Question

How do you factor `24r^2 - 10r - 21`?

Answer 1

`24r^2-10r-21` is of the form `ar^2+br+c` with `a=24`, `b=-10` and `c=-21`.

The discriminant of this quadratic is given by the formula:

`Delta = b^2-4ac = (-10)^2-(4xx24xx-21) = 100+2016 = 2116 = 46^2`

Being a positive perfect square, the equation `24r^2-10r-21 = 0` has two distinct rational roots, given by the formula:

`r = (-b+-sqrt(Delta))/(2a) = (10+-46)/48`

That is `r=-36/48=-3/4` and `r=56/48=7/6`.

Hence `24r^2-10r-21 = (4r+3)(6r-7)`

Answer 2

Factor f(x) = 24x^2 - 10x - 21 = (x - p)(x - q).
Use the new AC Method to factor trinomials. It is fast, systematic, no factoring, no solving binomials.
Converted trinomial `y' = x^2 - 10x - 504` = (x - p')(x - q').
Find p' and q' by composing factor pairs of a.c = -504. Proceed:... (-14, 16)(-18, 28). This last sum is 10 = -b. Then p' = 18 and q' = -28.

Then, `p = (p')/a = 18/24 = 3/4`, and `q = (q')/a = -28/24 = -7/6`

Factored form: `y = (x + 3/4)(x - 7/6) = (4x + 3)(6x - 7)`

Check by developing `y = 24x^2 - 28x + 18x - 21` . OK