Question

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

Answer 1

`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)`

Answer 2

`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))`