How do you find other zeros given one zero $f (x) = x^{3} - 8 x^{2} + 4 x + 48; 4$ $f(x)=x^3-8x^2+4x+48; 4$ ?

Question

27659 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-27T03:20:32

$- 2 and 6$ $-2 and 6$ .

$(x - 4)$ $(x-4)$ is a factor of f.

Upon division by (x - 4), the other factor of f is obtained as

$x^{2} - 4 x - 12 = (x - 6) (x + 2)$ $x^2-4x-12=(x-6)(x+2)$ .

So, the other roots are given by

$x - 6 = x + 2 = 0$ $x-6=x+2=0$

How do you find other zeros given one zero f(x)=x3−8x2+4x+48;4f(x)=x^3-8x^2+4x+48; 4?