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.