How do you find all the zeros of $f(x)=4x^3-12x^2-x+3$ ?

Question

5257 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-09T19:29:06

$f(x)$ has zeros: $1/2$ , $-1/2$ , $3$

Notice that the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

$f(x) = 4x^3-12x^2-x+3$

$=(4x^3-12x^2)-(x-3)$

$=4x^2(x-3)-1(x-3)$

$=(4x^2-1)(x-3)$

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

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

Hence the zeros of $f(x)$ are: $1/2$ , $-1/2$ , $3$

How do you find all the zeros of f(x)=4x^3-12x^2-x+3f(x)=4x^3-12x^2-x+3?