# Factorize the following algebraic expressions:(i) $y^2+5y-36$(ii) $(a^2-5a)^2-36$(iii) $(a+7)(a-10)+16$

Given:

The given expressions are:

(i) $y^2+5y-36$

(ii) $(a^2-5a)^2-36$

(iii) $(a+7)(a-10)+16$

To do:

We have to factorize the given algebraic expressions.

Solution:

Factorizing algebraic expressions:

Factorizing an algebraic expression implies 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.

(i) The given expression is $y^2+5y-36$.

We can factorize the given expression by splitting the middle term. Splitting the middle term means we have to rewrite the middle term as the sum or difference of the two terms.

$y^2+5y-36$ can be written as,

$y^2+5y-36=y^2+9y-4y-36$              [Since $5y=9y-4y$ and $y^2 \times (-36)=9y \times (-4y) =-36y^2$]

$y^2+5y-36=y(y+9)-4(y+9)$

$y^2+5y-36=(y+9)(y-4)$

Hence, the given expression can be factorized as $(y-4)(y+9)$.

(ii) The given expression is $(a^2-5a)^2-36$.

We can factorize the given expression by splitting the middle term. Splitting the middle term means we have to rewrite the middle term as the sum or difference of the two terms.

$(a^2-5a)^2-36$ can be written as,

$(a^2-5a)^2-36=(a^2-5a)^2-(6)^2$

Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize the given expression as,

$(a^2-5a)^2-36=(a^2-5a)^2-(6)^2$

$(a^2-5a)^2-36=(a^2-5a+6)(a^2-5a-6)$

Now,

$a^2-5a+6$ can be factorized by splitting middle term as,

$a^2-5a+6=a^2-3a-2a+6$              [Since $-5a=-3a-2a$ and $a^2 \times 6=-3a \times (-2a) =6a^2$]

$a^2-5a+6=a(a-3)-2(a-3)$

$a^2-5a+6=(a-3)(a-2)$.................(I)

$a^2-5a-6$ can be factorized by splitting middle term as,

$a^2-5a-6=a^2+a-6a-6$              [Since $-5a=a-6a$ and $a^2 \times (-6)=a \times (-6a) =-6a^2$]

$a^2-5a-6=a(a+1)-6(a+1)$

$a^2-5a-6=(a+1)(a-6)$.................(II)

Using (I) and (II), we get,

$(a^2-5a)^2-36=(a-6)(a-3)(a-2)(a+1)$

Hence, the given expression can be factorized as $(a-6)(a-3)(a-2)(a+1)$.

(iii) The given expression is $(a+7)(a-10)+16$.

We can factorize the given expression by splitting the middle term. Splitting the middle term means we have to rewrite the middle term as the sum or difference of the two terms.

$(a+7)(a-10)+16$ can be written as,

$(a+7)(a-10)+16=a(a-10)+7(a-10)+16$

$(a+7)(a-10)+16=a^2-10a+7a-70+16$

$(a+7)(a-10)+16=a^2-3a-54$

$(a+7)(a-10)+16=a^2-9a+6a-54$              [Since $-3a=-9a+6a$ and $a^2 \times (-54)=-9a \times 6a =-54a^2$]

$(a+7)(a-10)+16=a(a-9)+6(a-9)$

$(a+7)(a-10)+16=(a-9)(a+6)$

Hence, the given expression can be factorized as $(a-9)(a+6)$.

Updated on: 11-Apr-2023

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