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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: hard on February 23, 2009, 09:32:15 pm

Title: help
Post by: hard on February 23, 2009, 09:32:15 pm: Amm chapter 1 revision has this q:

sketch the graph of y = x^4 - 4x^3 + 3x^2 + 4x - 4

can someone help please.
Title: Re: help
Post by: Flaming_Arrow on February 23, 2009, 09:35:32 pm: to work out the intercepts, use trial and error to work out a factor then use long division to get the rest then you should be able to sketch it from there
Title: Re: help
Post by: hard on February 23, 2009, 09:42:26 pm: yer i worked out the long division and i got $(x-1)(x^3 - 3x^2+4)$

but from here i'm stuck. Stuck as in not sure how to factories the second bracket with calculator or quickly by hand.
Title: Re: help
Post by: /0 on February 23, 2009, 09:46:18 pm: $(-1)^3-3(-1)^2+4=0$ , so $(x+1)$ is another factor.
Title: Re: help
Post by: wombifat on February 23, 2009, 09:47:04 pm: it'd probably be faster to do it the long way by hand than to ask a question about it here
Title: Re: help
Post by: GerrySly on February 23, 2009, 09:49:27 pm: Quote from: hard on February 23, 2009, 09:42:26 pm
yer i worked out the long division and i got $(x-1)(x^3 - 3x^2+4)$

but from here i'm stuck. Stuck as in not sure how to factories the second bracket with calculator or quickly by hand.
Just use the remainder / factor thereom on the second bracket, -1 seems to work as a factor, so then you just have to divide $x^3 - 3x^2 + 4$ by $x+1$ , I seemed to get $(x-1)(x+1)(x^2 - 4x + 4)$ , which is easily factorised to $(x-1)(x+1)(x-2)^2$

There may be some errors there lol just throwing that out there
Title: Re: help
Post by: hard on February 24, 2009, 12:32:53 am: ah ye thanks guys
Title: Re: help
Post by: NE2000 on February 24, 2009, 04:11:14 pm: Quote from: wombifat on February 23, 2009, 09:47:04 pm
it'd probably be faster to do it the long way by hand than to ask a question about it here

agree lol
Title: Re: help
Post by: hard on February 24, 2009, 07:01:21 pm: can someone also help mem with this q:

$(10a^-^2b^-^3)^-2/5ab^3 * (2a^2b)^-^3$
Title: Re: help
Post by: Flaming_Arrow on February 24, 2009, 07:22:35 pm: $\frac{(10a^-^2b^-^3)^{-2}}{5ab^3 * (2a^2b)^-^3}$

$\frac{10^{-2}a^4b^6}{(5ab^3)*2^{-3}a^{-6}b^{-3}}$

make it positive

$\frac{2^3a^4b^6*a^6b^3}{(10^2*5ab^3)}$

$\frac{8a^{10}b^9}{500ab^3}$

$\frac{2a^{9}b^6}{125}$