Factor and remainder theorem problem

[chwala]

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##

 

[fresh_42]

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\;?##

 

[LCKurtz]

@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]

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##

 

[fresh_42]

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?

 

[chwala]

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

 

[fresh_42]

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.

 

[chwala]

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

 

[chwala]

@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

 

[LCKurtz]

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

 

[fresh_42]

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]

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...

 

[fresh_42]

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.