Thread: Factorise Equation

Factorise f(x) completely.

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

Originally Posted by gilagila

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)...

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))

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 ? If so, please do so, showing all steps.

Originally Posted by mr fantastic

No.

Can you factorise ? 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 ?

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 ....

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 ?

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

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

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.

Fix the mistake, post your answer.

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 ?

Originally Posted by gilagila

...Is it correct now ?

Yes, oui, si, ...

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...

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 ??

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.