# How do you factor 2y^2 + 29y + 90?

Mar 3, 2018

$\left(x + 10\right) \left(2 x + 9\right)$

#### Explanation:

When factorising a quadratic that has a coefficient $> 1$, we always multiply the coefficient by the constant given at the end of the quadratic.

$\therefore$ $2 {y}^{2} + 29 y + 90 \to 2 {y}^{2} + 29 y + 180$

When factorising, we always want to find two numbers that add to make the second number $\left(29\right)$ and multiply to make $\left(180\right)$. The best way to do this is to find the factors of the constant at the end of the quadratic $\left(180\right)$ and see if they can add to make the middle number $\left(29\right)$

As this quadratic has a $+$ and another $+$ this means both numbers will be positive.

$180 :$

$180 \times 1$ cannot make to $29$...

$90 \times 2$ cannot make $29$...

$60 \times 3$ cannot make $29$...

$45 \times 4$ cannot make $29$...

$36 \times 5$ cannot make $29$...

$30 \times 6$ cannot make $29$...

$20 \times 9$ CAN make $29$... so therefore we use these two numbers.

Removing the $29$ and plugging the $20$ and $9$ into the quadratic:

$2 {x}^{2} + 20 x + 9 x + 90$

Always remember to change the constant back to the original...

Split up into two parts:

$2 {x}^{2} + 20 x$ and $9 x + 90$

$2 {x}^{2} + 20 x \to 2 x \left(x + 10\right)$ taking out a factor of $2 x$

$9 x + 90 \to 9 \left(x + 10\right)$ taking out a factor of $9$

This leaves us with:

$2 x \left(x + 10\right) + 9 \left(x + 10\right)$ <- Notice that the brackets are the same, this is one of our solutions. The brackets should ALWAYS be the same...

$2 x \cancel{\left(x + 10\right)} + 9 \cancel{\left(x + 10\right)}$

This leaves us with our other bracket of $\left(2 x + 9\right)$

$\therefore$ the final answer is $\left(x + 10\right) \left(2 x + 9\right)$

We can always check by expanding out the brackets:

$2 {x}^{2} + 9 x + 20 x + 90 \to 2 {x}^{2} + 29 x + 90$

$\therefore$ $\left(x + 10\right) \left(2 x + 9\right)$ is correct...