How do you factor 24x^{4} + 22x^{3} - 10x^{2}?

2 Answers
Oct 8, 2016

2x^2(3x-1)(4x+5)

Explanation:

There is a color(blue)"common factor" of 2x^2 in all 3 terms.

rArr2x^2(12x^2+11x-5)

To factorise the quadratic in the bracket, use the a-c method.

That is consider the factors of - 60 which sum to + 11

These are + 15 and - 4

now write the quadratic expression as.

12x^2-4x+15x-5 and factorise in groups.

color(red)(4x)color(blue)((3x-1))color(red)(+5)color(blue)((3x-1))

Take out the common factor (3x - 1).

rArrcolor(blue)((3x-1))color(red)((4x+5))

rArr12x^2+11x-5=(3x-1)(4x+5)

Pulling it all together.

24x^4+22x^3-10x^2=2x^2(3x-1)(4x+5)

Oct 8, 2016

2x^2(x-1/3)(x+5/4)

Explanation:

In this question we are asked to factor that is to change this algebriac expression into factors .

First,let us check if there is common factor :

24x^4+22x^3-10x^2
=color(blue)2*12*color(blue)(x^2)x^2+color(blue)2*11*color(blue)(x^2)*x-5*color(blue)(2*x^2)

As it is shown in blue color the common factor is color(blue)(2*x^2)

color(blue)(2*x^2)color(red)((12x^2+11x-5))
Let us calculate delta for the expression color(red)(12x^2+11x-5) since we can't factor using polynomial identities.

Knowing the Quadratic formula of a quadratic equation color(green)(ax^2+bx+c=0)is

color(green)(delta=b^2-4ac)
Roots are:
color(green)((-b+sqrtdelta)/(2a))
color(green)((-b-sqrtdelta)/(2a))

delta=11^2-4*(12)(-5)=121+240=361
The roots are:
color(red)(x_1=(-11+sqrt361)/(2*12)=(-11+19)/24=8/24=1/3)
color(red)(x_2=(-11-sqrt361)/(2*12)=(-11-19)/24=-30/24=-5/4)
So,
color(red)(12x^2+11x-5)
=color(red)((x-1/3)(x+5/4))

24x^4+22x^3-10x^2
color(blue)(2*x^2)color(red)((12x^2+11x-5))
color(blue)(2*x^2)color(red)((x-1/3)(x+5/4))