Factorise fully 5x² - 45?

=5 * (x^2 - 9)
=5*(x+3)*(x-3)

5(x² - 9) = 5 (x - 3) (x + 3)

5x^2 - 45
= 5(x^2 - 9)
= 5(x + 3)(x - 3)

That was in the SG Maths exam yesterday if I remember correctly...
First you take out a common factor (in this case 5)
5(x^2 - 9)

This then becomes a difference of two squares, and can be factorised to:
5(x - 3)(x + 3)

What you did was a very good start - you just didn't finish the problem. The integer 5 is also a factor!

5(x-3)(x+3)

The factors are related to the roots

5x² - 45 = 0 // set = 0 to find the roots

5x² = 45 // add 45 each side

x² = 9 // divide each side by 5, and keep the 5 in your mind

x = +- 3 // take square root each side

So from the realationship between roots and factors

the factors of (x² - 9) are (x - 3)(x + 3)

Dont forget the 5 we divided by to get:

5(x - 3)(x + 3)

multiply out to test:
5(x - 3)(x + 3) = 5(x² - 9) = 5x² - 45

Hope this helps,

Peter

5x² - 45
take 5 as common
5 (x^2 - 9)
5 (x-3)(x+3) bcoz a^2- b^2 =(a+b)(a-b)

Common factor, then difference of two squares...

= 5(x^2 - 9)
=5(x + 3)(x - 3)

5(x^2-9) then using difference of two squares 5(x+3)(x-3)

Since it's 5(x-3)(x+3)

Putting in: (5x-15)(x+3) is correct, as you can multiply through the first bracket by the constant outside:

5(x-3) = 5x-15

Hope this settles your nerves :P