# How do you factor #12x^2-40x-32#?

##### 3 Answers

#### Explanation:

Given equation is

Equate

Cancel all the common terms so that we get a simpler equation (remember what you're cancelling too, for here it is

i.e

Now, use the factoring equation

where

So, we get

So, that means, the roots of the equation are

Now, factorizing the simpler equation is as follows

Remember that we simplified the original equation by dividing by

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

#### Explanation:

First take out common factor of 4.

hence

# 4(3x^2 - 10x - 8 ) # now require to factor the quadratic

# 3x^2 - 10x -8# To factor the quadratic :

# ax^2 + bx + c #

consider the factors of the product ac which also sum to give b , the coefficient of the x term.

Here a = 3 , b = -10 and c = -8product ac

# = (3xx-8) = -24# and factors of -24 are ± (1,2,3,4,6,8,12, 24 ). The required factors are +2 and -12 as they sum to - 10 , the middle term.

Now replace - 10x by -12x + 2xhence

# 3x^2 - 12x + 2x - 8# and factor

#[3x^2-12x = 3x(x-4)]" and "[ 2x - 8 = 2(x-4)]# there is a common factor of (x-4) → (x-4)(3x+2)

#rArr 12x^2-40x-32 = 4(x-4)(3x+2)#

#### Explanation:

Given expression is

From inspection we see that

So the expression becomes after

After having found the first factor now we need to find factors of

We need to split coefficient of second term into two parts so that the sum of parts is equal to the middle term and their product is equal to the product of coefficients first and third term.

We observe that product of coefficients of first and third term

Coefficient of middle term

Product

Since there is

Also the negative part should be greater than

Let me see if

The product of these two parts is equal to

Now pair first two, and pair last two. Find out the common factor in each pair. Must take care of

In the first pair we see that

We get two terms. Again observe that

Finally putting down all three factors together of equation (1)