Question

Factor the polynomial 6x4 + 24x3 − 72x2 completely by first factoring out the GCF, and then factoring the rest of the expression.

Answer 1

Answer:

After factorizing the given polynomial we get 6x² ( x + 6 ) ( x - 2 )

Step-by-step explanation:

Given Polynomial:

$6x^4+24x^3-72x^2$

we need to completely factorize the polynomial.

Consider,

$6x^4+24x^3-72x^2$

GCF = 6x², By taking it common out

$=6x^2(x^2+4x-12)$

$=6x^2(x^2+6x-2x-12)$

$=6x^2(x(x+6)-2(x+6))$

$=6x^2(x+6)(x-2)$

Therefore, After factorizing the given polynomial we get 6x² ( x + 6 ) ( x - 2 )

Answer 2

$64x^4+24x^3-72x^2=(8x^2)(8x^2)+(8x^2)(3x)-(8x^2)(9)\\\\=8x^2(8x^2+3x-9)=(*)\\-------------------------------\\Quadratic\ formula\\ax^2+bx+c\\\\8x^2+3x-9\\a=8,\ b=3,\ c=-9\\\Delta=b^2-4ac\to\Delta=3^2-4(8)(-9)=9+288=297=(9)(33)\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\to x_1=\dfrac{-3-\sqrt{(9)(33)}}{2(8)}=\dfrac{-3-3\sqrt{33}}{16}\\\\x_2=\dfrac{-b+\sqrt\Delta}{2a}\to x_2=\dfrac{-3+\sqrt{(9)(33)}}{2(8)}=\dfrac{-3+3\sqrt{33}}{16}\\--------------------------------$

$(*)=8x^2\left(x-\dfrac{-3-3\sqrt{33}}{16}\right)\left(x-\dfrac{-3+3\sqrt{33}}{16}\right)$