# [Maths HELP!] Factorising...

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1. ### OP Agjsdfd My title Member Level 2

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Hey guys,
got my calculator paper maths on monday.. anywayz.

I swear I knew how to factorise. Have been doing circular geomatry, pythagoras theorm and quadratic equation the whole day.. and now I totally forgot how to factorise

2p² - 4pq

Some tips and examples would be helpful.
I dont mind hard examples aswell.
Gotta learn it....

2. ### OP Agjsdfd My title Member Level 2

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Seeing the examples in the book looks quite simple, and am expanding them, works fine when I expand.. but cant even solve "2p² - 4pq"..

3. ### NightKry GBAtemp Regular Member Level 2

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I THINK this is right...

2p² - 4pq

=(2p)(p-2q)

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do this
2p² - 4pq
p(2p-4q)
=(2p)(p-2q)

5. ### OP Agjsdfd My title Member Level 2

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BOAH!
It looks right.

6. ### Issac Iᔕᔕᗩᑕ Supervisor GBAtemp Patron Level 16

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wouldn't that be 2p * (p-2q) ?

I don't know what you're supposed to factorise?

EDIT:

oups.. i was slow ^^ haha.. oh well.

what you gotta do: look for what you can remove in BOTH terms!

in this case where you got "2pp - 4pq" you have p in both places => p* (2p - 4q) p*(2p - 2*2q) => p2 (p-2q) right? are you following?

it's like: (ABCD + ABDF + ADFE)
what is in all of the above?
A is => A(BCD + BDF + DFE)
D is => AD(BC + BF + FE)

Not the best example perhaps... but do you follow?

AX + BY + CZ ? Cant do shit, unless A B and C are numbers... for example: 3X + 24Y + 543Z... what can you do? well everything is dividable with 3! => 3* (X + 8 Y + 181 Z)....

I hope this clarifies it a bit!

7. ### NightKry GBAtemp Regular Member Level 2

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there are only two ways to do a factoring equation (i think)

one is follow this rule and do it backwards:

(a+b)(a+b)= a² +2ab+b²
That's FOIL...

and the second way (which was your question) was to take out the like terms..
2p² - 4pq
the like terms would be 2, and p in each "term" (term= a constant and veriable.. i think)
so if you / both terms by 2p, you'd end up wi (p-2q)...
then you turn it back into the original equation, you'd have to times it back in hence the (2p)(p-2q)

8. ### OP Agjsdfd My title Member Level 2

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GOT IT!
I made a mistake with the like term of 2p². As we have -4pq, so that would mean I have p-2q left. so 2p(p-2q)

Makes sense!
Thanks...

9. ### NightKry GBAtemp Regular Member Level 2

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You're welcome! What grade are you in btw?

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11th.

11. ### distorted.freque I'm getting mixed signals here! Member Level 2

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...right...I was going to add my own little rant but NightKry's explanation suffices. Also, as far as I know, the FOIL method is used for a different kind of factorization problem...>_>

12. ### NightKry GBAtemp Regular Member Level 2

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Oh okay. I was wondering why i couldn't remember anything like quadratic equations.. *Whew* I've yet to learn it.

13. ### Issac Iᔕᔕᗩᑕ Supervisor GBAtemp Patron Level 16

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of course people can have trouble with these equations! it's nothing strange at all!
I for one STARTED to learn math in the 7th grade, since my teachers sucked before that... it's a shame.. eitherway: I'm studying at a university now, ending the second year on thursday, and it's a technology programme = math, programming, and more.
Now it hasn't been an easy way to get here, but I'm struggling the best I can with math, and i fully understand if something isn't clear to some!

14. ### OP Agjsdfd My title Member Level 2

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Ok, got another question, which am not sure about.

*) AT is a tangent at T to a circle, centre O.

OT = x cm, AT = (x+5) cm , OA = (x +8) cm.

a) show that x² - 6x -39 = 0

b) Solve the quation x² - 6x -39 = 0 to find the radius of the circle.

I did solve this type of questions in the class, but we had Rectangler shapes or Quadratical shapes. This is an exam style question. My process is right, but still not sure what I am suppose to do by the way...

By the way, this question belongs to chapter "Quadratic Equations"

15. ### Issac Iᔕᔕᗩᑕ Supervisor GBAtemp Patron Level 16

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x = a/2+/- sqrt((a/2)^2 + (b/2)^2)
or SOMETHING like that... i don't know it by heart

That's the answer to the b) question at last... or you could use this:
(And that form requires that it is x² - 6x -39 = 0

16. ### OP Agjsdfd My title Member Level 2

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I knw all methods of solving quadratic equation.
But am not sure what the question is really asking me..

1st method is factorising.
2nd is Completing the square => x²+px+q = (x + half of P)² - (half of p)² + q
3rd is the big quadratic formula(which I dont remember by head, as we are given that formula in the exam)

17. ### xcalibur Gbatemp's Chocolate Bear Member Level 5

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X= (b +/- sqrt(b^2 - 4ac))/2a

18. ### Hyperlisk_ GBAtemp Regular Member Level 2

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Is that a right triangle? If not, I can't think of how you would solve it...

19. ### OP Agjsdfd My title Member Level 2

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@Xcalibur
I knw, thats the big quadratic thingy, which I dont need to know at the moment.

@Hyperlisk_
The shape is not drawn accurately.

I defenitely need to make sure, that I understand how to solve this type of question. As their werent any of these questions in the non-calculator paper.

20. ### Issac Iᔕᔕᗩᑕ Supervisor GBAtemp Patron Level 16

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All it want's is for you to get x... it's actually quite easy...
now for the a)

pythagoras theorem or whatever the name is in english= X^2 + Y^2 = Z^2
where X= OT, Y = AT. Z = OA.

this is: x² + (x² + 10x + 25) = (x² + 16x + 64)
now move the right side over to the left by subtraction:
x² + (x² + 10x + 25) - (x² + 16x + 64) = 0
and solve it => x² - 6x - 39 = 0

b)

using the formula above:
(x - 3)² - (-3² + 39) = 0

(x - 3)² = -3² + 39
(sqrt() = square root of...)
x - 3 = sqrt(-3² + 39)

x - 3 = (+ or -) sqrt(9 + 39)

x = (+ or -)sqrt( 30 ) +3

and this gives the answer X = sqrt( 48 ) +3 (+ and not - since a radius can't be negative..)
(And maybe you can round sqrt(48) to 7.... since sqrt (49) is 7... i dont know... and that'd give the radius x = 10....

though it seems weird.. someone has to confirm this

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