Factorize $(x+1) (x+2) (x+3) (x+6) - 3x^2$.


Given:  $(x+1) (x+2) (x+3) (x+6) - 3x^2$.

To do: To factorize the given polynomial.

Solution:

$=( x+1)( x+2)( x+3)( x+6)-3x^2$

$=( x+1)( x+6)( x+3)( x+2)-3x^2$

$=( x^2+6x+1x+6)( x^2+3x+2x+6)-3x^2$

$=( x^2+6x+6+1x)( x^2+5x+6+x-x)-3x^2$

$=( x^2+6x+6+1x)( x^2+5x+x+6-x)-3x^2$

$=( x^2+6x+6+1x)( x^2+6x+6-x)-3x^2$

$=( x^2+6x+6)^2( +1x)(-x)-3x^2$

$=( x^2+6x+6)^2-x^2-3x^2$

$=( x^2+6x+6)^2-4x^2$

$=( x^2+6x+6)^2-( 2x)^2$

$=( x^2+6x+6+2x)( x^2+6x+6-2x)$       [On applying $a^2-b^2=( a+b)( a-b)$]

$=( x^2+8x+6)( x^2+6x-2x+6)$

$=( x^2+8x+6)( x^2+4x+6)$

$=x(x+8+\frac{6}{x})x(x+4+\frac{6}{x})$

$=x^2( x+8+\frac{6}{x})( x+4+\frac{6}{x})$

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Updated on: 10-Oct-2022

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