# Very Interesting Question on Division of Polynomials

1. Jul 1, 2009

### modulus

Very Interesting Question on Division of Polynomials!!

1. Question: 'When a polynomial f(x)= x^4 - 6x^3 + 16x^2 - 25x + 10 is divided by another polynomiall g(x)= x^2 - 2x + k, the remainder is x+a. Find the value of a k and a'.

2. Relevant equations

3. I tried solving it by using long division with f(x) as the dividend and g(x) as the divisor, and I figured out the remainder in terms of k and equated it with x+a; this was my first equation. After that, I used the realtion of dividend-remainder=quotient*divisor, taking the remainder in terms of 'a'. This was my second equation.

After that, I subtracted my first equation from my second, so that the 'a' cancelled out, and got an equation of purely 'k', but the equation still had the variable 'x' in it, so I'm stuck again.

I need help, is my approach to the question wrong?

2. Jul 1, 2009

### HallsofIvy

Staff Emeritus
Re: Very Interesting Question on Division of Polynomials!!

I don't understand why you are dealing with "two equations". Just divide $x^4- 6x^3+ 16x^2- 25x+ 10$ by $x^2- 2x+ k$. What remainder do you get? You should be able to determine k by setting the coefficient of x equal to 1 and then calculate a.

3. Jul 2, 2009

### modulus

Re: Very Interesting Question on Division of Polynomials!!

I don't quite understand what you mean by setting the coefficient of 'x' equal to one.

But, I divided f(x) by g(x) and got the following remainder:
2kx - 9x + 8k - k^2 + 10

4. Jul 2, 2009

### HallsofIvy

Staff Emeritus
Re: Very Interesting Question on Division of Polynomials!!

In other words, you got (2k- 9)x- (k^2- 8k- 10) and that must be equal to x+ a.
Okay, on one of those has "coefficient of x" equal to 2k- 9 and the other has coefficient 1. In order to be equal for all x, those coefficients must be the same: 2k- 9= 1. Then the constant terms, -k^2+ 8k+ 10 and a must be equal: -k^2+ 8k+ 10= a.

(However, I did not get "-k^2+ 8k+ 10" for the constant term. You might try dividing again to check.)

5. Jul 3, 2009

### jontyjashan

Re: Very Interesting Question on Division of Polynomials!!

After using long division, you should find
(x^4 - 6x^3 + 16x^2 - 25x + 10)/(x^2 - 2x + k) = x^2 - 4x + 8 - k + {(2k - 9)x + k^2 - 8k + 10}/{x^2 - 2x + k}

So the remainder is:

x + a = (2k - 9)x + k^2 - 8k + 10.

Equating like co-efficients of x gives

2k - 9 = 1 and k^2 - 8k + 10 = a.

Solving for k in the first equation gives:

2k = 10

k = 5.

Substituting into the second equation gives:

5^2 - 8. 5 + 10 = a

25 - 40 + 10 = a

a = -5.

So a = -5 and k = 5.

6. Jul 3, 2009

### jontyjashan

Re: Very Interesting Question on Division of Polynomials!!

Is my answer correct????????????????

7. Jul 3, 2009

### modulus

Re: Very Interesting Question on Division of Polynomials!!

Thanks