how do i factor this problem?

We are given to factor:

12x² + 40x + 32

First, observe that all three terms have 4 as a factor, so factor that out:

4(3x² + 10x + 8)

Now, for the quadratic factor, we need two factors of 3·8 = 24 whose sum is 10, so we need 6 (3·2) and 4 (1·4), and we have:

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

All the co-efficients in this question can be divided by 4.

So

(12x^2 + 40x +32)

can be written as

4(3x^2+ 10x + 8)

Then you need to factorize out the quadratic equation which is in the brackets (3x^2 + 10x + 8)

As it is 3x^2 and 3 has 2 factors (1 and itself) you know the x terms in the brackets will be 3 and 1.

As the equation deals with only positive terms, you also know the signs in each bracket will be positive.

8 has 2 pairs of factors: 1 & 8 and 4 & 2. You can find out which one you require by trial and error.

In this instance, the equation becomes (3x + 4) (x + 2)

Don't forget the original 4

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

First factor out the common term of 4.

4(3x^2 + 10x + 8)

Multiply the 3 and 8 inside the parentheses to get 24.

Now find two numbers that multiply to 24 and add up to 10. You should get 6 and 4. Divide both by 3 and reduce.

6/3 = 2

4/3

Those are the opposites of the zeros. When you write the factors, make the denominator the coefficient of x.

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

12x²+40x+32

4(3x^2 + 10x + 8) = 0

3x^2 + 10x + 8 = 0

(3x + 4)(x + 2)

First, find the largest number that will divide into all three terms. In this case, that would be 4. Once you divide each term by 4, think about what the other factor would be for each. SO..... 4(3xsquared + 10X + 8)

12x²+40x+32

=4(3x^2 +10x +8)

= 4(3x^2 +6x +4x+8)

= 4{3x(x+2) +4(x+2)}

=4(x+2)(3x+4) ................Ans