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# Resolve each of the following quadratic trinomials into factors:

(i) $11x^2-54x+63$

(ii) $7x-6x^2+20$

(iii) $3x^2+22x+35$

Given:

The given quadratic trinomials are:

(i) $11x^2-54x+63$

(ii) $7x-6x^2+20$

(iii) $3x^2+22x+35$

To do:

We have to factorize the given quadratic trinomials.

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 $11x^2-54x+63$.

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.

Here,

The coefficient of $x^2$ is $11$

The coefficient of $x$ is $-54$

The constant term is $63$

$11x^2-54x+63$ can be written as,

$11x^2-54x+63=11x^2-33x-21x+63$ [Since $-54x=-33x-21x$ and $11x^2 \times 63=-33x \times (-21x) $]

$11x^2-54x+63=11x(x-3)-21(x-3)$

$11x^2-54x+63=(11x-21)(x-3)$

**Hence, the given expression can be factorized as $(11x-21)(x-3)$.**

(ii) The given expression is $7x-6x^2+20$.

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.

Here,

The coefficient of $x^2$ is $-6$

The coefficient of $x$ is $7$

The constant term is $20$

$7x-6x^2+20$ can be written as,

$-6x^2+7x+20=-6x^2+15x-8x+20$ [Since $7x=15x-8x$ and $-6x^2 \times 20=15x \times (-8x) =-120x^2$]

$-6x^2+7x+20=-3x(2x-5)-4(2x-5)$

$-6x^2+7x+20=(-3x-4)(2x-5)$

**Hence, the given expression can be factorized as $(-3x-4)(2x-5)$.**

(iii) The given expression is $3x^2+22x+35$.

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.

Here,

The coefficient of $x^2$ is $3$

The coefficient of $x$ is $22$

The constant term is $35$

$3x^2+22x+35$ can be written as,

$3x^2+22x+35=3x^2+15x+7x+35$ [Since $22x=15x+7x$ and $3x^2 \times 35=15x \times 7x =105x^2$]

$3x^2+22x+35=3x(x+5)+7(x+5)$

$3x^2+22x+35=(3x+7)(x+5)$

**Hence, the given expression can be factorized as $(3x+7)(x+5)$.**