Division of polynomials?

Question

All the other questions I've been able to answer. These two on the other hand i've tried over and over just in case I made a minor mistake in the method but I'm doing something wrong and it never works out. first q: 2x^6-x^4+x^3+6x^2-5x, x+2. Answer's supposed to be 154. I keep getting 138? I did do the step of having 0x^5, and adding the +0 at the end to make it complete but its still not working out. Another thing that gets me is division using fractions which the book fails to explain properly. second q: x^5+3x^3-4x^2+6x-8, 3-2x. Thank you.

Answer 1

I got a remainder of 138 on the first problem also... Did you copy the problem right?

For the second question, you need to change 3-2x into something in the form of x + ____ or x - ____.

3-2x = 0
-2x = -3
x = 3/2
x - 3/2 = 0

Now, do the same thing you've done for all the other problems.

Answer 2

I get 138 for the remainder as well.

In fact
2x^6-x^4+x^3+6x^2-5x

= (x+2)(2x^5-4x^4+7x^3-13x^2+32x-69) + 138.
The answer is not 154. You are right.

Are you only finding the remainder, or are you supposed to do the long division as well? You could use the first answerer's approach and do synthetic long division for x - 3/2:

......1.........0.........3..........-4..........6............-8
......0....... 3/2......9/4........63/8.....93/16....567/32
.......1.........3/2.....21/4...... 31/8....189/32...311/32

However this means that the given polynomial is equal to
(x-3/2)*(x^4 +(3/2)x^3 +(21/4)x^2 +(31/8)x + 189/16) + 311/32

If the divisor is actually 3-2x, we must multiply that first factor by -2 and divide the second by -2, giving

(3-2x)(-(1/2)x^4-(3/4)x^3-(21/8)x^2-(31/16)x-189/32) + 311/32.

As you can see, changing it so the coefficient of x is 1 doesn't affect the remainder, but it does affect the quotient.

Answer 3

In the first case you are right .The remainder is 138In the secon case you can write
2-3x =-3(x-2/3) and the polynom
So the division is between

-1/3*(x^5 +3x^3 -4x^2+6x-8) and (x-2/3)

You can divide the parentesis in the same way you did with
(x+2) in the former case using 2/3 and multiplying the remainder by (-1/3)