# How do you find all the zeros of f(x) = 3x^2 − 4x − 15?

Apr 7, 2018

Zeros of $f \left(x\right)$ are $x = 3 \mathmr{and} x = - \frac{5}{3}$

#### Explanation:

$f \left(x\right) = 3 {x}^{2} - 4 x - 15 = 3 {x}^{2} - 9 x + 5 x - 15$

$= 3 x \left(x - 3\right) + 5 \left(x - 3\right) = \left(x - 3\right) \left(3 x + 5\right)$

Zeros of $f \left(x\right)$ are $x = 3 \mathmr{and} x = - \frac{5}{3}$ [Ans]

$x = 3 , \frac{- 5}{3}$

#### Explanation:

Using trial and error method $\left(x - 3\right)$ is a factor of $f \left(x\right)$
I.e$\left(x - 3\right) , x = 3$
$f \left(3\right) = 3 \left({3}^{2}\right) - 4 \left(3\right) - 15$
$= 3 \left(9\right) - 12 - 15$
$= 27 - 12 - 15$
$= 0$
Using long division of polynomial's
$\left(+\right) 3 x + 5$
$\sqrt[x - 3]{3 {x}^{2} - 4 x - 15}$
$\frac{\left(-\right) 3 {x}^{2} - 9 x}{5 x - 15}$
((-)5x-15)/(……)
Since $x$ is in the 2nd degree, the 2 factors of $f \left(x\right)$are $\left(x - 3\right)$ and $3 x + 5$
Equating factor to zero.
$x - 3 = 0 , 3 x + 5 = 0$
$x = 3 , 3 x = - 5$
$x = 3 , x = \frac{- 5}{3}$($\textcolor{b l u e}{z e r o ' s . O f . f \left(x\right)}$)