# 7th Class Mathematics Algebraic Expressions Factorization of the Polynomials

Factorization of the Polynomials

Category : 7th Class

### Factorization of the Polynomials

Let us recall that an algebraic expression that is expressed as the product of two or more expressions & each of these is a factor of the given algebraic expression. The process of writing a given algebraic expression as the product of two or more factors is called factorization.

Some of the common methods of factorization of the algebraic expressions are as follows:

•    Factorization by taking out a common factor.
•    Factorization by using identities
•    Factorization by regrouping the terms. Factorization of the Algebraic Expression $a{{x}^{2}}+bx+c$

Step 1:   Find the product of constant term and coefficient of ${{x}^{2}}$ i.e. ac.

Step 2:   Find factors of "ac".

Step 3:   Select the factors of 'ac' in such a way that addition or subtraction of factors must be equal to the coefficient of $x$ i. e "b".

Step 4:   If the product 'ac' is positive then both the factors are either positive or negative.

Step 5:   If the product 'ac' is negative then the two factors of 'ac' will have different sign. Find the factors of$~{{x}^{2}}+9x+18$

Solution:

Step 1: The product of constant term and coefficient of ${{x}^{2}}$ is 18

Step 2: The product is positive. Therefore, both the factors of 18 will be either be positive or negative.

Step 3: But the coefficient of $x$ is positive therefore, both the factors of 18 will be positive.

Step 4: Factors of 18 are $1\times 18,2\times 9$ and $3\times 6.$One pair whose sum is 9 should betaken.

Step 5: The required number are 3 & 6.

$\therefore {{x}^{2}}+9x+18={{x}^{2}}+(6+3)x+18$

$=(x+6)(x+3)$[By using identity $(x+a)(x+b)=$${{x}^{2}}+(a+b)x+ab]$

or ${{x}^{2}}+9x+18={{x}^{2}}+6x+3x\text{ }+18$ $=x(x+6)+3(x+6)=(x+6)(x+3).$

Taking out x from first two terms and 3 common from last two terms.  Factorize: ${{x}^{2}}-11x+30.$

(a)$(x-6)(x-5)$

(b)$(x-6)(x+5)$

(c)$~(7{{x}^{2}}-6)(8x-9)$

(d) $\left( 9{{x}^{2}}-62 \right)\left( 2x+3 \right)$

(e) None of these

Explanation

Let us find two numbers whose product is 30 & sum is $(-11)$ Since, the product is positive, therefore, either both the numbers will be positive or negative.

But the sum is negative hence, both the numbers will also be negative.

Required factors of 30 are $(-6)$ and $(-5).$

$\therefore {{x}^{2}}-11x+30={{x}^{2}}+\left\{ \text{(}-6\text{)}+(-5) \right\}\text{ }x+30$

$={{x}^{2}}-6x-5x+30=x(x-6)-5(x-6)=(x-6)(x-5)$ Factorize: ${{x}^{2}}+x-12$

(a)$(x+5)(x-3)$

(b) $(x-9)(x-52)$

(c)$(x+4)(x-3)$

(d) $(7x+{{5}^{2}})(6x-7)$

(e) None of these

Explanation

To find two numbers whose product is (- 12) & sum is 1.

Since the product is negative, one number will be positive & the other number will be negative.

Since the sum is positive, the numerically greater of two numbers will be positive.

So, the required factors are 4 & $(-3)$ ${{x}^{2}}+x-12={{x}^{2}}+\left\{ 4+(-3) \right\}x-12$

$={{x}^{2}}+4x-3x-12=x(x+4)-3(x+4)=(x+4)(x-3)$ If $x-\frac{1}{x}=7$ then the value of ${{x}^{4}}-\frac{1}{{{x}^{4}}}$ is:

(a) 2600

(b) 2599

(c) 2601

(d) 2401

(e) None of these Factorize: ${{(5a+4b)}^{2}}-{{\text{(}3a-2b\text{)}}^{2}}$

(a) $~4(a+3b)(4a+b)$

(b) $2(a+3b)(4a+b)$

(c) $3(a+3b)(4a+b)$

(d) $(a+3b)(4a+b)$

(e) None of these The factor of $4{{p}^{2}}+{{q}^{2}}+16-4pq+8q-16p$ is:

(a) $(-2p-q+4)(-2p+q+4)$

(b) $(2p+q+4)(2p+q+4)$

(c) $(-2p+q+4)(-2p+q+4)$

(d) $(-2p+q-4)(-2p+q-4)$

(e) None of these •   ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
•   ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
•   ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$
•   ${{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$or ${{\text{(}a+b\text{)}}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)$
•   ${{(a-b)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$ or ${{(a-b)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b)$
•   ${{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)$or $(a+b)({{a}^{2}}-ab+{{b}^{2}})$
•   ${{a}^{3}}-{{b}^{3}}={{(a-b)}^{3}}+3ab(a-b)$ or $(a-b)({{a}^{2}}+ab+{{b}^{2}})$
•    ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)$
•   If $(a+b+c)=0,$then ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$

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