Question

How do you factor `3x^2-2x-8`?

Answer 1

The solution is: `(3x+4)(x-2)`

The polynomial is of the type `ax^2+bx+c`.

There are two ways to factor.

The first one:

we have to find two numbers whose sum is `-2` (that is `c`) and whose product is `-24` (that is `a*c`).

`-24=-1*24`
`-24=-24*1`
`-24=-2*12`
`-24=-12*2`
`-24=-3*8`
`-24=-8*3`
`-24=-4*6`
`-24=-6*4`

The couple of numbers whose sum is `-2` is `-6,4`.

Now we have to "split" the monomial `-2x` in two parts: `-2x=-6x+4x`.

So our polynomial becomes:

`3x^2-6x+4x-8=3x(x-2)+4(x-2)=(x-2)(3x+4)`.

If it is known the way to solve an equation of 2° degree, here there is the second way to factor it:

We can imagine that this polynomial "becomes" an equation:

`3x^2-2x-8=0`, we can calculate the quantity

`Delta=b^2-4ac=4+96=100`,

and than we have to use this formula:

`x_(1,2)=(-b+-sqrtDelta)/(2a)=(2+-10)/6rArrx_1=-4/3;x_2=2`.

And now we have to use the formula:

`ax^2+bx+c=a(x-x_1)(x-x_2)`.

So:

`3x^2-2x-8=3(x+4/3)(x-2)=(3x+4)(x-2)`