# How do you factor  7y^2+19y+10 ?

Aug 8, 2015

$7 {y}^{2} + 19 y + 10 = \left(7 y + 5\right) \left(y + 2\right)$

#### Explanation:

If $7 {y}^{2} + 19 y + 10$ can be factored into two binomials:
$\textcolor{w h i t e}{\text{XXXX}}$$\left(a y + b\right) \left(c y + d\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= a c {y}^{2} + \left(a d + b c\right) y + b d$
then we need to find factors
$\textcolor{w h i t e}{\text{XXXX}}$$a , c$ of $7$
and
$\textcolor{w h i t e}{\text{XXXX}}$$b , d$ of $10$
such that
$\textcolor{w h i t e}{\text{XXXX}}$$a d + b c = 19$

The only integer factors of $7$ are
$\textcolor{w h i t e}{\text{XXXX}}$$\left(a , c\right) = \left(7 , 1\right)$
The (non-negative) integer factors of $10$ are
$\textcolor{w h i t e}{\text{XXXX}}$$\left(b , d\right) \in \left\{\begin{matrix}1 & 10 \\ 10 & 1 \\ 2 & 5 \\ 5 & 2\end{matrix}\right\}$

This gives us 4 possible combinations to check and we find:
$\textcolor{w h i t e}{\text{XXXX}}$$a d + b c = \left(7 \times 2\right) + \left(5 \times 1\right)$ gives the required $19$

So $\left(a y + b\right) \left(c y + d\right)$ becomes $\left(7 y + 5\right) \left(1 y + 2\right)$

Aug 8, 2015

Factor: $y = 7 {y}^{2} + 19 y + 10$

Ans: (7x + 5)(x + 2)

#### Explanation:

I use the new AC Method.
Converted $y ' = {y}^{2} + 19 y + 70 =$(y - p')(y - q').
p' and q' have same sign (Rule of signs)
Factor pairs of 70 --> (2, 35)(5, 14). This sum is 19 = b.
Then, p' = 5 and q' = 14.
Therefor, $p = \frac{p '}{a} = \frac{5}{7} ,$ and $q = \frac{14}{7} = 2$

Factored form: y = 7(y + 5/7)(y + 2) = (7y + 5)(y + 2).