Find the remainder of a polynomial

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Homework Help Overview

The discussion revolves around finding the remainder of a polynomial when divided by linear factors, specifically focusing on the polynomial f(x) = 2x^3 + 3x^2 - 6x + 1 and its division by (2x - 1) and (x + 2). Participants are exploring methods to solve for the roots of the polynomial equation 2x^3 + 3x^2 - 6x - 8 = 0.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using polynomial division and the Remainder Theorem to find remainders. Some express confusion about the original problem's requirements and the notation used. Others clarify the relationship between polynomial coefficients and potential roots.

Discussion Status

There is a mix of agreement and confusion regarding the methods used to find the remainders and solve the polynomial equation. Some participants have confirmed their calculations, while others are questioning the implications of the coefficients and the correctness of the original equation.

Contextual Notes

One participant notes a potential misunderstanding regarding the polynomial's coefficients and their significance in determining roots. There is also mention of a mistake in identifying the equation to be solved in part (ii), which has led to further clarification and exploration of the problem.

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

For example
[tex]x^2-4[/tex]
divided by
[tex]x+2[/tex]
is
[tex]x-2[/tex]
because
[tex](x-2)(x+2)=x^2-4[/itex]<br /> <br /> You should be doing the division with x's.[/tex]
 
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!
 
HallsofIvy said:
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?
 
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 2x2 - 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.
 
thanks for your help anyway!
 

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