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Factorise x^3 -9x

Hi -

Can someone show me how to factorise that completely please? Just came up on a C1 paper I'm doing. I know it should be simple, but I really don't know how to get further than x(x^2 - 9).

Thanks

Reply 1

Swayum

18

x^2 is x SQUARED.

9 is 3 SQUARED.

The negative sign means "DIFFERENCE".

So x^2 - 9 = square difference square. Oh look, there are TWO squares here.

Any bells ringing?

Reply 2

Miyavi

13

x(x-3)(x+3)

(x-3)(x+3) =
x^2 +3x -3x - 9 = x^2 - 9

Reply 3

Miyavi

13

by the way I worked it out in my head by using the rules that:

ax^2+bx+c = 0

Product of the roots = -c
Addition of the roots = b

Reply 4

marcusmerehay

17

Miyavi

by the way I worked it out in my head by using the rules that:

ax^2+bx+c = 0

Product of the roots = -c/a
Addition of the roots = b/a

corrected.

Reply 5

Swayum

18

marcusmerehay

corrected.

You're still wrong. It's product = c/a and sum = -b/a.

Reply 6

Darkest Knight

OP

12

Swayum

x^2 is x SQUARED.

9 is 3 SQUARED.

The negative sign means "DIFFERENCE".

So x^2 - 9 = square difference square. Oh look, there are TWO squares here.

Any bells ringing?

Ah yes it's the "difference of two squares" thing, but I still don't get it... :frown:

Reply 7

mfc20

11

Darkest Knight

Hi -

Can someone show me how to factorise that completely please? Just came up on a C1 paper I'm doing. I know it should be simple, but I really don't know how to get further than x(x^2 - 9).

Thanks

Funnily enough I did the exact same paper for my C1 mock this morning. Got that question right too! :biggrin:

Reply 8

Swayum

18

Darkest Knight

Ah yes it's the "difference of two squares" thing, but I still don't get it... :frown:

a^2 - b^2 = (a + b)(a - b)

Memorise this (you can prove it by expanding out the brackets on the right hand side). In your case, a = x and b = 3. It's a common result that you should be able to spot instantly.

Reply 9

Miyavi

13

Lawl yeah I forget divide by a as a lot of questions a=1.

Reply 10

DeanK2

Ahh yes - the product and sum of quadratic roots that is expected to be used for trivial factorisation problems.

OP - this will come in useful if you are looking for roots of the form a + bi , or k + y (where a + bi denotes an imaginary number) and k and y are two numbers that are not known.

However when it is clear (test some numbers - if you don't spot the difference of two squares, whack in say, x = 1, x = 2, xx= 3 - if you get zero, you have a root). You know that if (foor example 3 is a root), then you have a bracket that yields x = 3. This will only occur if the equation equals zero, and if you have (x - 3)(ax + b) = 0. [for this problem, of course with higher powers you could have (2x - 6)(ax^n-1 + bx^n-2 ... + z) = 0).

Reply 11

Darkest Knight

OP

12

Swayum

a^2 - b^2 = (a + b)(a - b)

Memorise this (you can prove it by expanding out the brackets on the right hand side). In your case, a = x and b = 3. It's a common result that you should be able to spot instantly.

Thanks, get it now :biggrin:

Reply 12

connor175

Hi just wondering what paper this was(the date)? And you should split the (x^2-9) into (x+3)(x-3) so you're final answer will be x(x+3)(x-3).

Factorise x^3 -9x

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