Find the remainder of a polynomial

[discombobulated]

7a) f(x0 = 2x³+3²-6x+1
Find the remainder when F(x) is divided by (2x-1)
b) (i) Find the remainder when f(x) is divided by (x+2)
(ii) Hence, or otherwise solve the equation: 2x³+3²-6x-8=0
giving your answer to two decimal places.

This is what I've done:
a) r= -1 (i did f(1/2) to get that)
b) (1) r = 9 ( i did f(-2) for that)
could someone please help me with part (ii)?

 

[NateTG]

I don't think you understand what the problem is asking for.

For example
$$x^2-4$$
divided by
$$x+2$$
is
$$x-2$$
because
[tex](x-2)(x+2)=x^2-4[/itex]

You should be doing the division with x's.

 

[HallsofIvy]

No, NateTG, what he did was perfectly feasible. To find the remainder, r, when polynomial P(x) is divided by (x-a) you CAN do the division, but it is simpler to use the fact that P(x)= (x-a)Q(x)+ r so that P(a)= r. Of course, dividing by x-1/2 will give the same remainder as dividing by 2x-1.

discombobulated, you twice wrote "2x³+3²-6x+1".
Didn't you really mean "2x³+ 3x²- 6x+ 1"?

For the third part, I'm not sure about the "Hence" but look at the coefficients. Those tell you that there is one very simple answer!

 

[discombobulated]

No, NateTG, what he did was perfectly feasible. To find the remainder, r, when polynomial P(x) is divided by (x-a) you CAN do the division, but it is simpler to use the fact that P(x)= (x-a)Q(x)+ r so that P(a)= r. Of course, dividing by x-1/2 will give the same remainder as dividing by 2x-1.
discombobulated, you twice wrote "2x³+3²-6x+1".
Didn't you really mean "2x³+ 3x²- 6x+ 1"?
For the third part, I'm not sure about the "Hence" but look at the coefficients. Those tell you that there is one very simple answer!

yes i know part a) and bi) are correct because i checked it by dividing too, but i find the other way easier. HallsofIvy, yes i did mean 2x³+ 3x²- 6x+ 1. (i should also point out that I'm not a, 'he'!) I don't understand what you're trying to say about the coefficients sorry.
I noticed that the remainder (9) has been taken away so the factors of 2x³+ 3x²- 6x-8 are (x+2) and (2x² - x-4)
so i tried this:
(x+2)((2x² - x-4)=0
(=-2
(2x² - x-4=0
then by completing the square i got:
x= 1.69 or x= -1.19
does this make sense?

 

[HallsofIvy]

Boy, did I mess up! I didn't notice that the equation you wanted to solve in (ii) was not the orginal equation! You are completely correct. Because you already new that 9 was the remainder when
2x³+ 3x²- 6x+1 was divided by (x+ 2), you knew that 2x³+ 3x²- 6x+1= (x+2)(something)+ 9.
Subtracting 9 from both sides tells you that
2x³+ 3x²- 6x- 8= (x+2)(something).
And that means that x= -2 is solution. Yes, that "something" is 2x<sup>2 - x+ 4. Solving that by completing the square tells you that
x= 1.69 and x= -1.19.

My "hint" was for solving the wrong equation! Notice that the coefficients in the original equation are 2, 3, -6, 1 and they add up to 0. That tells you that x= 1 is a solution. Of course, you could factor out x-1 to reduce to a quadratic and solve that, either by completing the square or using the quadratic formula.</sup>

 

[discombobulated]

thanks for your help anyway!