Question

How do you factor `5x^2+23x-10`?

Answer 1

`5x^2+23x-10=color(green)((5x-2)(x+5))`

Explanation:

We are hoping for integer values `a,b,c,d` such that
`(ax+b)(cx+d)=5x^2+23x-10`

Assuming `a` and `c` are positive
`b` and `d` must have opposite signs (since their product is negative)

and if `d` is the greater of `abs(d)` and `abs(b)`, then `d` must be greater than zero (since `23x` is positive):

#{: (underline("Factors of " 5),color(white)("XX"),underline("Factors of "-10),color(white)("XX"),underline("Cross product difference")), (1,,-1,,), (5,,10,,5), (" ",,,,), (1,,10,,), (5,,-1,,49), (" ",,,,), (1,,-2,,), (5,,5,,-5), (" ",,,,), (1,,5,,), (5,,-2,,23" ...we found it!") :}#

Answer 2

`5x^2+23x-10 = (5x-2)(x+5)`

Explanation:

We can use an AC method to factor the given quadratic:

`5x^2+23x-10`

Look for a pair of factors of `AC = 5*10 = 50` with difference `B=23`. (We look for a difference rather than a sum because the sign of the constant term is negative.)

The pair `25, 2` works.

Use this pair to split the middle term and factor by grouping:

`5x^2+23x-10 = 5x^2+25x-2x-10`

`color(white)(5x^2+23x-10) = (5x^2+25x)-(2x+10)`

`color(white)(5x^2+23x-10) = 5x(x+5)-2(x+5)`

`color(white)(5x^2+23x-10) = (5x-2)(x+5)`