# How do you factor 2t^2+7t+3?

Feb 2, 2015

Make $2 {t}^{2} + 7 t + 3$ equal to zero to form a quadratic equation.

Here's the general form of the quadratic equation

$a {x}^{2} + b x + c = 0$

You need to find two numbers that multiply to give $a \cdot c$ and that add to give $b$. For this example, the two numbers you need must multply to give $2 \cdot 3 = 6$ and add to give $\text{7}$.

To find them, just list all of the factors of $6$ and try to find a pair that satisfies the criteria given

$6 = 1 \cdot 2 \cdot 3$

In simple cases such as this one, these two numbers can be easily found to be $\text{6}$ and $\text{1}$, since

$6 + 1 = 7$ and $6 \cdot 1 = 6$

$2 {t}^{2} + 6 t + t + 3$
$\left(2 {t}^{2} + 6 t\right) + \left(t + 3\right) \implies 2 t \cdot \left(t + 3\right) + \left(t + 3\right)$, which can be written as
$\left(t + 3\right) \cdot \left(2 t + 1\right)$ $\to$ this is how the original equation factors.