Turn on thread page Beta
 You are Here: Home >< Maths

Core 4 Help watch

1. Part 3, I was able to do part 3 although I called 3x-1, 6x-2 instead but I was able to see that.

But the second part I simply cannot do. I thought the polynomial on the numerator 9x^3 + 21x^2 + 6x would probably go into 3x+1 as well.

Doing so gave me 3x^2 + 6x. I factorised this and got (x+2) and (x) as factors, I don't believe these are correct, so how would I do this?
2. (Original post by Dinasaurus)

Part 3, I was able to do part 3 although I called 3x-1, 6x-2 instead but I was able to see that.

But the second part I simply cannot do. I thought the polynomial on the numerator 9x^3 + 21x^2 + 6x would probably go into 3x+1 as well.

Doing so gave me 3x^2 + 6x. I factorised this and got (x+2) and (x) as factors, I don't believe these are correct, so how would I do this?
Doing this quickly in my head gives me

and

So:

What cancels with and what are you left with?
3. Equating coefficients?
4. (Original post by Dinasaurus)

Part 3, I was able to do part 3 although I called 3x-1, 6x-2 instead but I was able to see that.

But the second part I simply cannot do. I thought the polynomial on the numerator 9x^3 + 21x^2 + 6x would probably go into 3x+1 as well.

Doing so gave me 3x^2 + 6x. I factorised this and got (x+2) and (x) as factors, I don't believe these are correct, so how would I do this?
You think correcty. 3x+1 is indeed a factor of the numerator however , you are forgetting a number.

You need to express the numerator and denominator in terms of their factors and "cancel" appropriately.
5. (Original post by Kvothe the arcane)
You think correcty. 3x+1 is indeed a factor of the numerator however , you are forgetting a number.

You need to express the numerator and denominator in terms of their factors and "cancel" appropriately.
Yeah I know that part but it's getting the numerator as factors which is the problem for me, I used the grid method to work it out and I can't seem to get another number. I have tried once again with the grid method and I don't go any further than (3x^2+6x)

If I use the quadratic formula for this, (-6+-sqrt(36))/6

I only get 0 and -2.
6. (Original post by Dinasaurus)
Yeah I know that part but it's getting the numerator as factors which is the problem for me, I used the grid method to work it out and I can't seem to get another number. I have tried once again with the grid method and I don't go any further than (3x^2+6x)

If I use the quadratic formula for this, (-6+-sqrt(36))/6

I only get 0 and -2.
Zacken has factorized it nicely for you. Forget this complex problem. If you have , what are A and B? You've worked out B, A is all that is all you're missing.
7. (Original post by Dinasaurus)
Yeah I know that part but it's getting the numerator as factors which is the problem for me, I used the grid method to work it out and I can't seem to get another number. I have tried once again with the grid method and I don't go any further than (3x^2+6x)

If I use the quadratic formula for this, (-6+-sqrt(36))/6

I only get 0 and -2.

Using the quadratic formula on the quadratic factor gets you:

So the quadratic factorises as .
8. (Original post by Zacken)

Using the quadratic formula on the quadratic factor gets you:

So the quadratic factorises as .
Thank you! If I take out a factor of 3x+1 instead of 3x, I get 3x^2+6x, which gives me the roots -2 and 0 from the quadratic formula? Why does it not work this way?
9. (Original post by Dinasaurus)
Thank you! If I take out a factor of 3x+1 instead of 3x, I get 3x^2+6x, which gives me the roots -2 and 0 from the quadratic formula? Why does it not work this way?
It does work. If you take out a factor of you're left with , this has roots and meaning that .
10. (Original post by Zacken)
It does work. If you take out a factor of you're left with , this has roots and meaning that .

Would (x)(3x+1)(x+2)/ (3x-1)(3x+1)(x+2)

which cancels to x/(3x-1) not be correct then?

Sorry for asking so much, I see that the mark scheme says it is 3x on the top.

But if you take out a factor of 3x+1, then surely the other two linear factors are x and (x+2)?
11. (Original post by Dinasaurus)
Would (x)(3x+1)(x+2)/ (3x-1)(3x+1)(x+2)

which cancels to x/(3x-1) not be correct then?

Sorry for asking so much, I see that the mark scheme says it is 3x on the top.

But if you take out a factor of 3x+1, then surely the other two linear factors are x and (x+2)?
Ah, okay - I see your confusion. The problem is that any quadratic equation of the form will have those same roots, so you can't factorise it knowing just the roots. What you need to be able to do is just see:

.
12. (Original post by Zacken)
Ah, okay - I see your confusion. The problem is that any quadratic equation of the form will have those same roots, so you can't factorise it knowing just the roots. What you need to be able to do is just see:

.
So if I have no C value, I cannot use the quadratic formula?

Thank you for this explanation.
13. (Original post by Dinasaurus)
So if I have no C value, I cannot use the quadratic formula?

Thank you for this explanation.
You can use the quadratic formula and it will give you the linear factors. It is your job for you to multiply the entire thing by whatever you need to get it to look like the unexpanded version.

For example, if I gave you the quadratic formula will gives you roots as and , this allows you to factorise it as , but you now need to set to get the proper factorisation. In general, the will be the coefficient of the term.

Turn on thread page Beta

Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 15, 2016
Today on TSR

Edexcel C4 Maths Unofficial Markscheme

Find out how you've done here

1,125

students online now

Exam discussions

Poll
Useful resources

Make your revision easier

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

How to use LaTex

Writing equations the easy way

Study habits of A* students

Top tips from students who have already aced their exams

Create your own Study Planner

Never miss a deadline again

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE