We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
142
2
avatar

If h(x) = f(x)g(x) and f(x) = 2x +5, determine g(x).

 

I'm given h(x) = 10x\({^2}\) + 13x -30.

 

I did \({10x^2 + 13x - 30 \over 2x + 5}\) to isolate for g(x), but I tried factoring the numerator and it doesn't seem factorable?

 

How should I get the textbook answer of g(x) = 5x - 6?

 

Thank you! :)

 Jun 12, 2019

Best Answer 

 #1
avatar+8829 
+2
To factor   10x2 + 13x - 30 , 
let's split  13x  into two terms such that the product of their coefficients  =  (10)(-30)  =  -300

 

 

What two numbers add to  13  and multiply to  -300  ?     +25  and  -12 
So we can split the middle term like this:

 

 

10x2 + 13x - 30

 

=  10x2 + 25x - 12x  - 30

                                             Factor  5x  out of the first two terms.

=  5x(2x + 5) - 12x - 30

                                             Factor  -6  out of the last two terms.

=  5x(2x + 5) - 6(2x + 5)

                                             Factor  (2x + 5)  out of both remaining terms.

=  (2x + 5)(5x - 6)

 

Does that help answer your question? smiley

 Jun 12, 2019
 #1
avatar+8829 
+2
Best Answer
To factor   10x2 + 13x - 30 , 
let's split  13x  into two terms such that the product of their coefficients  =  (10)(-30)  =  -300

 

 

What two numbers add to  13  and multiply to  -300  ?     +25  and  -12 
So we can split the middle term like this:

 

 

10x2 + 13x - 30

 

=  10x2 + 25x - 12x  - 30

                                             Factor  5x  out of the first two terms.

=  5x(2x + 5) - 12x - 30

                                             Factor  -6  out of the last two terms.

=  5x(2x + 5) - 6(2x + 5)

                                             Factor  (2x + 5)  out of both remaining terms.

=  (2x + 5)(5x - 6)

 

Does that help answer your question? smiley

hectictar Jun 12, 2019
 #2
avatar
+1

Ah yes! Thank you!

Guest Jun 12, 2019

16 Online Users

avatar
avatar
avatar
avatar