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

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)#

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))#