How do you factor #x^3 + 3x^2 - 49x - 147=0# by grouping?

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Answer 1 · 2018-04-04T18:50:19

#(x+7)(x-7)(x+3)#

This polynomial has four terms. We should then make two groups of two, with each group having two terms that obviously have some common factor.

So, we group as follows:

#(x^3+3x^2)-(49x+147)=0#

We notice that out of #x^3+3x^2,# we can factor out #x^2, # yielding

#x^2(x+3)#

We notice that out of #49x+147,# we can factor out #49,# yielding

#49(x+3)#

We then have

#x^2(x+3)-49(x+3)=0#

This becomes

#(x^2-49)(x+3)=0#

We can factor further. #a^2-b^2=(a+b)(a-b),# so #x^2-49=(x+7)(x-7)#

The factored form is

#(x+7)(x-7)(x+3)#