How do you factor #12x^2-40x-32#?
Given equation is
Cancel all the common terms so that we get a simpler equation (remember what you're cancelling too, for here it is
Now, use the factoring equation
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
First take out common factor of 4.
# 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 = -8
# = (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 + 2x
# 3x^2 - 12x + 2x - 8#
#[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)#
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
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)