# Factor and remainder theorem problem

• chwala
In summary: And the problem is not to find ##f(x)##, but to find ##k##.No, even worse. Expand, and calculate ##(x^3-x^2(2k+2)+2(k^2+3k+1)-k^2-k ):(x-2)##. This happens if you apply the methods without thinking about the problem. And the problem is not to find ##f(x)##, but to find ##k##.No, even worse. Expand, and calculate ##(x^3-x^2(2k+2)+2(k^2+3k+1)-k^2-k ):(x-2)##. This happens if you apply the methods
chwala
Gold Member
Homework Statement
if## f(x)= x^3+ax^2+bx+c## and the roots of f(x) are ##1, k, k+1##

when ##f(x)## is divided by ##x-2## the remainder is ##20##.

show that ##k^2-3k-18=0##
Relevant Equations
factor/remainder theorem
##0=1+a+b+c##
##20=8+4a+2b+c##
it follows that,
##13=3a+b##
and,
##0=k^3+ak^2+bk+c##...1
##0=(k+1)^3+a(k+1)^2+(k+1)b+c##...2
subtracting 1 and 2,
##3k^2+k(3+2a)+14-2a=0##

chwala said:
Homework Statement:: if## f(x)= x^3+ax^2+bx+c## and the roots of f(x) are ##1, k, k+1##

when ##f(x)## is divided by ##x-2## the remainder is ##20##.

show that ##k^2-3k-18=0##
Homework Equations:: factor/remainder theorem

##0=1+a+b+c##
##20=8+4a+2b+c##
it follows that,
##13=3a+b##
and,
##0=k^3+ak^2+bk+c##...1
##0=(k+1)^3+a(k+1)^2+(k+1)b+c##...2
subtracting 1 and 2,
##3k^2+k(3+2a)+14-2a=0##
Firstly: Are you sure it isn't ##3k^2-k-18=0\;?##

Secondly: What have you calculated there?

Thirdly: What do you know, if a polynomial ##p(x)## has a zero (root) ##x=a\;?##

@chwala: You know the 3 roots and the leading coefficient is ##1##. So you should immediately be able to write the polynomial in factored form in terms of ##k##. Then set ##f(2)=20##. There's hardly anything to do.

chwala
fresh_42 said:
Firstly: Are you sure it isn't ##3k^2-k-18=0\;?##

Secondly: What have you calculated there?

Thirdly: What do you know, if a polynomial ##p(x)## has a zero (root) ##x=a\;?##

it is ##k^2-3k-18##
what i am trying to do there is solving the three factor equations and the remainder equation (simultaneous equations) in terms of ##k##

chwala said:
it is ##k^2-3k-18##
what i am trying to do there is solving the three factor equations and the remainder equation (simultaneous equations) in terms of ##k##
Ok, then I made a mistake somewhere, since I got ##3k^2-k-18=0##. But you haven't answered the important question: What does it mean if a polynomial ##p(x)## has a zero at ##x=a##? How many zeros does a polynomial of degree ##3## have at most?

in response to post ##3##,
i am getting,
##f(x)=x^3-2x^2(k-1)+x(k^2+3k+1)-k^2-k##

chwala said:
in response to post ##3##,
i am getting,
##f(x)=x^3-2x^2(k-1)+x(k^2+3k+1)-k^2-k##
I have a different polynomial. How did you get there? Probably a typo in the ##x^2## term.

this is question is a bit 'silly' just taken my time for no reason
The factors are## (x-1),(x-k)## and ##(x-k-1)##.
Therefore,
##(x-1)(x-k)(x-k-1)=0##
##x^3-x^2(2k+2)+2(k^2+3k+1)-k^2-k=0##
##f(x)=x^3-x^2(2k+2)+2(k^2+3k+1)-k^2-k##
using ##f(2)=20##
we have ## 20=8-8k-8+2k^2+6k+2-k^2-k##
giving as the required ## k^2-3k-18=0##
bingo Africa

fresh_42
LCKurtz said:
@chwala: You know the 3 roots and the leading coefficient is ##1##. So you should immediately be able to write the polynomial in factored form in terms of ##k##. Then set ##f(2)=20##. There's hardly anything to do.
Thanks, good brains there

chwala said:
this is question is a bit 'silly' just taken my time for no reason
The factors are## (x-1),(x-k)## and ##(x-k-1)##.
Therefore,
##(x-1)(x-k)(x-k-1)=0##

You mean ##p(x) = (x-1)(x-k)(x-k-1)##. And you can put ##x=2## in that and set it equal to ##20## without doing all that work below to multiply it out.

##x^3-x^2(2k+2)+2(k^2+3k+1)-k^2-k=0##
##f(x)=x^3-x^2(2k+2)+2(k^2+3k+1)-k^2-k##
using ##f(2)=20##
we have ## 20=8-8k-8+2k^2+6k+2-k^2-k##
giving as the required ## k^2-3k-18=0##
bingo Africa

chwala
LCKurtz said:
You mean ##p(x) = (x-1)(x-k)(x-k-1)##. And you can put ##x=2## in that and set it equal to ##20## without doing all that work below to multiply it out.
Don't complain, I even did the long division ...

chwala
LCKurtz said:
You mean ##p(x) = (x-1)(x-k)(x-k-1)##. And you can put ##x=2##in that and set it equal to ##20## without doing all that work below to multiply it out.
Yes, expand then substitute...

chwala said:
Yes, expand then substitute...
No, even worse. Expand, and calculate ##(x^3-x^2(2k+2)+2(k^2+3k+1)-k^2-k ):(x-2)##. This happens if you apply the methods without thinking about the problem.

chwala

## 1. What is the difference between the factor theorem and the remainder theorem?

The factor theorem states that if a polynomial P(x) has a root r, then (x-r) is a factor of P(x). The remainder theorem, on the other hand, states that the remainder when a polynomial P(x) is divided by (x-r) is equal to P(r). In other words, the remainder theorem is a consequence of the factor theorem.

## 2. How can I use the factor theorem to find the roots of a polynomial?

To use the factor theorem to find the roots of a polynomial, you can first factor the polynomial using techniques such as grouping or synthetic division. Then, set each factor equal to 0 and solve for the values of x that make each factor equal to 0. These values will be the roots of the polynomial.

## 3. Can the factor theorem be applied to all polynomials?

Yes, the factor theorem can be applied to all polynomials, as long as they are in standard form with integer coefficients. However, it may not always be easy to find the roots of a polynomial, especially if the polynomial has a high degree or complex roots.

## 4. How can I use the remainder theorem to evaluate a polynomial?

The remainder theorem can be used to evaluate a polynomial by dividing the polynomial by (x-r), where r is the value at which you want to evaluate the polynomial. The remainder of this division will be the value of P(r), which is the evaluation of the polynomial at x=r.

## 5. Are there any real-life applications of the factor and remainder theorem?

Yes, the factor and remainder theorem have various real-life applications in fields such as engineering, economics, and physics. For example, in engineering, these theorems are used to analyze electrical circuits and control systems. In economics, they are used to model supply and demand curves. In physics, they are used to solve problems related to motion and forces.

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