# Factorize the expression $(ax+by)^2+(bx-ay)^2$.

Given:

The given algebraic expression is $(ax+by)^2+(bx-ay)^2$.

To do:

We have to factorize the expression $(ax+by)^2+(bx-ay)^2$.

Solution:

Factorizing algebraic expressions:

Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution.

An algebraic expression is factored completely when it is written as a product of prime factors.

Here, we can factorize the expression $(ax+by)^2+(bx-ay)^2$ by grouping similar terms and taking out the common factors.

We can write $(ax+by)^2+(bx-ay)^2$ as,

$(ax+by)^2+(bx-ay)^2=(ax)^2+2(ax)(by)+(by)^2+(bx)^2-2(bx)(ay)+(ay)^2$                     [Since $(m+n)^2=m^2+2mn+n^2$ and $(m-n)^2=m^2-2mn+n^2$]

$(ax+by)^2+(bx-ay)^2=a^2x^2+2abxy+b^2y^2+b^2x^2-2abxy+a^2y^2$

$(ax+by)^2+(bx-ay)^2=a^2x^2+b^2y^2+b^2x^2+a^2y^2$

The terms in the given expression are $a^2x^2, b^2y^2, b^2x^2$ and $a^2y^2$.

We can group the given terms as $a^2x^2, b^2x^2$ and $b^2y^2, a^2y^2$

Therefore, by taking $x^2$ as common in $a^2x^2, b^2x^2$ and $y^2$ as common in $b^2y^2, a^2y^2$, we get,

$a^2x^2+b^2y^2+b^2x^2+a^2y^2=x^2(a^2+b^2)+y^2(a^2+b^2)$

Now, taking $(a^2+b^2)$ common, we get,

$a^2x^2+b^2y^2+b^2x^2+a^2y^2=(x^2+y^2)(a^2+b^2)$

Hence, the given expression can be factorized as $(x^2+y^2)(a^2+b^2)$.

Updated on: 06-Apr-2023

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