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What is the coefficient of x^3 when 24x^4 + 6x^3 + 4x^2 + 7x - 5 is multiplied by 6x^3 + 3x^2 - 3x + 4 and the like terms are combined?

 Apr 26, 2021
 #1
avatar+134 
+4

This is a pretty long question, but I think I'm up for answering it.

 

 

First, we need to multiply the two expressions given, and then combine like terms.

 

 

(24x^4 + 6x^3 + 4x^2 + 7x - 5) * (6x^3 + 3x^2 - 3x + 4)

 

I'm going to switch them around to make it easier:

 

(6x^3 + 3x^2 - 3x + 4) * (24x^4 + 6x^3 + 4x^2 + 7x - 5)

 

= 6x^3 (24x^4 + 6x^3 + 4x^2 + 7x - 5) + 3x^2 (24x^4 + 6x^3 + 4x^2 + 7x - 5) - 3x (24x^4 + 6x^3 + 4x^2 + 7x - 5) + 4 (24x^4 + 6x^3 + 4x^2 + 7x - 5)

 

= (144x^7 + 36x^6 + 24x^5 + 42x^4 - 30x^3) + (72x^6 + 18x^5 + 12x^4 + 21x^3 − 15x^2) + (−72x^5 − 18x^4 − 12x^3 − 21x^2 + 15x) + (96x^4 + 24x^3 + 16x^2 + 28x − 20)

 

= 144x^7 + 36x^6 + 24x^5 + 42x^4 - 30x^3 + 72x^6 + 18x^5 + 12x^4 + 21x^3 − 15x^2 −72x^5 − 18x^4 − 12x^3 − 21x^2 + 15x + 96x^4 + 24x^3 + 16x^2 + 28x − 20

 

= 144x^7 + 36x^6 + 72x^6 + 24x^5 + 18x^5 − 72x^5 + 42x^4 + 12x^4 − 18x^4 + 96x^4 - 30x^3 + 21x^3 − 12x^3 + 24x^3 − 15x^2 − 21x^2 + 16x^2 + 15x + 28x − 20

 

= 144x^7 + 108x^6 - 30x^5 + 132x^4 + 3x^3 - 20x^2 + 43x - 20

 

 

So, now let us figure out the coefficient of x^3. We can see in our final answer that we have 3x^3, so the coefficient would be 3.

 

Therefore, our answer is 3.

 

 

P.S. I certainly hope this helps, but as it was an extremely time-taking and quite difficult problem, my answer may be incorrect. I do believe I worked through it correctly, though, so I hope this helps :)

 Apr 27, 2021
 #2
avatar+114961 
+1

I think you have made this much more difficult and longer than necessary NotGuest.

 

You only need x^3 terms

These are the only terms that are relevant.      wink

 

\(6x^3*4=24x^3\\ 4x^2*-3x=-12x^3\\ 7x*3x^2=21x^3\\ -5*6x^3=-30x^3\\\)

Melody  Apr 28, 2021
 #3
avatar+134 
+2

That is another way to do it, but for me it was definitely easier to solve through the whole problem instead of going through each and every way to get a coefficient of three. It's just my opinion, though.

NotGuest  Apr 28, 2021
 #4
avatar+114961 
+1

Your way was not easier or more sensible. If you think about it you will work that out.

 

I just looked at each term in the first bracket and saw what I could pair it with from the second bracket.

Each of the 5 terms only had at most one pair, that is how it always must be.

Melody  Apr 28, 2021

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