LC question

Iwasfrozen

Hey guys could you help me out on this one ?

Let f(x) be a cubic polynomial such that
f(-1)=0, f(1)=0, and f(2)=5*f(0)
Find the third root of the equation f(x)=0

bnt

Are you sure you wrote the details correctly? Sure, it's possible to create a cubic polynomial that goes through all three given points. In fact, it's possible to create an infinite number of such polynomials: to nail down a single n-order polynomial, you need n+1 points. So you only have enough info there for a 2nd-order (quadratic) polynomial, which in this case happens to be a straight line on the x-axis! f(x) = 0x² + 0x + 0

Plus, as written, you already have been given three roots: [-1, 1, 2]. (Is that what they mean - just recognise a root when it's handed to you on a plate?) :pac:

Iwasfrozen

bnt wrote: »

Are you sure you wrote the details correctly? Sure, it's possible to create a cubic polynomial that goes through all three given points. In fact, it's possible to create an infinite number of such polynomials: to nail down a single n-order polynomial, you need n+1 points. So you only have enough info there for a 2nd-order (quadratic) polynomial, which in this case happens to be a straight line on the x-axis! f(x) = 0x² + 0x + 0

Plus, as written, you already have been given three roots: [-1, 1, 2]. (Is that what they mean - just recognise a root when it's handed to you on a plate?) :pac:

Oh stupid me !
Apologises, f(2)=5f(0) not 0. :P

bnt

Iwasfrozen wrote: »

Oh stupid me !
Apologises, f(2)=5f(0) not 0. :P

Do you mean f(2) = 5? But you still have only three points, so it still looks to me as if you can't derive a specific cubic polynomial from just three points. I can find a quadratic, but you already have both those roots.

Iwasfrozen

bnt wrote: »

Do you mean f(2) = 5? But you still have only three points, so it still looks to me as if you can't derive a specific cubic polynomial from just three points. I can find a quadratic, but you already have both those roots.

No, the three things we are given are:
f(-1)=0, f(1)=0 , and f(2)=5f(0)

bnt

Iwasfrozen wrote: »

No, the three things we are given are:
f(-1)=0, f(1)=0 , and f(2)=5f(0)

Do you mean f(2) = 5 times f(0)? Sorry - thought that might have been a typo. If so, I think I might see where this is going, though I'm not sure. (I didn't get my edumacation in Ireland, so I don't know what they teach you at LC level.)

My first guess is that that ratio f(2) = 5*f(0) is supposed to tell you something about the slope of the curve, and allow you to work out its derivative (which would be quadratic). (Anyone do it this way before? If I did, it was a long time ago.)
Then Integration would come in to play, which would give you a cubic polynomial with a constant of integration. You would use one of the points you were given to eliminate the constant, giving you the particular cubic polynomial, which you would solve to find the three roots.

Iwasfrozen

bnt wrote: »

Do you mean f(2) = 5 times f(0)? Sorry - thought that might have been a typo. If so, I think I might see where this is going, though I'm not sure. (I didn't get my edumacation in Ireland, so I don't know what they teach you at LC level.)

My first guess is that that ratio f(2) = 5*f(0) is supposed to tell you something about the slope of the curve, and allow you to work out its derivative (which would be quadratic). (Anyone do it this way before? If I did, it was a long time ago.)
Then Integration would come in to play, which would give you a cubic polynomial with a constant of integration. You would use one of the points you were given to eliminate the constant, giving you the particular cubic polynomial, which you would solve to find the three roots.

Hmm, no, the question is a part 1 on paper 1 so there isn't any differentiation or integration allowed. Only Algebra.

P.S Yes, f(2)=5(0) means f(2) = 5 times f(0). Sorry if that wasn't clear.

Fremen

Well, you know f(-1) and f(1) are roots, which means x+1 and x-1 are factors.
Since it's a cubic, there's just one root left. Let that root equal alpha.
How can you use the last piece of information you're given to deduce the value of alpha?

Eliot Rosewater

A very interesting question. I think the trick lies in the following:

The function intersects the x-axis at -1 and 1. But because f(2) = 5*f(0), both f(2) and f(0) are on the same side at of the x-axis. Either f(0) and f(2) are below the x-axis, or f(0) and f(2) are above it.

See?

Which means that the third root lies between -1 and 2. Sketch it out and see what I mean.

