Raelynn Velasquez

2023-03-06

How to factor quadratic equations box method?

Cooper Barker

For example: Factor $6{x}^{2}+7x-20$
You have $a{x}^{2}+bx+c$
Multiply ac which, in this problem is $\left(6\right)\left(-20\right)=-120$
Find two numbers whose product is $ac=-120$
and whose sum is $b=+7$
One number is positive and the other is negative because we want the product to be negative. The positive factor has a higher absolute value because we want the sum to be positive.
We start the list:
$\left(-1\right)\left(120\right)$ 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
$\left(-8\right)\left(15\right)$ the sum is 7
Write $6{x}^{2}+7x-20$ replacing $+7x$ with $-8x+15x$
$6{x}^{2}+7x-20=6{x}^{2}-8x+15x-20$ Now factor by grouping:
$\left(6{x}^{2}-8x\right)+\left(15x-20\right)=2x\left(3x-4\right)+5\left(3x-4\right)=\left(2x+5\right)\left(3x-4\right)$
The "factoring box" is a technique for grouping factors.
To use this box, you must first take out any factor that is common to all 3 terms. If we had started with $24{x}^{2}+28x-80$ we would have to first factor out the 4 to get $4\left(6{x}^{2}+7x-20\right)$
Place the first and last terms in the main diagonal (upper left and lower right). Then, insert the two terms we discovered above, -8x and +15x, in the remaining two locations.
$\left(\begin{array}{cc}6{x}^{2}& +15x\\ -8x& -20\end{array}\right)$
Notice that each row has common factors and each column has common factors
$\left(\begin{array}{cc}6{x}^{2}& +15x\end{array}\right)$ has common factor 3x
and $\left(\begin{array}{c}6{x}^{2}\\ -8x\end{array}\right)$ has common factor 2x
We'll write those factors on a new first row and a new left column:
$\left(\begin{array}{ccc}& 2x& +5\\ 3x& 6{x}^{2}& +15x\\ -4& -8x& -20\end{array}\right)$
The factors are the top row and the left column:
$\left(2x+5\right)\left(3x-4\right)$
Using a different box
If we had put the -8x and 15x in the other 2 places.
$\left(\begin{array}{cc}6{x}^{2}& -8x\\ +15x& -20\end{array}\right)$
$\left(\begin{array}{ccc}& 3x& -4\\ 2x& 6{x}^{2}& -8x\\ +5& +15x& -20\end{array}\right)$
The factors are $\left(3x-4\right)\left(2x+5\right)$

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