# How do you factor 6t^2+t-1?

May 3, 2018

$\left(3 t - 1\right) \left(2 t + 1\right)$

#### Explanation:

Use the rainbow method.

Multiply the outer coefficients. $6 \cdot - 1 = - 6$

Now find two values which when multiplied equal the new number, and when added equal the center coefficient, 1.

Our values are 3 and -2.

Change the $+ t$ to $+ 3 t$ and $- 2 t$

$6 {t}^{2} + 3 t - 2 t - 1$

Separate the expression into two sides.
$\left(6 {t}^{2} + 3 t\right) \mathmr{and} \left(- 2 t - 1\right)$

Factor the two expressions, and make sure the values inside the parentheses are equal.

$3 t \left(2 t + 1\right) \mathmr{and} - 1 \left(2 t + 1\right)$

We now have $3 t - 1$ and $2 t + 1$ as our factors.
$\left(3 t - 1\right) \left(2 t + 1\right)$