osula9a

2021-12-26

Factor each of the algebraic expressions completely. $12{x}^{3}-52{x}^{2}-40x$

usumbiix

Beginner2021-12-27Added 33 answers

We find factor of $12{x}^{3}-52{x}^{2}-40x$

We find factor of$12{x}^{3}-52{x}^{2}-40x$

Now$12{x}^{3}-52{x}^{2}-40x$

$=x(12{x}^{2}-52x-40)$

$=4x(3{x}^{2}-13x-10)$

$=4x(3{x}^{2}-15x+2x-10)$

$=4x(3x(x-5)+2(x-5))$

$=4x(x-5)(3x+2)$

Hence$12{x}^{3}-52{x}^{2}-40x=4x(x-5)(3x+2)$

We find factor of

Now

Hence

Karen Robbins

Beginner2021-12-28Added 49 answers

Factor 4x out of

Vasquez

Expert2022-01-08Added 669 answers

((12 * (x

((2

12x

Factoring 3x

The first term is, 3x

The middle term is, -13x its coefficient is -13.

The last term, "the constant", is -10

Step-1 : Multiply the coefficient of the first term by the constant 3 * -10 = -30

Step-2 : Find two factors of -30 whose sum equals the coefficient of the middle term, which is -13.

-30+1=-29

-15+2=-13

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -15 and 2.

3x

Step-4 : Add up the first 2 terms, pulling out like factors :

3x * (x-5)

Add up the last 2 terms, pulling out common factors:

2 * (x-5)

Step-5 : Add up the four terms of step 4:

(3x+2) * (x-5)

Which is the desired factorization

Answer: 4x * (x-5) * (3x+2)

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