# How do you factor the expression 10x^2-3x-1 ?

Dec 13, 2015

$\left(5 x + 1\right) \left(2 x - 1\right)$

#### Explanation:

Look for a pair of numbers whose PRODUCT is the product of the first and last numbers $\left(- 10\right)$ and sum is the middle number $\left(- 3\right)$.

The two numbers that meet these criteria are $- 5$ and $2$.

Rewrite the $- 3 x$ term in the expression as $- 5 x + 2 x$.

$10 {x}^{2} - 5 x + 2 x - 1$

Factor by grouping.

$\left(10 {x}^{2} - 5 x\right) + \left(2 x - 1\right)$

$5 x \left(2 x - 1\right) + 1 \left(2 x - 1\right)$

$\left(5 x + 1\right) \left(2 x - 1\right)$

Dec 13, 2015

Split the middle term (-3x) into two separate terms . Then, use grouping to complete the factoring.

#### Explanation:

Split the middle term into two separate terms so that the sum is still $- 3$ however, you want the product of the two new terms to equal $\left(10\right) \left(- 1\right) = - 10$. For this problem use $2$ and $- 5$

Here is the new expression ...

$10 {x}^{2} + 2 x - 5 x - 1$

Use grouping ...

$\left(10 {x}^{2} + 2 x\right) - \left(5 x + 1\right)$

$2 x \left(5 x + 1\right) - \left(5 x + 1\right)$

$\left(2 x - 1\right) \left(5 x + 1\right)$

hope that helped