Raelynn Velasquez

2023-03-06

How to factor quadratic equations box method?

Cooper Barker

Beginner2023-03-07Added 6 answers

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

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

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

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:

$(-1)\left(120\right)$ the sum is not +7

$(-2)\left(60\right)$ the sum is not +7

$(-3)\left(40\right)$ the sum is not +7

$(-4)\left(30\right)$ the sum is not +7

$(-5)\left(24\right)$ the sum is not +7

$(-6)\left(20\right)$ the sum is not +7

7 is not a factor

$(-8)\left(15\right)$ 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:

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

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 $4(6{x}^{2}+7x-20)$

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:

$(2x+5)(3x-4)$

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 $(3x-4)(2x+5)$

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

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

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:

$(-1)\left(120\right)$ the sum is not +7

$(-2)\left(60\right)$ the sum is not +7

$(-3)\left(40\right)$ the sum is not +7

$(-4)\left(30\right)$ the sum is not +7

$(-5)\left(24\right)$ the sum is not +7

$(-6)\left(20\right)$ the sum is not +7

7 is not a factor

$(-8)\left(15\right)$ 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:

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

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 $4(6{x}^{2}+7x-20)$

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:

$(2x+5)(3x-4)$

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 $(3x-4)(2x+5)$

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