# How do you factor the trinomial y^2-7y-30?

Jun 6, 2017

$\left(y + 3\right) \left(y - 10\right)$

#### Explanation:

${y}^{2} - 7 y - 30 = {y}^{2} - 10 y + 3 y - 30 = y \left(y - 10\right) + 3 \left(y - 10\right)$

$= \left(y - 10\right) \left(y + 3\right) = \left(y + 3\right) \left(y - 10\right)$[Ans]

Jun 6, 2017

$\left(y + 3\right) \left(y - 10\right)$

#### Explanation:

For a simpler trinomial like this quadratic, in the form ${y}^{2} + a y + b$, one can just look for a pair of numbers which multiply to get $b$ and sum to $a$. In this case, they must multiply to get $- 30$ and add to $- 7$, so the numbers are $3$ and $- 10$ because $3 \cdot - 10 = - 30$ and $3 - 10 = - 7$.
If you can't immediately see that then you can use another method like completing the square:
${y}^{2} - 7 y - 30 = 0$
${\left(y - \setminus \frac{7}{2}\right)}^{2} - {\left(\setminus \frac{7}{2}\right)}^{2} - 30 = 0$
${\left(y - \setminus \frac{7}{2}\right)}^{2} = \setminus \frac{169}{4}$
$y - \setminus \frac{7}{2} = \pm \sqrt{\setminus} \frac{169}{4} = \pm \setminus \frac{13}{2}$
$y = \setminus \frac{7 \pm 13}{2}$
$y = \setminus \frac{20}{2} \mathmr{and} \setminus \frac{- 6}{2} = 10 \mathmr{and} - 3$
So if our roots are 10 and -3, the factorisation of it is $\left(y - 10\right) \left(y - \left(- 3\right)\right) = \left(y - 10\right) \left(y + 3\right)$.
I hope I've explained that well enough.

Jun 6, 2017

$\left(y - 10\right) \left(y + 3\right)$

#### Explanation:

To factor this, you have to find 2 numbers such that $a + b = - 7$ and $a b = - 30$ The 2 numbers, $a$ and $b$ are $- 10$ and $3$.

Since ${y}^{2}$ can be expressed as $\left(y\right) \left(y\right)$, we get the following :

${y}^{2} - 7 y - 30 = \left(y - 10\right) \left(y + 3\right)$

Using the FOIL method, we can expand this to check

${y}^{2} + 3 y - 10 y - 30$