How do you factor 12x^2-40x-3212x240x32?

3 Answers
Mar 1, 2016

f(x)=4/3(x-4)(3x-2)f(x)=43(x4)(3x2)

Explanation:

Given equation is f(x)=12x^2-40x-32f(x)=12x240x32

Equate f(x)=0f(x)=0 so we get 12x^2-40x-32=012x240x32=0
Cancel all the common terms so that we get a simpler equation (remember what you're cancelling too, for here it is 44)
i.e 3x^2-10x-8=03x210x8=0

Now, use the factoring equation r_{1,2}=(-b+-sqrt{b^2-4ac})/(2a)r1,2=b±b24ac2a
where a=3,b=-10,c=-8a=3,b=10,c=8

So, we get r_{1,2}=(10+-sqrt{100+4*3*8})/(2*3)r1,2=10±100+43823
impliesr_{1,2}=(10+-sqrt{100+96})/6\impliesr_{1,2}=(10+-14)/6r1,2=10±100+966r1,2=10±146
\impliesr_{1,2}=(5+-7)/3r1,2=5±73

So, that means, the roots of the equation are r_1=4r1=4 and r_2=-2/3r2=23

Now, factorizing the simpler equation is as follows (x-4)(x-2/3)\implies(x-4)(3x-2)/3(x4)(x23)(x4)3x23

Remember that we simplified the original equation by dividing by 44? Now, multiply that back. That's the answer.

Mar 1, 2016

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

Explanation:

First take out common factor of 4.

hence 4(3x^2 - 10x - 8 ) 4(3x210x8)

now require to factor the quadratic 3x^2 - 10x -83x210x8

To factor the quadratic : ax^2 + bx + c ax2+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=(3×8)=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 - 83x212x+2x8

and factor[3x^2-12x = 3x(x-4)]" and "[ 2x - 8 = 2(x-4)][3x212x=3x(x4)] and [2x8=2(x4)]

there is a common factor of (x-4) → (x-4)(3x+2)

rArr 12x^2-40x-32 = 4(x-4)(3x+2)12x240x32=4(x4)(3x+2)

Mar 1, 2016

4(x-4)(3x+2)4(x4)(3x+2)

Explanation:

Given expression is

12x^2-40x-3212x240x32

From inspection we see that 44 can be factored out as each term can be divided by 44. We have the first factor.

So the expression becomes after 44 is factored out.
4(3x^2-10x-8)4(3x210x8) .......(1)

After having found the first factor now we need to find factors of
(3x^2-10x-8)(3x210x8) ........(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=3×8=24

Coefficient of middle term =-10=10.

Product -2424 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 andveandother +ve.+ve. We know that product of two positive numbers will always be positive.
Also the negative part should be greater than 1010.

Let me see if -12 and 212and2 are the two parts.
The product of these two parts is equal to -24 and24and sum is -10.10. Yes these do. Rewriting equation (2) with split terms we obtain

(3x^2-12x+2x-8)(3x212x+2x8)

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

(3x^2-12x)+(2x-8)(3x212x)+(2x8)

In the first pair we see that 3x3x is a common factor and in the second pair 22 is a common factor. Take common factor out of each pair and write as

3x(x-4)+2(x-4)3x(x4)+2(x4)
We get two terms. Again observe that (x-4)(x4) is a common factor of these. Taking it out as a common factor we rewrite as

(x-4)(3x+2)(x4)(3x+2) These are remaining two factor.
Finally putting down all three factors together of equation (1)

4(x-4)(3x+2)4(x4)(3x+2)