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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: squance on January 27, 2008, 10:56:35 am

Title: Polynomial HElp!
Post by: squance on January 27, 2008, 10:56:35 am: I can't seem to remember how to factorise some of theese so some help would be great!

1. (x+1)^3 - (2x-3)^3

2. x^3/8 - y^3

3. 64/a^3 - 125/b^3

4. 27 - (5-x)^3

5. 1 - 10a +25a^2

6. 3u^2 + 30uv + 75v^2

7. x^3 - x^2 -x +1

8. a^2 +2ab +b^2 -a
Title: Re: Polynomial HElp!
Post by: Collin Li on January 27, 2008, 11:09:05 am: Oh this will be fun :D

1) Using the "difference of perfect cubes" formula: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
$(x+1)^3 - (2x-3)^3 = \left[(x+1) - (2x-3)\right]\left[(x+1)^2 + (x+1)(2x-3) + (2x-3)^2]$

Expanding out all the terms, then collecting like terms:
$= (4-x)(x^2 + 2x + 1 + 2x^2 - x - 3 + 4x^2 - 12x + 9) = (4-x)(7x^2 - 11x + 7)$

Since this is methods, this is as far as you can factorise it. We can check this by looking at the discriminant of the second factor (the quadratic one):
$\Delta = (-11)^2 - 4\cdot 7\cdot 7 = 121 - 196 = -75 < 0$
Hence there are no solutions, and therefore no factors.

I would love to do the rest, but I'll leave it to someone else, because I've got to go and tutor right now. The first few more or less involve the same formula.
Title: Re: Polynomial HElp!
Post by: ice_blockie on January 27, 2008, 12:26:08 pm: I think coblin meant this formula $a^3-b^3 =(a-b)(a^2+ab+b^2)$ :) but all the rest is right
Title: Re: Polynomial HElp!
Post by: AppleXY on January 27, 2008, 12:32:01 pm: Yeah. He better correct it now or else I will :P But Collin's teaching is excellent. No doubt about that.
Title: Re: Polynomial HElp!
Post by: ice_blockie on January 27, 2008, 01:31:21 pm: The following two equations can be factorised using the grouping method.

$1 - 10a +25a^2$

$3u^2 + 30uv + 75v^2$

Question 5.
Step 1: Rewrite in order of powers. (not really necessary) $25a^2-10a+1$
Step 2: Evaluate $a \times c = 25 \times 1 = 25$
Step 3: Find the factors of $25$ which add up to $-10$ (middle term) i.e. $-5$ and $-5$ (Normally the factors will be half of the number, but if not use trial and error)
Step 4: Rewrite the original equation using the grouping approach: $25a^2-10a+1=25a^2-5a-5a+1=5a(5a-1)-(5a-1)=(5a-1)(5a-1)=(5a-1)^2$

NOTE: This is a particularly LONG method of factorizing. But Step 1, 2 and 3 can be done very quickly if you practice and can recognise the numbers.

The same method can be applied to question 6.

Question 6
Take out a common factor 3 $3u^2 + 30uv + 75v^2 = 3(u^2+10uv+25v^2)$

And then you have an equation that resembles question 5!!!!

$3(u^2+10uv+25v^2)=3(u^2+5uv+5uv+25v^2 )=3(u(u+5v)+5v(u+5v)=3(u+5v)(u+5v)=3(u+5v)^2$

EDIT: this is quite hard to explain (grouping method) can someone else explain it clearer?
Title: Re: Polynomial HElp!
Post by: ice_blockie on January 27, 2008, 01:53:07 pm: Question 7
$x^3 - x^2 -x +1$

Grouping Method again!!!
We group the terms together:(You sometimes need to rearrange terms, but not this one :))
$(x^3 - x^2)+(-x +1)=x^2(x-1)-1(x-1)=(x^2-1)(x-1)$

Expanding the Difference Of Perfect Squares (formula $a^2-b^2=(a-b)(a+b)$ )

