1. ## remainder theorem

what do i do for this

when f(x)= x^3+ax^2+bx-15

is divided by (x-1)
when divided by (x-2) there is remainder 35

find expressions for f(1) and f(2)

THEN

set up two simutltaneoues equations and solve to find a and b

then factorise f(x) completely

what i do???????????????

2. ## Re: remainder theorem

when f(x) is divided by (x-1), then what?

3. ## Re: remainder theorem

when f(x) is divided by (x-1), then what? there is no remainder

4. ## Re: remainder theorem

Originally Posted by Siron
when f(x) is divided by (x-1), then what?
there is no remainder

5. ## Re: remainder theorem

Originally Posted by kitobeirens
there is no remainder
If there is no remainder when f(x) is divided by (x-1) then you know that (x-1) is a factor of f(x) and that f(1)=0

The remainder theorem says that f(2) = 35 (see spoiler for more info on the remainder theorem)

Spoiler:
In more detail the remainder theorem says that $\displaystyle f(x) = q(x)g(x) + r(x)$ where r(x) is of a lower degree than g(x). g(x) itself is the divisor: (x-2) in your case, q(x) the quotient and r(x) the remainder. Subbing this into what you're told: $\displaystyle f(x) = q(x)(x-2) + r(x)$ and since r(x) is of lower degree than g(x) then r(x) is of degree 0 which is just a constant: $\displaystyle r(x) = r$.

At the point 2 (where x-2 =0): $\displaystyle f(2) = r$ which in your case is 35

6. ## Re: remainder theorem

Originally Posted by e^(i*pi)
If there is no remainder when f(x) is divided by (x-1) then you know that (x-1) is a factor of f(x) and that f(1)=0

The remainder theorem says that f(2) = 35 (see spoiler for more info on the remainder theorem)

Spoiler:
In more detail the remainder theorem says that $\displaystyle f(x) = q(x)g(x) + r(x)$ where r(x) is of a lower degree than g(x). g(x) itself is the divisor: (x-2) in your case, q(x) the quotient and r(x) the remainder. Subbing this into what you're told: $\displaystyle f(x) = q(x)(x-2) + r(x)$ and since r(x) is of lower degree than g(x) then r(x) is of degree 0 which is just a constant: $\displaystyle r(x) = r$.

At the point 2 (where x-2 =0): $\displaystyle f(2) = r$ which in your case is 35

so what is a and b ??

7. ## Re: remainder theorem

No, that isn't the way it works. $\displaystyle e^{i\pi}$ has given you the theory in order to enable you to deduce a and b.

$\displaystyle f(2)=35$ and
$\displaystyle f(1)=0$

Now it's your turn to do something.