Algebra (Cubic Polynomial) Help!

natrisioga

Hi,

Just wondering if someone could help me solve this question, or at least point me in the right direction.

f(x) is a cubic polynomial
f(-1)=0
f(1)=0
f(2)=5f(0)

i) Find the equation f(x).

ii) Now find 3 roots of f(x)=0

Thanks

Thomas_S_Hunterson

Well by the factor theorem, if f(a)=0 then (x-a) is a factor.

rainbowtrout

i've just noticed that this is a question on the Leaving Cert Higher Level Maths Mock. I can only assume you haven't done the paper yet.

I don't agree with people trying to get hold of the papers beforehand let alone look for the answers. If you can't do this now with your book in front of you how are you going to do it in June?

LeixlipRed

rainbowtrout wrote: »

i've just noticed that this is a question on the Higher Level Maths Mock.

There are numerous mock papers afaik (think about it, hundreds of schools all taking their exams on different days) . And the questions are hardly original either

rainbowtrout

LeixlipRed wrote: »

There are numerous mock papers afaik (think about it, hundreds of schools all taking their exams on different days) . And the questions are hardly original either

No there aren't. I'm a teacher and I correct/set mock papers. Check out the Leaving Cert Mocks thread. There are only 2 companies producing mock papers. It's cheating anyway

Pherekydes

rainbowtrout wrote: »

No there aren't. I'm a teacher and I correct/set mock papers. Check out the Leaving Cert Mocks thread. There are only 2 companies producing mock papers. It's cheating anyway

Hold off on the allegations of cheating. If there are only two companies producing them and you are setting mock papers, does that mean you are working for one of the companies? Or do schools set their own papers?

If teachers are coaching students to pass/get high points then they typically give questions that were on previous exams, do they not?

If the teacher in question gives the mock paper in advance then it's not cheating on the part of the pupils.

rainbowtrout

Slow coach wrote: »

Hold off on the allegations of cheating. If there are only two companies producing them and you are setting mock papers, does that mean you are working for one of the companies? Or do schools set their own papers?

Some schools set their own papers. I have set papers in one particular subject (not maths) for one company but I teach full time.

Slow coach wrote: »

If teachers are coaching students to pass/get high points then they typically give questions that were on previous exams, do they not?

If the teacher in question gives the mock paper in advance then it's not cheating on the part of the pupils.

True, and some teachers do. But a student should still be able to do the question themselves/attempt it themselves when it's on their mock paper as that's what they will be doing in June. It is after all a test to see how much they know

Fremen

Are you sure you meant f(0) = f(2)? did you mean f(2) = 0?
If the earlier condition holds, there's no unique polynomial which satisfies it: there are many possible solutions.

Otherwise, Sean_K gave you all the hints you need.

Rainbowtrout, you might as well argue that a student should be able to attempt everything themselves, since they should know the curriculum if they'll be doing it in june.

natrisioga

In relation to cheating, my teacher handed out a worksheet with revision questions that she thought we would need to know how to do for the mocks. This was one of the questions...I couldn't do it so I'm looking for help.

Sorry I did make a mistake it is f(2)=5f(0)

Any help would really be appreciated!

Thomas_S_Hunterson

natrisioga wrote: »

Any help would really be appreciated!

See my first reply.

Fremen

Sorry I did make a mistake it is f(2)=5f(0)

Any help would really be appreciated!

Nope, that still leaves the polynomial underdefined. You need four distinct values on the curve to solve it completely, since a cubic polynomial is represented as

ax^3 + bx^2 + cx + d = 0

Subbing in four values of x will give you four equations in four unknowns.

Are you sure it wasn't f(2) = 5 = f(0)?

MathsManiac

Fremen beat me to it while I was typing, but here's what I was going to say:

Even with the mistake now rectified, this is a very poor question. There is still no unique answer to part (i), although there is a unique answer to part (ii).

Secondly, f(x) is not an equation, it's a function (or an expression, strictly speaking), so part (i) is phrased poorly.

Leaving all that aside, I don't think the factor theorem is the easiest way to do this, since you'll need to then do algebraic long division or some fancy factorising instead of just doing some easy substitution. The following would be my hint to the OP:

Let f(x)= a x^3 + b x^2 + c x + d.

Each of the three conditions you were given can be used (by subbing in appropriately) to give you three equations contaning a, b, c and d.

Three equations is not enough to find four unknowns, which is why part (i) is poor. You can find b, c, and d in terms of a, and then write the function with an "a" still in it. You'll still be able to do part (ii). I suspect this outcome wasn't intentional, so if you want, just assume a=1, and find values for b, c and d from that.

natrisioga

Fremen wrote: »

Are you sure it wasn't f(2) = 5 = f(0)?

It is definitely f(2)=5f(0)

natrisioga

I took your advice and went and did some simultaneous equations but I have just ended up with 0 as an answer, maybe I have a sign wrong?

f(x)=ax^3+bx^2 +cx+d

f(-1)=0
= -a+b-c+d=0

f(1)=0
=a+b+c+d=0

f(2)=5f(0)
=8a+4b+2c+d=5d
=8a+4b+2c-4d=0
=4a+2b+c-2d=0

1. -a+b-c+d=0
2. a+b+c+d=0
4. 2b+2d=0

3. 4a+2b+c-2d=0
1. -1a+1b-c+1d=0
5. 3a+3b-d=0

3. 4a+2b+c-2d=0
2. 1a+1b+c+1d=0 (-1)
3. 4a+2b+c-2d=0
2. -a-1b-1c-1d=0
6. 3a+1b-3d=0 (-1)
5. 3a+3b-d=0
6. -3a-b+3d=0
5. 3a+3b-1d=0
7. 2b+2d=0

4. 2b+2d=0 (-1)
7. 2b+2d=0
4. -2b-2d=0
7. -2b-2d=0
0=0

MathsManiac

Your work is pretty good in that you're doing the right kind of thing to solve such sim. eqns. The reason you're getting 0=0 is that you have only 3 eqns for 4 unknowns. The best you can hope for is to get everything in terms of one letter.

From your equation 4, you get d=-b

Similarly, if you had subtracted eqs 1 and 2 instead of adding them, you'd have got c = -a.

Now we've eliminated two variables. If you sub. the d=-b and the c = -a into your equation 3: (4a+2b+c-2d=0), you get an equation in a and b alone:
4a+2b-a+2b=0, which gives you b= (-3/4)a.

Now you've written the whole thing in terms of a, so the function can be written as:

f(x)= a x^3 + (-3/4)a x^2 + (-a) x + (3/4)a

or, if you prefer, factoring out in front:

f(x) = (a/4)*( 4 x^3 -3 x^2 - 4 x +3)

This is as good as the answer gets for part (i). That is, for ANY value of a, the above function satisfies the conditions given in the question.

But the value of "a" won't affect the roots, so you can still do part (ii)

f(x) = 0 => 4 x^3 -3 x^2 - 4 x +3 = 0

You should be able to solve this, especially since 2 of the roots were given in the question!

LeixlipRed

rainbowtrout wrote: »

No there aren't. I'm a teacher and I correct/set mock papers. Check out the Leaving Cert Mocks thread. There are only 2 companies producing mock papers. It's cheating anyway

Cheating? Ridiculous comment. The question isn't original. The person who set it didn't have an epiphany and design some new question for the paper. I'm sure you could find an identical question on a previous LC or mock LC paper. As a teacher you should know that. Having only two mock papers as you claim means this sort of thing will happen