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# C1 question - bit confused Watch

1. Hey this is a question for C1 edexcel.

How do i factorise this?

y = 3x + 2x^2 - x^3 = x(3+2x -x^2) << get this bit

Dont know how to factorise (3+2x -x^2)

The two numbers which make -3 are 3 and -1. But i dont know how to put them in the brackets because when x is postive i know it looks like (x )(x ) but dont know how it should look for this?

y=(3-x)(1+x) (this is the answer) why do they have a postive 1 when the numbers are 3 and MiNUS 1 ?

Thanks
2. (Original post by jayseanfan)
Hey this is a question for C1 edexcel.

How do i factorise this?

y = 3x + 2x^2 - x^3 = x(3+2x -x^2) << get this bit

Dont know how to factorise (3+2x -x^2)

The two numbers which make -3 are 3 and -1. But i dont know how to put them in the brackets because when x is postive i know it looks like (x )(x ) but dont know how it should look for this?

y=(3-x)(1+x) (this is the answer) why do they have a postive 1 when the numbers are 3 and MiNUS 1 ?

Thanks
I suggest that you take a minus sign out of the bracket, so that you get -x(x^2-2x-3) and carry on from there.
3. (Original post by tiny hobbit)
I suggest that you take a minus sign out of the bracket, so that you get -x(x^2-2x-3) and carry on from there.
thanks but how about if the quesiton was factorise(3+2x -x^2)
4. (Original post by jayseanfan)
thanks but how about if the quesiton was factorise(3+2x -x^2)
make it equal to 0 , multiply through by -1 to get x^2 - 2x - 3 which factorises to (x+1) (x-3) ??
5. Reverse it. It now looks like (c+-x)(c+-x) where c is a constant, for example 3, and +- is plus or minus x depending on the signs.

You're looking for numbers that make 3, not -3, so 3 and 1. You make the x in the first bracket negative so that you get 3x-x which is 2x when you multiply each bracket. You also need to make it negative because you need to get a -x^2.

Hope this helps lol, you'll get used to factorising quadratics like that eventually.
6. (Original post by jayseanfan)
thanks but how about if the quesiton was factorise(3+2x -x^2)
to make -x^2 you have to have (x+a)(-x+b) where a and b are constants

so from there:

(-x+3)(x+1)= -x^2-x+3x+3

this simplifies to: -x^2+2x+3

they just have the factors in another format in the answer but its exactly the same thing

Hope this helps happy to help if you have more questions
7. (Original post by mathew551)
Reverse it. It now looks like (c+-x)(c+-x) where c is a constant, for example 3, and +- is plus or minus x depending on the signs.

You're looking for numbers that make 3, not -3, so 3 and 1. You make the x in the first bracket negative so that you get 3x-x which is 2x when you multiply each bracket. You also need to make it negative because you need to get a -x^2.

Hope this helps lol, you'll get used to factorising quadratics like that eventually.
thanks for the reply, but im still a little confused. Why are we look for numebers whcih make 3 and not -3?

dont you do a X c which in this case is -1 x 3 = -3 ?
8. You're trying to make C = 3
as Ax^2 + Bx + C is the general form. Basically 3x1 = 3, but as A must be negative, one of the x's must therefore be negative. As you must then find what adds to make B, which should equal 2, the bracket with 3 in must have a -x in, as when you then expand it, it will be the required quadratic.
Try expanding the answer and you should see how it works!
9. (Original post by jayseanfan)
thanks but how about if the quesiton was factorise(3+2x -x^2)
There's no reason that you can't have a -1 in front of the brackets.

In spite of all of the other offers, I still suggest that you take out a factor of -1.

So 3 + 2x - x^2 becomes -1(x^2 -2x-3)

Now apply your rules to the bracket: factors of -3 are -3 and 1 (to give -2x)

-1(x-3)(x+1)

If you don't like the -1 at the front, put it back into one of the brackets e.g. make x-3 into -x+3 or 3-x

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