+0  
 
0
329
4
avatar

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

Guest Jul 20, 2017
 #1
avatar+19661 
+1

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

 

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

 

laugh

heureka  Jul 20, 2017
 #2
avatar
+1

How did you do it?

Guest Jul 20, 2017
 #3
avatar
+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
avatar+87336 
+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 )

 

 

cool cool cool

CPhill  Jul 20, 2017

16 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.