How do you factor given that #f(1)=0# and #f(x)=4x^3-4x^2-9x+9#?

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Answer 1 · 2016-09-01T00:44:03

#4x^3-4x^2-9x+9=(x-1)(2x+3)(2x-3)#

As in #f(x)=4x^3-4x^2-9x+9#, #f(1)=0#, one of the factors is #(x-1)#.

Now dividing #4x^3-4x^2-9x+9# by #(x-1)#, as

#4x^3-4x^2-9x+9#

= #4x^2(x-1)-9(x-1)#

= #(x-1)(4x^2-9)#

= #(x-1)((2x)^2-3^2)#

Now using identity #a^2-b^2=(a+b)(a-b)#, the above can be factorized as

#(x-1)(2x+3)(2x-3)#