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

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Answer 1 · 2016-09-01T00:44:03

$4 x^{3} - 4 x^{2} - 9 x + 9 = (x - 1) (2 x + 3) (2 x - 3)$ $4x^3-4x^2-9x+9=(x-1)(2x+3)(2x-3)$

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

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

$4 x^{3} - 4 x^{2} - 9 x + 9$ $4x^3-4x^2-9x+9$

= $4 x^{2} (x - 1) - 9 (x - 1)$ $4x^2(x-1)-9(x-1)$

= $(x - 1) (4 x^{2} - 9)$ $(x-1)(4x^2-9)$

= $(x - 1) ({(2 x)}^{2} - 3^{2})$ $(x-1)((2x)^2-3^2)$

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

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

How do you factor given that f(1)=0f(1)=0 and f(x)=4x3−4x2−9x+9f(x)=4x^3-4x^2-9x+9?