Question

find all zeros of the function $f(x)=6x^3 - 19x^2 - 65x + 50$. enter the zeros separated by commas.
question help: video written example

Explanation:

Step1: Test rational root candidates

Using Rational Root Theorem, possible roots are $\pm1, \pm2, \pm5, \pm10, \pm25, \pm50, \pm\frac{1}{2}, \pm\frac{5}{2}, \pm\frac{25}{2}, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{5}{3}, \pm\frac{10}{3}, \pm\frac{25}{3}, \pm\frac{50}{3}, \pm\frac{1}{6}, \pm\frac{5}{6}, \pm\frac{25}{6}$.
Test $x=5$: $f(5)=6(5)^3 -19(5)^2 -65(5)+50=750-475-325+50=0$. So $x=5$ is a root.

Step2: Factor out $(x-5)$

Use polynomial division or synthetic division to divide $6x^3-19x^2-65x+50$ by $(x-5)$.
Synthetic division:

$$\begin{array}{r|rrrr} 5 & 6 & -19 & -65 & 50 \\ & & 30 & 55 & -50 \\ \hline & 6 & 11 & -10 & 0 \end{array}$$

Result: $6x^2+11x-10$

Step3: Factor quadratic polynomial

Factor $6x^2+11x-10$:
Find two numbers: $15$ and $-4$ (since $15\times(-4)=-60$, $15-4=11$)
Rewrite: $6x^2+15x-4x-10=3x(2x+5)-2(2x+5)=(3x-2)(2x+5)$

Step4: Solve for remaining roots

Set each factor to 0:
$3x-2=0 \implies x=\frac{2}{3}$
$2x+5=0 \implies x=-\frac{5}{2}$

Answer:

$-\frac{5}{2}, \frac{2}{3}, 5$