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

Factor each of the algebraic expressions completely. $12{x}^{3}-52{x}^{2}-40x$
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usumbiix
We find factor of $12{x}^{3}-52{x}^{2}-40x$
We find factor of $12{x}^{3}-52{x}^{2}-40x$
Now $12{x}^{3}-52{x}^{2}-40x$
$=x\left(12{x}^{2}-52x-40\right)$
$=4x\left(3{x}^{2}-13x-10\right)$
$=4x\left(3{x}^{2}-15x+2x-10\right)$
$=4x\left(3x\left(x-5\right)+2\left(x-5\right)\right)$
$=4x\left(x-5\right)\left(3x+2\right)$
Hence $12{x}^{3}-52{x}^{2}-40x=4x\left(x-5\right)\left(3x+2\right)$
Not exactly what you’re looking for?
Karen Robbins
$12{x}^{3}-52{x}^{2}-40x$
Factor 4x out of $12{x}^{3}-52{x}^{2}-40x$
$4x\left(3{x}^{2}-13x-10\right)$
$4x\left(3x+2\right)\left(x-5\right)$
Not exactly what you’re looking for?
Vasquez

((12 * (x${}^{3}$))-(2${}^{2}$ * 13x${}^{2}$))-40x
((2${}^{2}$ * 3x${}^{2}$)-(2${}^{2}$ * 13x${}^{2}$))-40x
12x${}^{3}$-52x${}^{2}$-40x=4x * (3x${}^{2}$-13x-10)
Factoring 3x${}^{2}$-13x-10
The first term is, 3x${}^{2}$ its coefficient is 3.
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${}^{2}$ - 15x + 2x - 10
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)