given 6x^2 + kx-5=(2x-1)(3x+5), what is the value of k?

[eddy2017]

Hi to all teachers. I do not know if i am allowed to post because i was banned for some time. but I apologize to everyone i hurt with my words. i have continued to study on my own. i talked to the moderator for a while and he wanted to help me. i am really busy with work but keep trying my best for math on my own.
I am in doubt as to how to initiate this exercise?

 

[Otis]

I am in doubt as to how to initiate this exercise?

Multiply the factors on the right-hand side. Compare the result to the left-hand side.

 

[eddy2017]

distributing (2x-1)(3x+5)
6?2+7?−5
this is the result for right hand side after distributing by foiling the right hand side,
now comparing
6x^2 + kx-5= 6?^2+7?−5
i see that 6x^2 is in both sides. i'm going to cancel them and see what happens
kx-5= 7x-5
i suppose i have to isolate the k, right?
let me try to solve for k on my own and show you then.
i think this is easy now,
the two -5's cancel out leaving me with
kx=7x
kx/x = 7x/x
k= 7
what do you think, Otis?. and thanks.

 

[Otis]

i suppose i have to isolate the k, right?
... what do you think, Otis?

That's one way, but you don't "have" to do that. My suggestion is to compare the two quadratic polynomials.

6x^2 + kx - 5
6x^2 + 7x - 5

The value of k is obvious, by inspection.

 

[JeffM]

Perfect, eddy

(I am not disagreeing with Otis. Once you have seen enough such problems, you do not have to go through the steps.)

 

[eddy2017]

Thank youuu!.