is to write the equation in the correct form. To be in the correct form, you must remove all parentheses from each side of the equation by distributing, combining all like terms and finally set the equation equal to zero. Use factoring strategies to factor the problem. Use the zero product property and set each factor containing a variable equal to zero. Solve each factor that was set equal to zero by getting the x on one side and the answer on the other side.