# What are all the possible rational zeros for y=21x^2+11x-40 and how do you find all zeros?

Aug 19, 2016

The Zeroes are, $x = - \frac{5}{3} , \mathmr{and} , \frac{8}{7}$.

#### Explanation:

To find the zeroes of $y = 21 {x}^{2} + 11 x - 40$, we have to factorise it.

We observe that,

$21 \times 40 = \left(7 \cdot 3\right) \times \left(5 \cdot 8\right) = \left(7 \cdot 5\right) \times \left(3 \cdot 8\right) = 35 \times 24$

and, $35 - 24 = 11$.

$\therefore y = \underline{21 {x}^{2} + 35 x} - \underline{24 x - 40}$

$= 7 x \left(3 x + 5\right) - 8 \left(3 x + 5\right)$

$= \left(3 x + 5\right) \left(7 x - 8\right)$

Hence, the Zeroes are, $x = - \frac{5}{3} , \mathmr{and} , \frac{8}{7}$.