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# helpppp!!! 6q^2 - 7qr - 24r^2 ?????

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helpppp!!!     6q^2 - 7qr - 24r^2

Guest Jul 20, 2017
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#1
+18369
+1

6q^2 - 7qr - 24r^2

$$6q^2 - 7qr - 24r^2 = 6(q-\frac83r) (q+\frac32r)$$

heureka  Jul 20, 2017
#2
+1

How did you do it?

Guest Jul 20, 2017
#3
+2

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)

Guest Jul 20, 2017
#4
+75333
+1

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 )

CPhill  Jul 20, 2017

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