Factorising...

[Hooligan]

I've been trying to factorise this question, which works one way, but not another, and not too sure why... anyone know?

Question: Factorise

Method 1: Use Cross-Multiplication
2x   5
2x   -1
; correct, and works.

Method 2: Use Quadratic Formula
Answers from using Quadratic Formula: or
, however, when we expand this, it .
If we inspect the expanded version of , we can see it is off by a factor of 4, so the answer should be to make it
Why does the step before we inspect, it does not equal the original equation, when we just solved the solutions to the equation?

Method 3: An extension of using Quadratic Formula
After obtaining answers from using Quadratic Formula;












Method 2 is the one that has been driving me nuts for hours... why doesn't it work without the inspection? (I know that I'll be off if I don't 'inspect' but why when using the quadratic formula it doesn't just spit out the full answer, but produces an answer which is off by a certain factor?) 

[TrueTears]

By the fundamental theorem of algebra:

for some constant ''.

Then you just have to find .

Since the LHS and RHS are identical you can sub in or any value of to work out .

when

[Hooligan]

By the fundamental theorem of algebra.

for some constant ''.

Then you just have to find a.

Since the LHS and RHS are identical you can sub in x = 0 or any value of x to work out a.

Oh really? I have never heard of it... so when we factorise using the Quadratic Formula, we should never forget the 'factor'? Interesting...

Thanks TT!!

[TrueTears]

Never forget the constant!!

Remember the fundamental theorem of algebra states that every polynomial in has at least one complex zero which means every polynomial can be factored like:

Which means the constant is simply just the coefficient of the leading degree.

[zzdfa]

"Why does the step before we inspect, it does not equal the original equation, when we just solved the solutions to the equation?"

because there are infinitely many quadratic equations with those 2 roots. so you cannot determine the quadratic just from the roots. the quadratic formula only gives you the roots, hence not enough information.

[Hooligan]

Thanks guys...

One further question; so if I want to have the factorised form, in the form: (ax+b)(cx+d), then the only way to do it is to use the cross multiplication, method?

[Hooligan]

Where did kamil's post, go?

[TrueTears]

Thanks guys...

One further question; so if I want to have the factorised form, in the form: (ax+b)(cx+d), then the only way to do it is to use the cross multiplication, method?

Any form you have written above is correct (with the constant) ie, is just as factorised as

However it may vary depending on the context of the question, if the question asked you to have only integer values in your factors then would be the way to go and etc.

[zzdfa]

 you can just see that 4=2^2 and then write

4(x+5/2)(x-1/2)=2(x+5/2)*2(x-1/2)=(2x+5)(2x-1)
edit TT beat me

[Hooligan]

Thanks guys...

One further question; so if I want to have the factorised form, in the form: (ax+b)(cx+d), then the only way to do it is to use the cross multiplication, method?

Any form you have written above is correct (with the constant) ie, is just as factorised as

However it may vary depending on the context of the question, if the question asked you to have only integer values in your factors then would be the way to do and etc.

Oh, I see... so technically the book should list both of these answers, but because the one without the constant at the front 'looks neater' in that it has no yucky fractions and the constant, they priotise the neater one as the correct answer... INTERESTING...

[TrueTears]

lol they're exactly the same thing.

Taking a factor of a out of each bracket leaves so I guess it's implied both are acceptable

[Hooligan]

lol they're exactly the same thing.

Taking a factor of a out of each bracket leaves so I guess it's implied both are acceptable

Oh yeah! How cool! ^_^

[Ilovemathsmeth]

Method 2 gives the same result as method 1, however it has taken out the common factor of 2 to simplify the expression further. Both would be accepted in an exam. =)

[Hooligan]

Cheers guys! I understand it a whole lot better! xx