factorise

Hedgecutter

I need to factorise this but i'm getting stuck 3x^2-12x+9=0

3(x^2-4x+3)=0
x=-1 x=-3

Don't think im right and not sure which way to go from here

JIdontknow

Hedgecutter wrote: »

I need to factorise this but i'm getting stuck 3x^2-12x+9=0

3(x^2-4x+3)=0
x=-1 x=-3

Don't think im right and not sure which way to go from here

I think you're right just maybe the signs are wrong, if you sub back in:

-1: 3(-1)^2 - 12(-1) + 9
Any number squared is positive so

3 - - 12 + 9 would be 24, BUT using +1 it gives 3 -12 + 9 = -9 + 9 = 0.

-3: 3(-3)^2 - 12(-3) + 9
27 + 36 + 9 is 72 BUT
+3 gives 27 -36 + 9 which is -9 + 9=0.

So I'm getting 1, 3 as the factors.

Hedgecutter

JIdontknow wrote: »

I think you're right just maybe the signs are wrong, if you sub back in:

-1: 3(-1)^2 - 12(-1) + 9
Any number squared is positive so

3 - - 12 + 9 would be 24, BUT using +1 it gives 3 -12 + 9 = -9 + 9 = 0.

-3: 3(-3)^2 - 12(-3) + 9
27 + 36 + 9 is 72 BUT
+3 gives 27 -36 + 9 which is -9 + 9=0.

So I'm getting 1, 3 as the factors.

Could I use quadratic formula here.

TheBody

Hedgecutter wrote: »

I need to factorise this but i'm getting stuck 3x^2-12x+9=0

3(x^2-4x+3)=0
x=-1 x=-3

Don't think im right and not sure which way to go from here

Is the question to SOLVE the quadratic equation [latex]3x^2-12x+9=0[/latex]?

If it is, the first thing I would do is divide both sides by 3 to make the numbers easier to deal with. We would get:

[latex]x^2-4x+3=0[/latex]

You could then factorise to get

[latex](x-3)(x-1)=0[/latex]

This imples that [latex]x=3[/latex] or [latex]x=1[/latex]

TheBody

You could use the quadratic root formula but it's a bit on the slow side in comparison to the bracket method.

Hedgecutter

Thanks guys. Started maths course this year and its been 20 years sense I did anything like this. forgotten some of the basics.

TheBody

Your welcome.

Keep the questions coming and we will help you any way we can.

Hedgecutter

TheBody wrote: »

Your welcome.

Keep the questions coming and we will help you any way we can.

Thanks again and I will stick up a question im doing at the moment. More finding the centroid of the area, see what you think.

JIdontknow

You might know already but there's the abc formula which sometimes helps with factorising.

-B ± square root of (B^2 - 4AC) / 2A ( if I remember it correctly!)

The x squared part is a, the x part is b and the number alone is c,

5X^2 + 10X + 20
A=5
B=10
C=20

Hedgecutter

JIdontknow wrote: »

You might know already but there's the abc formula which sometimes helps with factorising.

-B ± square root of (B^2 - 4AC) / 2A ( if I remember it correctly!)

The x squared part is a, the x part is b and the number alone is c,

5X^2 + 10X + 20
A=5
B=10
C=20

Is there any situations you can't use that formula.

TheBody

It will always work provided that it's a quadratic equation you are dealing with. It is particularly useful with the roots are not "nice numbers".

I wouldn't forget about the bracket method though. Factorising is an EXTREEMLY useful skill to have and you will need it in other situations.

MathsManiac

Also, you seem to be confusing factorising with solving. These are two different things (although in the case of quadratic expressions or equations, you can use one as a route to doing the other).

If the question is to solve the equation 3x^2 - 12x + 9 = 0, then the answer is x = 1 and x = 3.

If, on the other hand, the question is to factorise the expression 3x^2 - 12x + 9, then the answer is 3(x-3)(x-1).

The correct answer to one of these questions is not the correct answer to the other.

bnt

Hedgecutter wrote: »

Is there any situations you can't use that formula.

Depends on what you mean by "can't". A quick test to determine whether a quadratic formula has real roots is to calcluate b²-4ac. If the result is negative, it has no real roots. But the quadratic formula still "works".