# Factorise Equation

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• October 2nd 2011, 11:09 PM
gilagila
Factorise Equation
$
f(x) = 2(x-3)(x+2)^2 + 2(x+2)(x-3)^3
$

Factorise f(x) completely.

Mind to tell me how I going to start to solve this ?
• October 2nd 2011, 11:10 PM
Prove It
Re: Factorise Equation
Quote:

Originally Posted by gilagila
$
f(x) = 2(x-3)(x+2)^2 + 2(x+2)(x-3)^3
$

Factorise f(x) completely.

Mind to tell me how I going to start to solve this ?

Start by taking out the common factors 2, (x - 3) and (x + 2)...
• October 3rd 2011, 01:04 AM
gilagila
Re: Factorise Equation
Quote:

Originally Posted by Prove It
Start by taking out the common factors 2, (x - 3) and (x + 2)...

is it look like this ?

f(x)=2(x-3)(x^2+4x+4) + 2(x+2)(x-3)(x^2-6x+9)

=2((x^3+4x^2+4x)+(-3x^2-12x-12))+2((x^2-3x+2x-6)(x^2-6x+9))
• October 3rd 2011, 01:54 AM
mr fantastic
Re: Factorise Equation
Quote:

Originally Posted by gilagila
is it look like this ?

f(x)=2(x-3)(x^2+4x+4) + 2(x+2)(x-3)(x^2-6x+9)

=2((x^3+4x^2+4x)+(-3x^2-12x-12))+2((x^2-3x+2x-6)(x^2-6x+9))

No.

Can you factorise $2ab^2 + 2ba^3$ ? If so, please do so, showing all steps.
• October 3rd 2011, 02:01 AM
gilagila
Re: Factorise Equation
Quote:

Originally Posted by mr fantastic
No.

Can you factorise $2ab^2 + 2ba^3$ ? If so, please do so, showing all steps.

Sorry sir, not understand what you mean, mind to explain ? TQ

2ab(b+a^2) or just stop like that only ?
• October 3rd 2011, 03:56 AM
mr fantastic
Re: Factorise Equation
Quote:

Originally Posted by gilagila
Sorry sir, not understand what you mean, mind to explain ? TQ

2ab(b+a^2) or just stop like that only ?

2ab(b+a^2) is correct.

Now note that a = x - 3 and b = x + 2 in the question that you posted ....
• October 3rd 2011, 04:46 AM
gilagila
Re: Factorise Equation
Quote:

Originally Posted by mr fantastic
2ab(b+a^2) is correct.

Now note that a = x - 3 and b = x + 2 in the question that you posted ....

ooh...ic

meaning that

f(x)=2(x-3)(x+2)^2 + 2(x+2)(x-3)^3

=2ab^2 + 2ba^3 = 2ab(b+a^2)

=2(x-3)(x+2)[(x+2)+(x-3)^3]

Factorise f(x) completely
=2(x-3)(x+2)[(x+2)+(x-3)^3]

or any other format ?
• October 3rd 2011, 05:05 AM
CaptainBlack
Re: Factorise Equation
Quote:

Originally Posted by gilagila
ooh...ic

meaning that

f(x)=2(x-3)(x+2)^2 + 2(x+2)(x-3)^3

=2ab^2 + 2ba^3 = 2ab(b+a^2)

=2(x-3)(x+2)[(x+2)+(x-3)^3]

Factorise f(x) completely
=2(x-3)(x+2)[(x+2)+(x-3)^3]

or any other format ?

No, you still have (x-3)^3 inside the right most bracket, it should be squared not cubed. When you have corrected that expand the right most bracket and verify that it can(not) be further factorised.

CB
• October 3rd 2011, 06:45 AM
gilagila
Re: Factorise Equation
Quote:

Originally Posted by CaptainBlack
No, you still have (x-3)^3 inside the right most bracket, it should be squared not cubed. When you have corrected that expand the right most bracket and verify that it can(not) be further factorised.

CB

mind to show me how to do it, not realy sure how to do
• October 3rd 2011, 12:51 PM
mr fantastic
Re: Factorise Equation
Quote:

Originally Posted by gilagila
mind to show me how to do it, not realy sure how to do

Oh come on! You were shown exactly how to do the question. You made a careless mistake and that was also pointed out to you. Fixing the error is simple - you were told exactly what to do. Sorry, but you have been given more than enough help here to get the final answer.

• October 3rd 2011, 04:09 PM
gilagila
Re: Factorise Equation
f(x)=2(x-3)(x+2)^2 + 2(x+2)(x-3)^3

=2ab^2 + 2ba^3 = 2ab(b+a^2)

=2(x-3)(x+2)[(x+2)+(x-3)^2]

Factorise f(x) completely
=2(x-3)(x+2)[(x+2)+(x-3)^2]

oops, sorry for the reply. Not noticed that I had changed the squared to cube, now I rectified.

Is it correct now ?
• October 3rd 2011, 04:33 PM
Wilmer
Re: Factorise Equation
Quote:

Originally Posted by gilagila
...Is it correct now ?

Yes, oui, si, ...
• October 3rd 2011, 04:41 PM
Prove It
Re: Factorise Equation
Quote:

Originally Posted by gilagila
f(x)=2(x-3)(x+2)^2 + 2(x+2)(x-3)^3

=2ab^2 + 2ba^3 = 2ab(b+a^2)

=2(x-3)(x+2)[(x+2)+(x-3)^2]

Factorise f(x) completely
=2(x-3)(x+2)[(x+2)+(x-3)^2]

oops, sorry for the reply. Not noticed that I had changed the squared to cube, now I rectified.

Is it correct now ?

It's correct, but it's not complete. You now need to expand everything contained in the square brackets, collect like terms, and see if that factorises as well...
• October 3rd 2011, 06:35 PM
gilagila
Re: Factorise Equation
Quote:

Originally Posted by Prove It
It's correct, but it's not complete. You now need to expand everything contained in the square brackets, collect like terms, and see if that factorises as well...

Factorise f(x) completely
=2(x-3)(x+2)[(x+2)+(x-3)^2]

= 2(x-3)(x+2)[(x+2)+(x^2-6x+9)]

= 2(x-3)(x+2)(x^2-5x+11)

Is that the final answer, or still need to factorise the (x^2-5x+11) ?

Wondering how to factorise that, (x - p)(x-q)
p+q=5, pq=11 ??
• October 3rd 2011, 06:38 PM
Prove It
Re: Factorise Equation
Quote:

Originally Posted by gilagila
Factorise f(x) completely
=2(x-3)(x+2)[(x+2)+(x-3)^2]

= 2(x-3)(x+2)[(x+2)+(x^2-6x+9)]

= 2(x-3)(x+2)(x^2-5x+11)

Is that the final answer, or still need to factorise the (x^2-5x+11) ?

Wondering how to factorise that, (x - p)(x-q)
p+q=5, pq=11 ??

If you check the discriminant (-5)^2 - 4(1)(11) = 25 - 44 = -19, since this is negative that means it does not factorise over the reals. So yes, that's the final answer.
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