Finding the Value of a in a Polynomial Function Using Remainder Theorem

[gibguitar]

Polynomial functions... find "a"

Homework Statement

When ax³ - 4x² + 5x - 3 is divided by (x+2) and (x-1), the remainders are equal. Find a.

Don't know where to start. A little help? Hints?

 

[eumyang]

Use the Remainder Theorem: if a polynomial f(x) is divided by x - k, then the remainder is f(k). Use it twice and set the two expressions equal.

 

[Ray Vickson]

Homework Statement

When ax³ - 4x² + 5x - 3 is divided by (x+2) and (x-1), the remainders are equal. Find a.

Don't know where to start. A little help? Hints?

Hint: carry out the procedure of polynomial division. If you don't know how, there are numerous resources available to show you the steps. See, eg., http://www.sosmath.com/algebra/factor/fac01/fac01.html or
http://en.wikipedia.org/wiki/Polynomial_long_division .

RGV

 

[gibguitar]

Okay thanks. I guess the question now is, how to do I do long division of polynomials with two variables?

 

[Ray Vickson]

Okay thanks. I guess the question now is, how to do I do long division of polynomials with two variables?

You don't. The only "variable" is x. The unknown quantity 'a' is not regarded as a 'variable', even though you don't know what it is yet! Look at it this way: you could plug in some value of 'a', such as a = 1, then carry out the two divisions. You could plug in another value such as a = 5.8 and carry out divisions again. For most values of 'a' you get different remainders when you carry out the two divisions, but for some very special value of values of 'a' the two remainders are the same. You are supposed to find that value of 'a'. Notice that 'a' is never a variable in any of the divisions---only x is.

RGV

 

[gibguitar]

You don't. The only "variable" is x. The unknown quantity 'a' is not regarded as a 'variable', even though you don't know what it is yet! Look at it this way: you could plug in some value of 'a', such as a = 1, then carry out the two divisions. You could plug in another value such as a = 5.8 and carry out divisions again. For most values of 'a' you get different remainders when you carry out the two divisions, but for some very special value of values of 'a' the two remainders are the same. You are supposed to find that value of 'a'. Notice that 'a' is never a variable in any of the divisions---only x is.

RGV

Ahh... makes sense... well, I could do trial and error and finally come up with an answer, but that would take too long, I'm assuming there's a better way. I did what you said and when I divided:

(x+2) into x^3-4x^2+5x-3 = x^2-6x+17 remainder: -37
(x-1) into x^3-4x^2+5x-3 = x^2-3x+2 remainder: -1

So obviously, a ≠ 1.

Through trial and error, I found the answer to be a = -3
Both gave me remainders of -5 after dividing by (x+2) and (x-1)

What is the quicker (correct) method of getting a = -3?

 

[Ray Vickson]

The quick way is to just keep 'a' unevaluated until the very end. In other words, in your division you will get quantities that have 'a' in them.

RGV

 

[gibguitar]

I actually got the answer using remainder theorem! I will put the answer here for future reference in case anyone in the future needs help!

Remainder when divided by x+2 is f(-2)
Remainder when divided by x-1 is f(1)

f(-2) =a(-2)³-4(-2)²+5(-2)-3
f(-2) = -8a -16 -10 -3
= -8a -29

f(1) = a(1)³-4(1)²+5(1)-3
f(1) = a -4 +5 -3
= a -2

a -2 = -8a -29
9a = -27

a = -3Thanks for all the help guys!