Factorising :confused:

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 :confused:

If anyone could show working out for this question with the answer so I can try and apply it myself hopefully figure it out.

Reply 1

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?

Reply 2

Shakaa

OP

8

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 :s-smilie:

Reply 3

*MJ*

2

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. :smile:

Reply 4

FranticMind

18

Answer below

Spoiler

Reply 5

LetoKynes

14

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 :smile:

Reply 6

etnies

9

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 :s-smilie:

As aforementioned make it quadratic , then find the routs and factorise :yep:

Reply 7

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

:smile:

Reply 8

Shakaa

OP

8

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

:smile:

Thats got me sorted now thanks :smile:

Reply 9

steve2005

17

john1234567

well you can start by making y=0

No, you can't assume

y=0

Reply 10

steve2005

17

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

:smile:

Expand the brackets and you don't get

2-3x-2x^2

so the factorisation is not correct. The correct factorisation is

(x+2)(1-2x)

Reply 11

etnies

9

Shakaa

Thats got me sorted now thanks :smile:

The correct and simpler way to factorise this equations is as it follows:

0=2-3x-2x^{2}

1)Move everything to the left and you will get

2x^{2} + 3x - 2 = 0

2)Solve it as quadratic ->

D= 25

then

x_1,x_2 = \frac{-3\pm5}{4}=(-2),\frac{1}{2}

These are the routs and you have

(x+2)(x-\frac{1}{2}) = (x+2)(2x-1)

steve2005

No, you can't assume

y=0

Yes,we can (sounds like Obama :biggrin:

)

Reply 12

steve2005

17

etnies

The correct and simpler way to factorise this equations is as it follows:

0=2-3x-2x^{2}

1)Move everything to the left and you will get

2x^{2} + 3x - 2 = 0

2)Solve it as quadratic ->

D= 25

then

x_1,x_2 = \frac{-3\pm5}{4}=(-2),\frac{1}{2}

These are the routs and you have

(x+2)(x-\frac{1}{2}) = (x+2)(2x-1)

Yes,we can (sounds like Obama :biggrin:

)

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

y=0

then the equation can be solved by equating each factor to zero.

However, the original question was simply factorise.

Reply 13

Weeman5872

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.

Factorising :confused:

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