Factor:12w^2+10w-8
Here's one just like it. Use it as a model to
factor yours by:
40x²+50x-15
First factor out the common factor of 5
5(8x²+10x-3)
Now to factor what is in the parentheses
Multiply the absolute values of the coefficient of the first
term and last term
8·3 = 24
Think of two positive integers that have product 24 and
which have DIFFERENCE of the absolute value of the coefficient
of the middle term, 10. [Note: it's DIFFERENCE when the last
sign is -, but it's SUM when it's +]
These as 12 and 2.
Now use 12 and 2 to rewrite the middle term + 10x.
+10x = +12x - 2x
So replace +10x by +12x - 2x
5[8x²+12x-2x-3]
Now factor by grouping.
Factor the first two terms by taking out 4x
5[4x(2x+3)-2x-3]
Now factor the last two terms. You can take out a -1.
[Always factor out a negative number if the third term is
negative
5[4x(2x+3)-1(2x+3)]
Now factor out (2x+3)
5[(2x+3)(4x-1)]
Now dispense with the brackets.
5(2x+3)(4x-1)
The answer to your problem is either
2(3x+4)(2x-1) or 2(2x-1)(3x+4)
Edwin
AnlytcPhil@aol.com