Math Is Fun Forum

greengoop

Member

Registered: 2008-03-12

Posts: 1

PSF Method Factorisation

I'm having trouble with the PSF Method of factorisation so far I have figured out that it is as follows (using the example in my book):

Question 1 (My questions are in brackets next to the area they refer to)

Factorise using the PSF Method

4y² + 4y - 3

P = Product of first and last terms = -12y² (typo)

S = Sum (what does "Sum" refer to?) or middle term = 4y

F = Factors of P that give S = 6y, -2y

∴ 4y² + 4y - 3 = 4y² + 6y - 2y - 3

= 2y(2y + 3) - 1(2y + 3) [How did it go from this: "2y(2y + 3) - 1(2y + 3)" to this "(2y + 3)(2y - 1)"?????]

=(2y + 3)(2y - 1)

Question 2 I think this question is just IMPOSSIBLE

2a² + 11a² + 5

P = 10a²

S =  11a²

F = Factors of P that give S = x, y

x = ????

y = ????

Last edited by greengoop (2008-05-13 11:25:31)

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If you think dogs can't count, try putting three dog biscuits in your pocket and then giving Fido only two of them

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: PSF Method Factorisation

P = Product of first and last terms = -12y

I think you mean -12y².

S = Sum (what does "Sum" refer to?) or middle term = 4y

I'm not completely certain, but I think it's the value that your 2 factors have to add up to, or "sum" to.

= 2y(2y + 3) - 1(2y + 3) [How did it go from this: "2y(2y + 3) - 1(2y + 3)" to this "(2y + 3)(2y - 1)"?????]

They're using the distributive property.  It can be confusing at first, so use replacement variables to visualize what's going on.

Let's use u = 2y + 3.  The the left side of the equation becomes 2yu - 1u.  Using the distributive property you can see that this is equal to (2y - 1)u.  Now, substitute 2y + 3 back in for u and you get (2y - 1)(2y + 3).  The trick is to treat everything inside the parentheses as a single variable.

Question 2 I think this question is just IMPOSSIBLE

2a² + 11a² + 5

P = 10a²

S =  11a²

F = Factors of P that give S = x, y

x = ????

y = ????

There are 2 possible factor pairing that can give 10a²: 2a*5a and a*10a.  Only one of these adds up to S, or 11a², and that is a*10a.  Thus, we get an expression that looks like this:

2a² + 10a + a + 5 = 2a(a + 5) + 1(a  +5) = (2a + 1)(a + 5).

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Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: PSF Method Factorisation

Is this what you are looking for?

For you question 2, I have another method

2a² + 11a + 5

2a           1

a             5

1xa +2ax5= 11a
1x5=5

2a² + 11a + 5= (2a+1)(a+5)

Anyway, you have to find Aa² + Ba + C

E        G

F         H
Such that GF+EH=B,  GH=C , EF=A

Last edited by Dragonshade (2008-05-13 04:03:56)