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Question

Find the square root of [(x2+5x+6)(x2+7x+10)(x2+8x+15)]

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Solution

Given [(x2+5x+6)(x2+7x+10)(x2+8x+15)]
Lets factorize, each polynomial.
Consider x2+5x+6=x2+3x+2x+6 [Since, 5x=3x+2x]
=x(x+3)+2(x+3)
x2+5x+6=(x+3)(x+2)
Similar way, consider (x2+7x+10)=x2+5x+2x+10
[Since, 7x=5x+2x]
=x(x+5)+2(x+5)
(x2+7x+10)=(x+5)(x+2)
Now, consider (x2+8x+15)=x2+5x+3x+15
=x(x+5)+3(x+5)
(x2+8x+15)=(x+5)(x+3)
Thus, given expression can be written as (x2+5x+6)(x2+7x+10)(x2+8x+15)=(x+3)(x+2)(x+5)(x+2)(x+5)(x+3)
=(x+2)2(x+3)2(x+5)2
(x2+5x+6)(x2+7x+10)(x2+8x+15)=[(x+2)(x+3)(x+5)]2
Taking square root on both sides, we get
(x2+5x+6)(x2+7x+10)(x2+8x+15)=[(x+2)(x+3)(x+5)]2
(x2+5x+6)(x2+7x+10)(x2+8x+15)=(x+2)(x+3)(x+5)


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