Iwasfrozen

Eliot Rosewater wrote: »

A very interesting question. I think the trick lies in the following:

The function intersects the x-axis at -1 and 1. But because f(2) = 5*f(0), both f(2) and f(0) are on the same side at of the x-axis. Either f(0) and f(2) are below the x-axis, or f(0) and f(2) are above it.

See?

Which means that the third root lies between -1 and 2. Sketch it out and see what I mean.

I'm sorry, I don't. How can you be sure that f(2) and f(0) are both on the same side of the x-axis ?

Fremen

He (she?) is right, but that information isn't necessary to answer the question. It's because one is five times the other, so they have the same sign.

Big hint:
You know what all but one of the factors are. When in doubt, multiply out.
(Acutally, that's terrible advice, but it rhymes, and in this case it works)

Eliot Rosewater

Iwasfrozen wrote: »

I'm sorry, I don't. How can you be sure that f(2) and f(0) are both on the same side of the x-axis ?

If f(0) is below the x-axis it is negative. 5 times a negative number is also negative. Therefore f(2)=5*f(0) is also negative. f(2) is also below the x-axis.

If f(0) is above the x-axis it is positive. 5 times a positive number is also positive. Therefore f(2) is also positive. etc


Freemen is right though, go with his advise. I got a little excited about my analytical observation (that proves to be true when one gets the answer)

Iwasfrozen

Fremen wrote: »

He (she?) is right, but that information isn't necessary to answer the question. It's because one is five times the other, so they have the same sign.

Big hint:
You know what all but one of the factors are. When in doubt, multiply out.
(Acutally, that's terrible advice, but it rhymes, and in this case it works)

Thanks.

So when I multiply my factors out I get:
(x + 1)(x - 1) = x^2 - 1.

So would I then be right in saying (x^2 - 1)( ______ ) = ax^3 + bx^2 + cx + d ?

Iwasfrozen

Eliot Rosewater wrote: »

Freemen is right though, go with his advise. I got a little excited about my analytical observation (that proves to be true when one gets the answer)

No problem, I get like that sometimes also.

Fremen

nah, you're forgetting the (x+alpha) factor

Edit:
so what you have is
f(x) = (x+1))(x-1)(x+alpha)

or

f(x) = x^3 + alpha x^2 - x - alpha

so now what do you know about f?

ZorbaTehZ

You don't have to multiply out because the constant in a cubic polynomial is the product of the roots, and from f(2)=5*f(0) you know that (-1)*(1)*x=5 thus the third root is 5.

Iwasfrozen

Fremen wrote: »

nah, you're forgetting the (x+alpha) factor

Edit:
so what you have is
f(x) = (x+1))(x-1)(x+alpha)

or

f(x) = x^3 + alpha x^2 - x - alpha

so now what do you know about f?

Haha, I new about the (x+alpha) I just didn't put it in. :P

So now we have:
x^3 + alpha x^2 - x - alpha = ax^3 + bx^2 + cx + d

Take x^3:
x^3 = ax^3
a = 1

Take x^2:
alpha x^2 = bx^2
alpha = b

Take x:
- x = cx
-1 = c

Take R:
- alpha = d
- b = d.

This is as far as I got in my copybook, but now I'm positively stuck.

ZorbaTehZ

Actually mine above is incorrect, I thought you had f(2)=0 also, looks like you're going to have to multiply it out

Fremen

you don't need to equate it to ax^3 + bx^2 + cx + d. You have this nice expression for f, and you know f(-1)=0, f(1)= 0 and f(2) = 5f(0). Sooo...?

(Nah ztz, you don't need to, but it makes it more obvious)

Ignore this until you've solved the problem :
Bnt was right when he said that f isn't uniquely determined. I've been assuming the coefficient of x^3 is 1, but you get a correct answer if you multiply across by any constant c. Bit of a dud question, they should have stated that the coefficient of x^3 is 1.

Iwasfrozen

Fremen wrote: »

you don't need to equate it to ax^3 + bx^2 + cx + d. You have this nice expression for f, and you know f(-1)=0, f(1)= 0 and f(2) = 5f(0). Sooo...?

Alright, so you want me to work only with x^3 + alpha x^2 - x - alpha.
could I perhaps take alpha out of both to get x^3 + alpha(x^2 - 1) - x ?

then I would take x out of the other two munbers and get x( x^2 - 1)

So that would leave me with:
X^2 - 1 = 0
x = sqrt(1)

and:
x + alpha = 0
x = - alpha.

