# How do you factor completely 14a^2 - 68a + 48?

$14 {a}^{2} - 68 a + 48 = 2 \left(7 a - 6\right) \left(a - 4\right)$

#### Explanation:

We have:

$14 {a}^{2} - 68 a + 48$

We can first factor out the Largest Common Factor of 14, 68, and 48:

• $14 = 2 \times 7$
• $68 = 2 \times 2 \times 17$
• $48 = 2 \times 2 \times 2 \times 2 \times 3$

The Largest Common Factor is 2:

$2 \left(7 {a}^{2} - 34 a + 24\right)$

Now we look for factors in the form of $\left(a x + b\right) \left(c x + d\right)$ where

• $a c = 7$
• $b d = 24$
• $a d + b c = - 34$

We know the factors for $7 = 7 \times 1$, so let's set $a = 7 , c = 1$

The factors for $24 = - 1 \times - 24 , - 2 \times - 12 , - 3 \times - 8 , - 4 \times - 6$ (because $a d + b c$ is a negative number and we've set $a c$ to be positive, we need these factors to be negative).

Let's try some factors and trial and error our way into this:

$a \textcolor{w h i t e}{00000} b \textcolor{w h i t e}{0000} c \textcolor{w h i t e}{00000} \mathrm{dc} o l \mathmr{and} \left(w h i t e\right) \left(000000\right) a \mathrm{dc} o l \mathmr{and} \left(w h i t e\right) \left(00000\right) b c \textcolor{w h i t e}{000} a d + b c$

$7 \textcolor{w h i t e}{000} - 8 \textcolor{w h i t e}{000} 1 \textcolor{w h i t e}{000} - 3 \textcolor{w h i t e}{000} - 21 \textcolor{w h i t e}{000} - 8 \textcolor{w h i t e}{000} - 29$

$7 \textcolor{w h i t e}{000} - 6 \textcolor{w h i t e}{000} 1 \textcolor{w h i t e}{000} - 4 \textcolor{w h i t e}{000} - 28 \textcolor{w h i t e}{000} - 6 \textcolor{w h i t e}{000} - 34$

This gives us:

$2 \left(7 a - 6\right) \left(a - 4\right)$