How do you find all zeros of the function $f (x) = 3 x^{3} + x^{2} - 48 x - 16$ $f(x)=3x^3+x^2-48x-16$ ?

Question

2135 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-01T03:11:25

Factor $f (x)$ $f(x)$ to find that $f (x) = 0$ $f(x) = 0$ if and only if $x \in {- 4, - \frac{1}{3}, 4}$ $x in {-4, -1/3, 4}$

$f (x) = 3 x^{3} + x^{2} - 48 x - 16$ $f(x) = 3x^3+x^2-48x-16$

$= 3 x (x^{2} - 16) + x^{2} - 16$ $= 3x(x^2-16)+x^2-16$

$= (3 x + 1) (x^{2} - 16)$ $=(3x+1)(x^2-16)$

$= (3 x + 1) (x + 4) (x - 4)$ $=(3x+1)(x+4)(x-4)$

Thus $f (x) = 0$ $f(x) = 0$ if and only if $3 x + 1 = 0$ $3x+1 = 0$ or $x + 4 = 0$ $x+4 = 0$ or $x - 4 = 0$ $x-4 = 0$ .

Solving, this gives us

$3 x + 1 = 0 \Rightarrow x = - \frac{1}{3}$ $3x+1 = 0 => x = -1/3$

$x + 4 = 0 \Rightarrow x = - 4$ $x+4 = 0 => x = -4$

$x - 4 = 0 \Rightarrow x = 4$ $x-4 = 0 => x = 4$

Thus the solution set to $f (x) = 0$ $f(x) = 0$ is $x \in {- 4, - \frac{1}{3}, 4}$ $x in {-4, -1/3, 4}$

How do you find all zeros of the function f(x)=3x3+x2−48x−16f(x)=3x^3+x^2-48x-16?