How do you factor quadratic equations box method?

Mar 29, 2015

Here's an example:

Example: Factor $6 {x}^{2} + 7 x - 20$

You have $a {x}^{2} + b x + c$

Multiply $a c$ which, in this problem is $\left(6\right) \left(- 20\right) = - 120$

Fins two numbers whose product is $a c = - 120$
and whose sum is $b = + 7$

Because we want the product to be negative, one number is positive and the other is negative. Because we want the sum to be positive, the positive factor is the one with the greater absolute value.

We start the list:
$\left(- 1\right) \setminus \left(120\right)$ the sum is not $+ 7$
$\left(- 2\right) \left(60\right)$ the sum is not $+ 7$
$\left(- 3\right) \left(40\right)$ the sum is not $+ 7$
$\left(- 4\right) \left(30\right)$ the sum is not $+ 7$
$\left(- 5\right) \left(24\right)$ the sum is not $+ 7$
$\left(- 6\right) \left(20\right)$ the sum is not $+ 7$
$7$ is not a factor
$\left(- 8\right) \left(15\right)$ STOP the sum is $7$

Write $6 {x}^{2} + 7 x - 20$ replacing $+ 7 x$ with $- 8 x + 15 x$
(Or by $+ 15 x - 8 x$ either will work.)

$6 {x}^{2} + 7 x - 20 = 6 {x}^{2} - 8 x + 15 x - 20$ Now factor by grouping:

$\left(6 {x}^{2} - 8 x\right) + \left(15 x - 20\right) = 2 x \left(3 x - 4\right) + 5 \left(3 x - 4\right) = \left(2 x + 5\right) \left(3 x - 4\right)$

The "factoring box" is a way of doing the factoring by grouping.

To use this box, you must first take out any factor that is common to all 3 terms. (If we had started with $24 {x}^{2} + 28 x - 80$ we would have to first factor out the $4$ to get $4 \left(6 {x}^{2} + 7 x - 20\right)$

Put the first and last terms in main diagonal (upper left and lower right) Then put in the two terms we found above $- 8 x$ and $+ 15 x$ in the other 2 places.

$\left(\begin{matrix}6 {x}^{2} & + 15 x \\ - 8 x & - 20\end{matrix}\right)$

Notice that each row has common factors and each column has common factors (in some problems the only common factor is $1$)

$\left(\left(6 {x}^{2} , + 15 x\right)\right)$ has common factor $3 x$

and $\left(\begin{matrix}6 {x}^{2} \\ - 8 x\end{matrix}\right)$ has common factor $2 x$

We'll write those factors on a new first row and a new left column:

$\left(\begin{matrix}\null & 2 x & + 5 \\ 3 x & 6 {x}^{2} & + 15 x \\ - 4 & - 8 x & - 20\end{matrix}\right)$

(The $-$ is used where both are $-$.)
The factors are the top row and the left column:

$\left(2 x + 5\right) \left(3 x - 4\right)$

Using a different box

If we had put the $- 8 x$ and $15 x$ in the other 2 places.

$\left(\begin{matrix}6 {x}^{2} & - 8 x \\ + 15 x & - 20\end{matrix}\right)$

$\left(\begin{matrix}\null & 3 x & - 4 \\ 2 x & 6 {x}^{2} & - 8 x \\ + 5 & + 15 x & - 20\end{matrix}\right)$

The factors are $\left(3 x - 4\right) \left(2 x + 5\right)$