1. ## hey guys

hey guys. New user here obviously. Hoping to bring my math grades up with help from you guys.

I was trying to ask a question before but I'm not really sure which section this belongs in.

I need to graph this equation

Y=x^3-9x^2+27x-27

I'm assuming I need to factor it first but I can't figure it out.

2. ## Re: hey guys

Originally Posted by ForeverConfused

I need to graph this equation

Y=x^3-9x^2+27x-27

I'm assuming I need to factor it first but I can't figure it out.
$x^3-9x^2+27x-27$

well we know that $(-3)^3-=-27$ so let's try $x-3$ as a factor

$\dfrac{x^3-9x^2+27x-27}{x-3}=x^2-6x+9$

and it's fairly easily recognized that

$x^2-6x+9=(x-3)^2$

so

$x^3-9x^2+27x-27=(x-3)^3$

So this is basically $x^3$ but shifted over 3 to the right. You should be able to draw a quick sketch of $x^3$.

Do this but center it at $x=3$

3. ## Re: hey guys

what do you mean by

" let's try x−3 as a factor"

sorry it's been about a year since I last did any maths

4. ## Re: hey guys

Originally Posted by ForeverConfused
what do you mean by

" let's try x−3 as a factor"

sorry it's been about a year since I last did any maths
I mean assume that $x^3-9x^2+27x-27=f(x) (x-3)$

where we don't know $f(x)$ yet. We can find it by dividing $x^3-9x^2+27x-27$ by $(x-3)$ and seeing if it comes out even.

If it does (it does) we know that we were right in guessing $(x-3)$ as a factor.

If you haven't had math in ages why do you have to graph this function?

5. ## Re: hey guys

Well it's summer break and my parents told me to take advanced functions during the break so it will be easier for me when school starts again

6. ## Re: hey guys

this is the way I did it and got the correct answer for this question and the next

oh and for the future what section should I post my problems in?

y=x^3 - 9x^2 - 27x - 27
=(x^3 - 27)+ (9x^2 - 27x)
=(x^3-27) -9x (x - 3)
=(x^3 - 3^3) -9x (x - 3)
=(x-3)(x^2 + 3x + 9) - 9x(x - 3)
=(x-3)(x^2 + 3x +9 - 9x)
=(x-3)(x^2 - 6x +9)
=(x-3)(x-3)(x-3)
=(x-3)^3

7. ## Re: hey guys

Looks good to me, good job!

Pre-Calc would be a likely forum.

-Dan