Odd problem

[JackMD]

When 3a² – 5a + c is divided by a + k, the quotient is 3a + 1 and the remainder is 3. Find c and k.

Can someone break this down for me struggling to figure it out? I don't understand where to start nor what to do after that.
Thank you in advance.

 

[Steven G]

Think about how you would divide 17 by 5.

When 17 is divided by 5 the quotient is 3 and the remainder is 2. This is exactly in the language of your post. Do you agree with my answer of quotient being 3 and the remainder is 2?

How would you check my result? You would multiply 5 by 3, then add 2 and see if you get 17.

Now you do the check and see if you can find c and k.

Alternatively you can use synthetic division.

 

[JackMD]

I am still a tad confused, I have to finish this problem tonight but, how would I use synthetic division to find my solution? Seeing how we are already given the quotient what would be the point in doing so?

 

[tkhunny]

Really big hint:

27 / 6 = 4 + 3/6 <== This should look a lot like your problem, but a numerical example.

27 = 4*6 + 3 <== This should look a lot like your solution, but a numerical example.

 

[Steven G]

Come on and try. Multiply the equivalent to 3 by the equivalent to 5 and then add the remainder 2.

So you need to compute (a+k)(3a+1) + 3 and set this equal to 3a^62-5a+c. Then equate coefficients.

 

[Dr.Peterson]

When 3a² – 5a + c is divided by a + k, the quotient is 3a + 1 and the remainder is 3. Find c and k.

Can someone break this down for me struggling to figure it out? I don't understand where to start nor what to do after that.
Thank you in advance.

As I understand it, "a" is taken as the variable in a polynomial (think "x"), so the equation you get when you write out the meaning of the given facts is an equation that must be true for all a -- so you can equate coefficients of a. If this were just about specific numbers, it couldn't be solved.

That may be what you are missing.

 

[JackMD]

So it would be 3x²-5x+c / x+k and that would be = to 3x+1? If I solve that

It would be

k=8x+1−3x²−c/x
c=−3x³−kx+8x2+x
and I cannot figure out what x would = to.

I really am stuck on this part.

 

[Steven G]

I am sorry but I can't help you if you don't do what I ask you to do.

Simply compute (a+k)(3a+1) + 3 and set this equal to 3a^2-5a+c.


For the next step here is an example: Find k and c. If 7a^2 + 4ka +9 = 7a^2 + 12a + c then 4k = 12, so k=3 and c=9.

 

[Steven G]

So it would be 3x²-5x+c / x+k and that would be = to 3x+1? If I solve that

It would be

k=8x+1−3x²−c/x
c=−3x³−kx+8x2+x
and I cannot figure out what x would = to.

I really am stuck on this part.

You are NOT being asked to find out what x ( or a) equals. You are being asked to find c and k. Please read the problem.

 

[Dr.Peterson]

So it would be 3x²-5x+c / x+k and that would be = to 3x+1? If I solve that

No. The remainder is 3, and you can't ignore it.

Are you familiar with the "division algorithm", in which we expression division as

Dividend = Divisor * Quotient + Remainder ?​

That's the form you want to write it in. You wrote it as an explicit division, but omitted the remainder.

Now, as has been suggested, you could alternatively carry out the division (3x²-5x+c) / (x+k), using either long division or synthetic division; the coefficients you get will be expressions, not just numbers, so it might be confusing to you; that's what I don't recommend it. But it may be instructive.

In both methods, you need to use the fact that two polynomials are equal (that is, the same polynomial, not just the same number) when all coefficients are equal. So you'll be setting coefficients equal to one another.

 

[JackMD]

I figured it out thank you all.

c = 1, k = -2

I do appreciate the help.

 

[Oiler]

As a closing remark:
[MATH]3a^2-5a+c=(a+k)(3a+1)+3=3a^2+a+3ka+k+3=3a^2+a(1+3k)+(k+3)[/MATH]hence,
[MATH]1+3k=-5[/MATH][MATH]k+3=c[/MATH][MATH]\therefore k=-2, c=1[/MATH]