Is the third factor x = sqrt(1) ?

RoundTower

Fremen wrote: »

Bit of a dud question, they should have stated that the coefficient of x^3 is 1.

why? they didn't ask you to find f!

Fremen

Ok, so you had this polynomial which you knew nothing about, other than the three bits of information

f(1) = 0
f(-1) = 0
5f(0) = f(2)

So the first two bits of information tell you that

f(x) looks like (x-1)(x+1)(x + alpha).

You haven't used the last bit yet. You can see f as a rule which says
"plug in X, get out X^3 + alpha X^2 - X - alpha"

so f(Jimmydog) = Jimmydog^3 +(alpha Jimmydog^2) - Jimmydog - alpha

So how are we gonna use this last bit of information?
Plug in 0, get out something. Plug in 2, get out something else.

why? they didn't ask you to find f!

Oh. Yeah, that was a bit of a brainfart.

Iwasfrozen

Fremen wrote: »

Ok, so you had this polynomial which you knew nothing about, other than the three bits of information

f(1) = 0
f(-1) = 0
5f(0) = f(2)

So the first two bits of information tell you that

f(x) looks like (x-1)(x+1)(x + alpha).

You haven't used the last bit yet. You can see f as a rule which says
"plug in X, get out X^3 + alpha X^2 - X - alpha"

so f(Jimmydog) = Jimmydog^3 +(alpha Jimmydog^2) - Jimmydog - alpha

So how are we gonna use this last bit of information?
Plug in 0, get out something. Plug in 2, get out something else.

Ahhhh !

So we have:
f(x) = x^3 + alpha x^2 - x - alpha

that means:
f(2) = 8 + 4*alpha - 2 - alpha
f(0) = 0 + 0 - 0 - alpha.

f(2) = 5*f(0)
8 + 4*alpha - 2 - alpha = -5*alpha
6 = -8*alpha
alpha = -6/8 = -3/4

Is this correct ?

Eliot Rosewater

Yup! Now you just have to find the root.

Fremen

You tell me.
Sub in alpha = -3/4 into the expression for f and see if it agrees with the information you were given. If it does, you know you have the right expression for f, and you can easily find the third value for x at which f(x) = 0.

It's a really useful skill for the LC to find ways to check your answers like that. Besides, I could easily have screwed up the calculation for f.

Iwasfrozen

Fremen wrote: »

You tell me.
Sub in alpha = -3/4 into the expression for f and see if it agrees with the information you were given. If it does, you know you have the right expression for f, and you can easily find the third value for x at which f(x) = 0.

It's a really useful skill for the LC to find ways to check your answers like that. Besides, I could easily have screwed up the calculation for f.

ok.

f(x) = x^3 + alpha*x^2 - x - alpha
f(x) = x^3 - 3/4*x^2 - x + 3/4 = 0
f(1) = 1 - 3/4 - 1 + 3/4 = 0 True.

Ok so it works out.
But now how do I find the third solution for x?

Fremen

So you knew

f(x) = (x-1)(x+1)(x+alpha)

then you solved for alpha.
You want to find the values of x for which

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

What values are these?

Iwasfrozen

Fremen wrote: »

So you knew

f(x) = (x-1)(x+1)(x+alpha)

then you solved for alpha.
You want to find the values of x for which

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

What values are these?

Aaaaah,

x - 3/4 = 0
x = 3/4.

Thanks for all of your help Fremen, couldn't have done it without you bro.

Fremen

No borra

MathsManiac

Two minor further points are worth mentioning, I think:

Firstly, it's a bit odd to call the third factor (x+alpha), because in this context, you would usually use the letter alpha to represent the root. Therefore, it would have been more natural to make the last factor (x - alpha).

Secondly, you would probably be penalised in the LC exam for making the assumption that the coefficient of x is equal to 1. Even though it doesn't affect the answer, you are making incorrect assertions along the way. As already pointed out, it is not true that the function has to be f(x) = (x-1)(x+1)(x-alpha). It could be any constant multiple of this. Therefore you should really write it as f(x)=k(x-1)(x+1)(x-alpha). In this case, the k will cancel out when you write f(2)=5f(0) and so will not affect the answer. Alternatively, you could write the function as f(x) = (x-1)(x+1)(ax+b). In this latter case, you end up with b=-3a/4, which allows you to find the root.