Question

How to solve `8x^2−11x−10` by the factorization method?

Answer 1

`(8x+5)(x-2)`

Explanation:

Trying to factor by splitting the middle term

Step 1: Multiply the coefficient of the first term by the constant
`8 xx -10` = `-80`

The first term is, `8x^2` its coefficient is `8` .
The middle term is, `-11x` its coefficient is `-11` .
The last term, "the constant", is `-10`

Step 2: Find two factors of `-80` whose sum equals the middle term, which is `-11`

`-40 + 2 = -38`
`-20 + 4 = -16`
`-16+ 5 = -11` -----> Correct!

Step 3 : Rewrite the equation by splitting the middle term using the two factors found in step 2 above, `-16` and `5`

`8x^2-16x+5x-10`

`8x(x-2) + 5(x-2)`

`(8x+5)(x-2)` -----> Final result (factors)

Answer 2

The equation set equal to `0` gives
`x=2 orx = -5/8`

Explanation:

If you want this to be SOLVED, then there needs to be an EQUATION first.

Without an equation the expression can only be simplified or factorised.

I will assume you intended `8x^2-11x-10=0`

Find the factors of the quadratic trinomial:

Find factors of `8 and 10` whose products subtract to give `11`

`" "8 and "10"`
`" "darr " "darr`
`" "1" "2" "rarr 8xx 2 = 16`
`" "8" "5" "rarr1xx5 =ul5`
`color(white)(xxxxxxxxxxxxxxx.xx)11" "larr` subtract to get `11`

We have the correct factors, now add the signs:
`" "8 and "-10"`
`" "darr " "darr`
`" "1" "-2" "rarr 8xx -2 = -16`
`" "8" "+5" "rarr1xx+5 =+ul5`
`color(white)(xxxxxxxxxxxxxxxxxxx.xx)-11" "larr` get `-11`

`(x-2)(8x+5)=0" "larr` here are the two factors

Either factor could be equal to `0`

If `x-2=0" "rarr x=2`

If `8x+5=0" "rarr 8x =-5 " "rarr x = -5/8`