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Factorize the following

$(i)4x^{2}+20x+25$

$(ii)9y^{2}-66yz+121z^{2}$

$(iii)\left ( 2x+\frac{1}{3} \right )^{2}-\left ( x-\frac{1}{2} \right )^{2}$

Answers (1)

$(i)\left ( 2x+5 \right )\left ( 2x+5 \right )$

Solution

Given, $4x^{2}+20x+25$

We can write the given expression as:

$(2x)^{2}+(5)^{2}+2(2x)(5) \; \; \; \; \cdots (i)$

Now, we know that

$(a+b)^{2}=a^{2}+b^{2}+2ab$

Comparing equation (i) with the above identity, we have:

a = 2x, b = 5

So,

$\\(2x)^{2}+(5)^{2}+2(2x)(5) =(2x+5)^{2}\\ =(2x+5)(2x+5)$

$(ii)\left ( 3y-11z \right )\left ( 3y-11z \right )$

Solution

Given, $9y^{2}-66yz+121z^{2}$

We can write the given expression as:

$(3y)^{2}+(11z)^{2}-2(3y)(11z) \; \; \; \; \cdots (i)$

Now, we know that

$(a-b)^{2}=a^{2}+b^{2}-2ab$

Comparing equation (i) with the above identity, we have:

a = 3y, b = 11z

So,

$\\=(3y)^{2}+(11z)^{2}-2(3y)(11z)=(3y-11z)^{2}\\ =(3y-11z)(3y-11z)$

$(iii)\left ( x+\frac{5}{6} \right )\left ( 3x-\frac{1}{6} \right )$

Solution

Given, $\left ( 2x+\frac{1}{3} \right )^{2}-\left ( x-\frac{1}{2} \right )^{2}$

Now, we know that

$a^{2}-b^{2}=(a-b)(a+b)$

So,

$\\ \left ( 2x+\frac{1}{3} \right )^{2}-\left ( x-\frac{1}{2} \right )^{2}=\left ( 2x+\frac{1}{3}- \left ( x-\frac{1}{2} \right )\right ).\left ( 2x+\frac{1}{3}+ \left ( x-\frac{1}{2} \right )\right )\\ =\left ( 2x+\frac{1}{3}- x+\frac{1}{2} \right ).\left ( 3x+\frac{2-3}{6} \right )\\ =\left ( x+\frac{2+3}{6} \right )\left ( 3x-\frac{1}{6} \right )\\ =\left ( x+\frac{5}{6} \right )\left ( 3x-\frac{1}{6} \right )$

Hence the factorized form is: $\left ( x+\frac{5}{6} \right )\left ( 3x-\frac{1}{6} \right )$ .

Posted by

infoexpert24

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