# How do you factor 6x^2 + 7x - 20 = 0?

Mar 27, 2015

A quadratic equation is simply another way of solving a problem if the solution cannot be factored logically.

Let’s say we have the equation ${x}^{2} + 2 x - 3$ for example. This equation could be solved logically using the factors of the first and last terms.

To begin, we can state the factors of the first term, ${x}^{2}$. Imagine there’s an invisible 1 in front of the ${x}^{2}$, therefore the factors are 1, because only $1 \cdot 1 , \mathmr{and} - 1 \cdot - 1$ will multiply to get one. Then we can analyze the third term, $- 3$. The factors of $- 3$ are either $1 \cdot - 3 , \mathmr{and} - 1 \cdot 3$.

Now we can check and see if any of the factors can combine in order to get a $+ 2$, the middle term (don’t worry about the x’s, those will carry over). Recall $1 = - 1 , 1$, and $- 3 = 1 , - 1 , 3 , - 3$

From our factors we can use a -1 and a 3 to get +2. Therefore,
$\left(x + 3\right) \left(x - 1\right) = 0$ is our derived factorization. Then plug in the values to make the statement true, -3 or 1 will both result in an answer of 0 and our the possible values for x.

However , when the logical factorization seen above is not possible, we can plug our numbers into the quadratic equation .

$a {x}^{2} + b x + c$ is the standard way we view an equation. Using the values from the equation above, $a = 1 , b = 2 , \mathmr{and} c = - 3$.

After our a, b, and c values are found we can plug them into the actual quadratic equation.

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Note : This equation may look intimidating, but as long as you follow factoring rules, you should have no problem. It’s totally normal to come out with an answer containing square roots.

Mar 29, 2015

Here's a somewhat mechanical approach:

To factor $6 {x}^{2} + 7 x - 20$

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

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

Find 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:

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

$6 {x}^{2} + 7 x - 20 = 0$
$\left(2 x + 5\right) \left(3 x - 4\right) = 0$
$2 x + 5 = 0$ or $3 x - 4 = 0$

$x = \frac{- 5}{2}$ or $x = \frac{4}{3}$

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 $4 \left(6 {x}^{2} + 7 x - 20\right) = 0$)

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 othe 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:

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

Jun 26, 2015

We can factorize it easily.

#### Explanation:

$6 {x}^{2} + 7 x - 20$
$= 6 {x}^{2} + \left(15 - 8\right) x - 20$
$= 6 {x}^{2} + 15 x - 8 x - 20$
$= 3 x \left(2 x + 5\right) - 4 \left(2 x + 5\right)$
$= \left(2 x + 5\right) \left(3 x - 4\right)$

We need to multiply the coefficient of ${x}^{2}$ which is 6 with constant which is 20 and then we get 120. Then we need to find two numbers which would be equal to 120 when multiplied and equal to 7 when subtracted. Such numbers are 15 & 8 because (15-8)=7 and (15 x 8)=120.
The we put "(15-8)" before the the variable $x$ and multiply it with $x$ and we get $6 {x}^{2} + 15 x - 8 x - 20$ . Then we take $3 x$ in common from the first half and then we take $- 4$ in common from the 2nd half. We get the answers.