# What are all the possible rational zeros for y=12x^2-28x+15 and how do you find all zeros?

Sep 2, 2016

Zeros of $y = 12 {x}^{2} - 28 x + 15$ are $x = \frac{5}{6}$ and $x = \frac{3}{2}$.

#### Explanation:

Zeros of a function $f \left(x\right)$ are those values of $x$, for which $f \left(x\right) = 0$.

As $y = 12 {x}^{2} - 28 x + 15$ can be factorized, let us do so by splitting the middle term $- 28$ in two parts so that their product is 12×15=180. It is apparent that these are $- 18$ and $- 10$.

Hence $y = 12 {x}^{2} - 28 x + 15$

= $12 {x}^{2} - 18 x - 10 x + 15$

= $6 x \left(3 x - 3\right) - 5 \left(2 x - 3\right)$

= $\left(6 x - 5\right) \left(2 x - 3\right)$

It is apparent that if either $6 x - 5 = 0$ or $2 x - 3 = 0$, $y = 12 {x}^{2} - 28 x + 15$ will be zero.

Hence, zeros of $y = 12 {x}^{2} - 28 x + 15$ are $x = \frac{5}{6}$ and $x = \frac{3}{2}$.