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# C1 factorisation

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1. Can someone explain how to factorise this please:

x^3-x^2-8x+12=0 in the form (x-2)^2 (x+p) where p is a positive constant
2. (x^3-x^2-4^2)-16+12=0
=-4
?
3. Consider the constant in the left hand side of the equation.
4. Use the factor theorem.
5. Expand (x-2)^2 and then use algebraic division with the result!
6. (Original post by Anon07079191)
Expand (x-2)^2 and then use algebraic division with the result!
(Original post by NotNotBatman)
Use the factor theorem.
(Original post by TimGB)
Consider the constant in the left hand side of the equation.
(Original post by Oblivion99)
(x^3-x^2-4^2)-16+12=0
=-4
?

Oh i got it guys. Thank you
7. (Original post by Anon07079191)
Expand (x-2)^2 and then use algebraic division with the result!
(x-2)(x-2)(x+3).

P is 3.

I expanded (x-2)^2 and divided x^3-x^2-8x+12 by it.

Then I found x+3.
8. (Original post by YeSand)
(x-2)(x-2)(x+3).

P is 3.

I expanded (x-2)^2 and divided x^3-x^2-8x+12 by it.

Then I found x+3.
yeah, thanks
9. (Original post by YeSand)
(x-2)(x-2)(x+3).

P is 3.

I expanded (x-2)^2 and divided x^3-x^2-8x+12 by it.

Then I found x+3.
Edit.

(x-2)(x-2)(x+3)

is the same as

(x-2)^2 (x+3)
10. (Original post by YeSand)
(x-2)(x-2)(x+3).

P is 3.

I expanded (x-2)^2 and divided x^3-x^2-8x+12 by it.

Then I found x+3.
Yeah, sounds good. In hindsight you could've expanded (X-2)^2 and the equated the two equations to see that (X-2)^2 ends in a +4 and therefore them remaining factor must end in +3 to have a constant of +12 in the answer!
11. Check out AZ TUTORING ON Youtube. They have a lot of past paper maths exam questions amongst others

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