$(x^2-1) = (x-1)(x+1)$ Therefore $(x^2-1)(x-1)=(x-1)(x+1)(x-1)=(x+1)(x-1)^2$

Question 8
$a^2 + 2ab + b^2 - a$

The first part of this equation resembles $(a+b)^2=a^2+2ab+b^2$

So we factorise that part:
$[a^2 +2ab +b^2] - a = (a+b)^2-a$

We will use the “Difference of Perfect Squares” formula. (using the fact $(\sqrt{a})^2 =a$ )
$(a+b)^2-(\sqrt{a})^2 = (a+b-\sqrt {a})(a-b+\sqrt {a})$
Title: Re: Polynomial HElp!
Post by: Mao on January 27, 2008, 02:06:58 pm: The grouping method is really just the reverse of FOIL (first, outer, inner, last) expansion

$(ax+b)(cx+d)=(ac)x^2+(ad+bc)x+bd$

as for the grouping method, when we multiply the first coefficient with the constant... we get $ac \times bd = abcd$
and then the two numbers we're looking for needs to multiply to the product of the first coefficient and the constant (i.e. $abcd$ ), and need to add to the second coefficient ( $ad+bc$ ), hence by trail and error, we arrive at the two numbers, namely $ad$ and $bc$

reverse foil is as follows:

$(ac)x^2+(ad+bc)x+bd$

$= (ac)x^2 + (ad)x + bc(x) + bd$

$= ax(cx+d) + b(cx+d)$

$= (ax+b)(cx+d)$

as can be seen, for this method to work the second coefficient need to be broken down correctly for reverse FOIL to happen, hence why we first need to find these two numbers.

this is what ice_blockie referred to as the "grouping method"

it may seem complex here, but practical application of this method usually involve equations a lot simpler:

$(x+a)(x+b) = x^2 + (a+b)x + ab$
so as soon as you find out a and b, you can factorise without having to reverse FOIL

$(x+a)^2 = x^2 + (2a)x + a^2$
you can get "a" by just looking at it

$(ax+b)^2 = a^2 x^2 + (2ab)x + b^2$
also fairly simple mental arithmetics

other methods that are also useful:
Complete the square (ice blockie question 8)
Remainder theorem (havent seen around)
Synthetic method of polynomial factorisation (totally forgot how that works *looks up in wikipedia*)

so yeah.... :D
Title: Re: Polynomial HElp!
Post by: Ahmad on January 27, 2008, 02:17:17 pm: Here's an exercise. Using some of these ideas, find x and y:

$4xy + 5y^2 + x^2 + 1 + 2y = 0$
Title: Re: Polynomial HElp!
Post by: ice_blockie on January 27, 2008, 02:38:39 pm: $5y^2+4xy+x^2+1+2y=0$
$4y^2+y^2+4xy+x^2+1+2y=0$
$(2y+x)^2+(y+1)^2=0$

Let $a = (2y+x)^2$
$b = (y+1)^2$

Since a and b are squares, $a \geqslant 0$ and $b \geqslant 0.$

Only $0 + 0 = 0$ fits this condition so therefore

$y = -1$ and $x = 2$
Title: Re: Polynomial HElp!
Post by: Ahmad on January 27, 2008, 02:40:17 pm: Right. :)
Title: Re: Polynomial HElp!
Post by: Collin Li on January 27, 2008, 02:40:33 pm: Quote from: ice_blockie on January 27, 2008, 12:26:08 pm
I think coblin meant this formula $a^3-b^3 =(a-b)(a^2+ab+b^2)$ :) but all the rest is right

Thank you, I have corrected this now.
Title: Re: Polynomial HElp!
Post by: Mao on January 27, 2008, 02:40:58 pm: Quote from: Ahmad on January 27, 2008, 02:40:17 pm
Right. :)
ur such a cool mod ahmad :P ;D
Title: Re: Polynomial HElp!
Post by: Ahmad on January 27, 2008, 02:41:59 pm: Why thank you, lol. :D
Title: Re: Polynomial HElp!
Post by: dcc on January 27, 2008, 04:07:47 pm: grouping method? that seems like an overly long way of writing out what can be recognized as a perfect way :|