# How do you factor 2d^2-7d-15?

Nov 17, 2015

$\left(2 d + 3\right) \left(d - 5\right)$.

#### Explanation:

This may be factorized as a trinomial to obtain the factors

$\left(2 d + 3\right) \left(d - 5\right)$.

To obtain this you look at the coefficient of the first term, 2, and see what its factors are. There is only 1 option, $1 \times 2$.

Then look at the coefficient of the last term, 15, and there are a 2 options here namely :
$1 \times 15 \mathmr{and} 3 \times 5$.

You now need to combine the options such that the sum of the factors when cross-multiplied give you the middle term of $- 7$.

The only option is $\left(\begin{matrix}2 & 3 \\ 1 & - 5\end{matrix}\right)$ since $1 \times 3 - 2 \times - 5 = - 7$

and so therefore the factors are the rows of the matrix shown.

Nov 17, 2015

(2d+3)(d−5)

#### Explanation:

Factor $2 {d}^{2}$$7 d$$15$ ?

To factor a quadratic function with a leading coefficient greater than 1, do the following steps:

• Multiply the leading coefficient and the constant:

$2 \cdot$-$15 =$-$30$

• What two numbers would multiply to give you -$30$ and add to give you -$7$ ?

$3 \cdot$-$10 =$-$30$
$3 +$-$10 =$-$7$

So, +$3$ & -$7$.

• Since we multiplied the constant by 2, divide the numbers you got above by 2:

+3/2 & -10/2

$- \frac{10}{2} =$-$5$ so, one of the factors is $\left(d - 5\right)$
but $+ \frac{3}{2}$ leaves you with an improper fraction, right? (Yes.)
So, instead of $\left(d + 3\right)$ we're going to have $\left(2 d + 3\right)$.

• Final factors:

$\left(2 d + 3\right) \left(d - 5\right)$