# 9th Class Mathematics Polynomials Factorization

Factorization

Category : 9th Class

### Factorization

The following are the different methods for factorization of a polynomials.

Taking Out Common Factor

If every term of a polynomial has a common factor then we divide each term by the common factor and multiply it by the remaining.

Factories: $7{{(a+b)}^{2}}-14(a+b)$

(a)$(a+b)]7(a+b)-14]$

(b) $7[{{(a+b)}^{2}}-2(a+b)]$ (c) $7(a+b)(a+b-2)$

(d) $7(a+b)(a+b-2a)$

(e) None of these

Explanation:

Here, 7 and $(a+b)$ is a common factor. Therefore, $7{{(a+b)}^{2}}-14(a+b)=7(a+b)(a+b-2)$

By Grouping

In this method, we group a given expression in such a way that we have a common factor. This method is effective for the expression in which it is not possible to take out common factor directly.

Factories:  ${{a}^{2}}{{y}^{2}}+(a{{y}^{2}}+1)y+a$

(a) $({{a}^{2}}y+1)(ay+1)$

(b) $(ay+y)({{a}^{3}}y+1)$

(c) $({{a}^{2}}y+1)(a+y)$

(d) $({{a}^{2}}{{y}^{2}}+a)(a+y)$

(e) None of these

Explanation:

${{a}^{2}}{{y}^{2}}+(a{{y}^{2}}+1)y+a={{a}^{2}}{{y}^{2}}+a{{y}^{3}}+y+a=$${{a}^{2}}{{y}^{2}}+a+a{{y}^{3}}+y$ $=a(a{{y}^{2}}+1)+y(a{{y}^{2}}+1)+(a+y)$

By Making a Trinomial of a Perfect Square

Adjust the trinomial in such a way that there must be a common factor.

The factor of ${{\left( 3x-\frac{1}{y} \right)}^{2}}-8\left( 3x-\frac{1}{y} \right)+16+\left( z+\frac{1}{y}-2x \right)\left( 3x-\frac{1}{y}-4 \right)$

(a)$\left( 3x-\frac{1}{y}-4 \right)(x+z-4)$

(b) $\left( 3x-\frac{1}{y}-4 \right)(3x+z-4)$

(c) $\left( 3x-\frac{1}{y}-4 \right)(2x+3z-4)$

(d) $\left( 3x-\frac{1}{y}-4 \right)(x+3z-4)$

(e) None of these

Explanation:

${{\left( 3x-\frac{1}{y} \right)}^{2}}-8\left( 3x-\frac{1}{y} \right)+16+\left( z+\frac{1}{y}-2x \right)\left( 3x-\frac{1}{y}-4 \right)$

$={{\left( 3x-\frac{1}{y} \right)}^{2}}-2\times 4\left( 3x-\frac{1}{y} \right)+{{4}^{2}}+\left( z+\frac{1}{y}-2x \right)\left( 3x-\frac{1}{y}-4 \right)$

$={{\left( 3x-\frac{1}{y}-4 \right)}^{2}}+\left( z+\frac{1}{y}-2x \right)+\left( 3x-\frac{1}{y}-4 \right)$

$=\left( 3x-\frac{1}{y}-4 \right)\left[ 3x-\frac{\bcancel{1}}{\bcancel{y}}-4+z+\frac{\bcancel{1}}{\bcancel{y}}-2x \right]$

$=\left( 3x-\frac{1}{y}-4 \right)(x-4+z)$

By Using Different Formulae

By the inspection of given expression use suitable formula (Identifies) and factories it.

The factor of ${{y}^{6}}+4{{y}^{2}}-1$ is

(a) $({{y}^{2}}+y-1)({{y}^{4}}-{{y}^{3}}+2{{y}^{2}}+y+1)$

(b) $({{y}^{2}}+y-1)({{y}^{4}}-{{y}^{3}}+y+1)$

(c) $(y+{{y}^{3}}-1)({{y}^{4}}-{{y}^{3}}+{{y}^{2}}+y+1)$

(d) $({{y}^{2}}+y-1)({{y}^{4}}-{{y}^{3}}+{{y}^{2}}+y-1)$

(e) None of these

Explanation:

${{y}^{6}}+4{{y}^{3}}-1={{y}^{6}}+{{y}^{3}}+3{{y}^{3}}-1)={{y}^{6}}+{{y}^{3}}+{{(-1)}^{3}}+3{{y}^{3}}$

$={{({{y}^{2}})}^{3}}+({{y}^{3}})+{{(-1)}^{3}}-3.({{y}^{2}}).y(-1)$

$=({{y}^{2}}+y-1)[{{({{y}^{2}})}^{2}}+{{y}^{2}}+{{(-1)}^{2}}-{{y}^{2}}.y-y(-1)-(-1)\times {{y}^{2}}$

[Using Formula:${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(a+b+c)$ $({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)]$

$=({{y}^{2}}+y-1)({{y}^{4}}+{{y}^{2}}-1-{{y}^{3}}+y+{{y}^{2}})$ $=({{y}^{2}}+y-1)({{y}^{4}}+{{y}^{3}}+2{{y}^{2}}+y-1)$

The general form of a quadratic polynomial is $a{{x}^{2}}+bx+c$, where $a\ne 0$, such expressions are factories as following:

Step 1: Take "a" common from whole expression, i.e.$a\left( {{x}^{2}}+\frac{b}{a}x+\frac{c}{a} \right)$

Step 2: Factories the expression $a\left( {{x}^{2}}+\frac{b}{a}x+\frac{c}{a} \right)$ by converting it as the different of two squares.

The factor of $-10{{a}^{2}}+31a-24$ is______.

(a) $\text{(2a}-\text{3})(\text{5a}-\text{8)}$

(b) $\text{(3}-\text{2a})(\text{8}-\text{5a)}$

(c) $\text{(2a}+\text{3})(\text{5a}-\text{8})$

(d) $\text{-(2a}-\text{3})(\text{5a}-\text{8})$

(e) None of these

Explanation:

The given expression is $-\text{1}0{{a}^{\text{2}}}+\text{31a}-\text{24}$

Step 1: Taking -10, common from whole expression then the expression will be$-10\left( {{a}^{2}}-\frac{31}{10}a+\frac{24}{10} \right)$

Step 2:  $-10\left( {{a}^{2}}-\frac{31}{10}a+\frac{24}{10} \right)$

$=-10\left[ {{a}^{2}}-2.\frac{31}{20}a+{{\left( \frac{31}{20} \right)}^{2}}-{{\left( \frac{31}{20} \right)}^{2}}+\frac{24}{10} \right]$

$=-10\left[ {{a}^{2}}-2.\left( \frac{31}{20} \right)a+{{\left( \frac{31}{20} \right)}^{2}}-\frac{961-960}{400} \right]=$$-10\left[ {{\left( a-\frac{31}{20} \right)}^{2}}-\frac{1}{400} \right]$

$=-10\left[ {{\left( a-\frac{31}{20} \right)}^{2}}-{{\left( \frac{1}{20} \right)}^{2}} \right]=$$-10\left[ \left( a-\frac{31}{20}+\frac{1}{20} \right)\left( a-\frac{31}{20}-\frac{1}{20} \right) \right]$

$=-10\left( a-\frac{3}{2} \right)\left( a-\frac{8}{5} \right)=-(2a-3)(5a-8)$

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