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

Question

3559 views around the world

You can reuse this answer
Creative Commons License

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

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

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