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

3 Answers
Mar 1, 2016

Answer:

#f(x)=4/3(x-4)(3x-2)#

Explanation:

Given equation is #f(x)=12x^2-40x-32#

Equate #f(x)=0# so we get #12x^2-40x-32=0#
Cancel all the common terms so that we get a simpler equation (remember what you're cancelling too, for here it is #4#)
i.e #3x^2-10x-8=0#

Now, use the factoring equation #r_{1,2}=(-b+-sqrt{b^2-4ac})/(2a)#
where #a=3,b=-10,c=-8#

So, we get #r_{1,2}=(10+-sqrt{100+4*3*8})/(2*3)#
#impliesr_{1,2}=(10+-sqrt{100+96})/6\impliesr_{1,2}=(10+-14)/6#
#\impliesr_{1,2}=(5+-7)/3#

So, that means, the roots of the equation are #r_1=4# and #r_2=-2/3#

Now, factorizing the simpler equation is as follows #(x-4)(x-2/3)\implies(x-4)(3x-2)/3#

Remember that we simplified the original equation by dividing by #4#? Now, multiply that back. That's the answer.

Mar 1, 2016

Answer:

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 = -8

product 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 + 2x

hence # 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)#

Mar 1, 2016

Answer:

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

Explanation:

Given expression is

#12x^2-40x-32#

From inspection we see that #4# can be factored out as each term can be divided by #4#. We have the first factor.

So the expression becomes after #4# is factored out.
#4(3x^2-10x-8)# .......(1)

After having found the first factor now we need to find factors of
#(3x^2-10x-8)# ........(2)

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 #=3xx-8=-24#

Coefficient of middle term #=-10#.

Product #-24# is the key to finding the the split.

Since there is #-# sign in front of this product, it tells us that the two parts must be such that one is #-ve and#other #+ve.# We know that product of two positive numbers will always be positive.
Also the negative part should be greater than #10#.

Let me see if #-12 and 2# are the two parts.
The product of these two parts is equal to #-24 and# sum is #-10.# Yes these do. Rewriting equation (2) with split terms we obtain

#(3x^2-12x+2x-8)#

Now pair first two, and pair last two. Find out the common factor in each pair. Must take care of #-ve# sign in front of any term while pairing and inserting brackets.

#(3x^2-12x)+(2x-8)#

In the first pair we see that #3x# is a common factor and in the second pair #2# is a common factor. Take common factor out of each pair and write as

#3x(x-4)+2(x-4)#
We get two terms. Again observe that #(x-4)# is a common factor of these. Taking it out as a common factor we rewrite as

#(x-4)(3x+2)# These are remaining two factor.
Finally putting down all three factors together of equation (1)

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