+26367 6q^2 - 7qr - 24r^2
\( 6q^2 - 7qr - 24r^2 = 6(q-\frac83r) (q+\frac32r) \)
Factor the following:
6 q^2 - 7 q r - 24 r^2
6 q^2 - 7 q r - 24 r^2 = 6 q^2 - 7 q r - 24 r^2:
6 q^2 - 7 q r - 24 r^2
The coefficient of q^2 is 6 and the coefficient of r^2 is -24. The product of 6 and -24 is -144. The factors of -144 which sum to -7 are 9 and -16. So 6 q^2 - 7 q r - 24 r^2 = 6 q^2 - 16 q r + 9 q r - 24 r^2 = 3 q (2 q + 3 r) - 8 r (2 q + 3 r):
3 q (2 q + 3 r) - 8 r (2 q + 3 r)
Factor 2 q + 3 r from 3 q (2 q + 3 r) - 8 r (2 q + 3 r):
Answer: | (2q + 3r ) (3q - 8r)
+128474 6q^2 - 7qr - 24r^2
To factor this kind of problem, first multiply the first coefficient abd last coefficient =
6 * -24 = -144
Now....we're looking for 2 factors the multiply to 144 and sum to the midle coefficient, -7
A little trial and error produces -16 and 9
Now, we can write the middle term as a combo of these
So we have
6q^2 - 16qr + 9qr - 24r^2 factoring each pair of terms, we have
2q (3q - 8r) + 3r (3q -8r ) and (3q -8r ) is common ro both terms......so write this and then the other two terms in a separate set of parentheses........so we have
(3q -8r ) ( 2q + 3r )
