How do you factor #4x ^ { 4} + 64x ^ { 3} + 60x ^ { 2} = 0#?

1 Answer
Shwetank Mauria
Mar 11, 2017

#4x^2(x+15)(x+1)=0# and and #x=0,-15# or #-1#

Explanation:

Generally we factorize a polynomial, but what has been given is an equation. As equations are solved using factorization, I intent that way and expect that tis is desired by the questioner.

As there are three monomials in the polynomial on the left and their HCF is #4x^2#

As such #4x^4+64x^3+60x^2=0# can be written as

#4x^2(x^2+16x+15)=0#

or #4x^2(x^2+x+15x+15)=0#

or #4x^2[x(x+1)+15(x+1)]=0#

or #4x^2(x+15)(x+1)=0#

Hence either #x=0# or #x+15=0# or #x+1=0#

and #x=0,-15# or #-1#

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