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\[h(x)=k(x+3)(x-2)(x-3)\]

3=k(2+3)(2-2)(2-3)

3=k

Looks alright so far.

is there more to or is it all ? :S

Er., wait, no

In your solving for k, you need to put in 0 for x if you're using the y-intercept.

would that make k zero ?

No, what you did made 3=0.

ohh i see

"3=k(2+3)(2-2)(2-3)"
3=k(5)(0)(-1) -> 3=0

i cancelled out k too :/

(2,3) is not a solution to the equation, but (0,3) is.

is dont see where you're getting (2,3) from

You plugged in x=2 and set y=3 in your attempt to solve for k.

So, I don't see where *you* got (2,3) from. ;-)

ohhhh ! i'm supposed to set x as 0 right ?

Yes, because you know the y-intercept is 3.

ohh okay i see it now :$

You could use any other known point, but that one is the most convenient.

so k would be 1/6

and h(x) = 1/6 (x+3)(x-2)(x-3)

never mind
h(x) = k(x+3)(x-2)(x-3)
h(0)=3
3=k(3)(-2)(-3)
3=18k

Yep, it's all good.

Thank-you once again :D

if you're not busy, can you help me with one last question

I think you know what you're doing, just be careful and don't rush through it.

Oh, sure. I think I can handle one more.

thanks :)

Find the equation of a cubic polynomial that gives a remainder of 3, when divided by (x+2)

how would i approach this problem ?

Do you remember the remainder theorem?

yes, the factor is equal to zero

Sort of..

but how would i use this info to make an equation, this is like working backwards from the norm :/

Yes, a lot of what you're doing here is using concepts to 'unsolve' equations.

If dividing by (x+2) gives a remainder of 3, then you know (-2,3) is a solution to the equation.

yes that's true

It wants a cubic function, so what does an equation for one of those look like?

like x^3-3x^2-4x-5

Sure, something like that, so to be more general, start with
\[y=ax^3+bx^2+cx+d\]

so something like \[y=x^3+x^2+x+d \] ?

I don't see why not.

hmm but "d" has to be a specific number right

Yep, that's what you're solving for.
Me, I went with y=ax^3, then solved for a.

(but I'm lazy like that)

ohh that way a would be -3/8 right

Uh huh.

\[:. y=\frac{ -3 }{ 8 }x^3\]

Looks beautiful.

thank you so much for your helppp, you're amazing :D

My pleasure.