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# Factorising :confused: watch

1. I can factorise or at least I assumed I could untill this question came up:

Factorise the following function:

Y = 2 - 3x - 2x^2

Has me well and truly

If anyone could show working out for this question with the answer so I can try and apply it myself hopefully figure it out.
2. well you can start by making y=0

2 - 3x - 2x^2 = 0

then you can move everything over to the other side

2x^2 + 3x - 2 = 0

do you know what to do from here?
3. (Original post by john1234567)
well you can start by making y=0

2 - 3x - 2x^2 = 0

then you can move everything over to the other side

2x^2 + 3x - 2 = 0

do you know what to do from here?

I thought I did I mean its a similar method I use for completing the square by reversing the equation but after having tried it on this question my answer seems to be wrong
4. Smiley fail.

Take it all over to the other side if you are really that confused.

You can factorise it by quadratic formula/completing the square if you are really that confused (solve and then put x - ans into a bracket).

Factorising just requires, a pencil, rubber and a lot of practice.
5. Answer below

Spoiler:
Show
(2x-1)(x+2)
6. (Original post by john1234567)
well you can start by making y=0

2 - 3x - 2x^2 = 0

then you can move everything over to the other side

2x^2 + 3x - 2 = 0

do you know what to do from here?
If you do this, you have to put a -ve sign after you factorise it because y isn't 0
7. (Original post by Shakaa)
I thought I did I mean its a similar method I use for completing the square by reversing the equation but after having tried it on this question my answer seems to be wrong

As aforementioned make it quadratic , then find the routs and factorise
8. ok i'll show you my method

a = 2
b = 3
c = -2

start off by finding ac

ac in this case is -4

now find two integers m and n such that m + n = ac

in this case, m = 4 and n = -1

so rewrite the equation as 2x^2 + 4x - x - 2 = 0

now imagine you've split the LHS down the middle, so you're left with

2x^2 + 4x
and
- x - 2

factorise 2x^2 + 4x to get 2x(x + 2)

the (x + 2) will be the bracket for - x - 2 as well, so write it there

- x - 2 = s(x + 2)

find a value of s that will make that equation true; in this case, -1

so we now have 2x(x + 2) + (-1)(x + 2) = 0

the (x + 2) is common in both terms, so

(x + 2)(2x - 1) = 0

9. (Original post by john1234567)
ok i'll show you my method

a = 2
b = 3
c = -2

start off by finding ac

ac in this case is -4

now find two integers m and n such that m + n = ac

in this case, m = 4 and n = -1

so rewrite the equation as 2x^2 + 4x - x - 2 = 0

now imagine you've split the LHS down the middle, so you're left with

2x^2 + 4x
and
- x - 2

factorise 2x^2 + 4x to get 2x(x + 2)

the (x + 2) will be the bracket for - x - 2 as well, so write it there

- x - 2 = s(x + 2)

find a value of s that will make that equation true; in this case, -1

so we now have 2x(x + 2) + (-1)(x + 2) = 0

the (x + 2) is common in both terms, so

(x + 2)(2x - 1) = 0

Thats got me sorted now thanks
10. (Original post by john1234567)
well you can start by making y=0
No, you can't assume
11. (Original post by john1234567)
ok i'll show you my method

a = 2
b = 3
c = -2

start off by finding ac

ac in this case is -4

now find two integers m and n such that m + n = ac

in this case, m = 4 and n = -1

so rewrite the equation as 2x^2 + 4x - x - 2 = 0

now imagine you've split the LHS down the middle, so you're left with

2x^2 + 4x
and
- x - 2

factorise 2x^2 + 4x to get 2x(x + 2)

the (x + 2) will be the bracket for - x - 2 as well, so write it there

- x - 2 = s(x + 2)

find a value of s that will make that equation true; in this case, -1

so we now have 2x(x + 2) + (-1)(x + 2) = 0

the (x + 2) is common in both terms, so

(x + 2)(2x - 1) = 0

Expand the brackets and you don't get so the factorisation is not correct. The correct factorisation is
12. (Original post by Shakaa)
Thats got me sorted now thanks
The correct and simpler way to factorise this equations is as it follows:

1)Move everything to the left and you will get
2)Solve it as quadratic -> then
These are the routs and you have

(Original post by steve2005)
No, you can't assume
Yes,we can (sounds like Obama )
13. (Original post by etnies)
The correct and simpler way to factorise this equations is as it follows:

1)Move everything to the left and you will get
2)Solve it as quadratic -> then
These are the routs and you have

Yes,we can (sounds like Obama )
The original question was

and obviously your solution is incorrect. It's always best to look at the original question rather than subsequent posts.

Obviously, if you are told that then the equation can be solved by equating each factor to zero.

However, the original question was simply factorise.
14. This is my method:

y = 2 - 3x - 2x^2

Look for factors of the constant term and the x^2 coefficient that will add up to make 3 I write it in a table:

1 2 | 1
-------
2 1 | 2

The factors of the constant term on the left and factors of x^2 on the right. Then multiply them diagonally and see which two give the x coefficiant. As you can see you can have 2 x 2 and 1 x 1 which when added together can give 3 or 5.

so now you know its:

2 | 1
-----
1 | 2

You write a backet, first with the top two numbers (reading from left to right) and second with bottom two)

So you get (2 x)(1 2x)

If you expand the x terms, you get x and 4x. Since we want -3x you make 4x negative and x positive, so you put the signs in the brackets so that you when you expand them, you get -4x + x. In this case it would be (2 + x)(1 - 2x)

Hope that helped in some way.

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