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Maths Polynomial Questions - HSN forum

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Maths Polynomial Questions


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#1 Joel

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Posted 28 November 2005 - 05:28 PM

Hi,

I'm stuck with quite a few questions:


4. Given that x[^2] + 1 is a factor of px[^3] + qx[^2] + px + q, find the other factor.


8. Construct a cubic equation with roots -power1.gifpowerslash.gifbase3.gif, power1.gifpowerslash.gifbase3.gif and power1.gifpowerslash.gifbase2.gif in the form ax[^3] + bx[^2] + cx + d = 0 where a, b, c and d are integers.


10. f(x) = 5x[^3] - x[^2] - 1

a) Find the remainder when f(x) is divided by 2x - 1.
I got remainder -0.625.

b) Write f(x) in the form (2x - 1)Q(x) + R where Q(x) is a quadratic expression and R is a constant.

c) Write f(x) in the form k[(2x - 1)(ax[^2] + bx + c) + d] where a, b, c and d are integers and k is rational.


12. f(x) = 4x[^3] + 4x[^2] - 21x + 9

a) Find the remainder when f(x) is divided by x - 1.
I got remainder -4.

b) The coefficient of x in f(x) can be altered so that when the new f(x) is divided by x - 1 the remainder is zero. Find the new coefficient.


Any help would be much appreciated,

Thanks.

#2 Paul

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Posted 28 November 2005 - 10:33 PM

just thought i'd try the first question i looked at:

"10. f(x) = 5x[^3] - x[^2] - 1

a) Find the remainder when f(x) is divided by 2x - 1.
I got remainder -0.625."

I also got -0.625

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#3 Joel

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Posted 28 November 2005 - 10:52 PM

Thanks Paul, do you know how to answer any of the other questions?

#4 George

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Posted 28 November 2005 - 10:57 PM

These seem to be slightly tricky questions!

For parts b and c of q10, you need to bear in mind that you've actually divided by x-\frac12 (ie you started with \frac12 on the left of the table for synthetic division).

i.e. you get f(x) = (x-\frac12)(\text{quadratic from bottom row}) + \text{remainder}

To give it in the form required, you'll need to double the (x-\frac12) - so you need to halve the quadratic (ie multiply each coefficient by 1/2).

Hope that helps a little bit smile.gif

#5 George

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Posted 28 November 2005 - 11:03 PM

For question 4, I'm not sure how to explain it!

You want px^3 + qx^2 + px + q = (x^2 + 1)( \text{something} )

Now, that something must be of the form ax + b, so that the ax term will multiply with the x^2 term to give you the px^3.

From that, you should be able to deduce what the 'a' is (ax \times x^2 = px^3 wink.gif ), and then do something similar for 'b'.



For 12, I also get -4 for the remainder in part a.

For part b, you basically need to find p in f(x) = 4x^3 + 4x^2 + px + 9 so that (x-1) is a factor - there's an example of a question like this in our notes for Polynomials and Quadratics smile.gif .

#6 Joel

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Posted 28 November 2005 - 11:15 PM

I think I understand now.

The bottom row from the synthetic division for 10. b) was: 5, 1.5, 0.75, -0.625.

So would the answer be:

f(x) = (x - 1/2)(5x^2 + 1.5x + 0.75) -0.625
(x2)= (2x - 1)(2.5x^2 + 0.75x + 0.375) -0.625

Is that correct?

#7 Joel

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Posted 28 November 2005 - 11:20 PM

I understand question 12 now, thanks.

Not sure about question 4 though...

#8 Dave

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Posted 28 November 2005 - 11:34 PM

if you have one factor

you can find the other by synthetically dividing the equation by the factor

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#9 George

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Posted 28 November 2005 - 11:42 PM

QUOTE(Joel @ Nov 28 2005, 11:15 PM)
I think I understand now.

The bottom row from the synthetic division for 10. b) was: 5, 1.5, 0.75, -0.625.

So would the answer be:

f(x) = (x - 1/2)(5x^2 + 1.5x + 0.75) -0.625
(x2)= (2x - 1)(2.5x^2 + 0.75x + 0.375) -0.625

Is that correct?

View Post


Yep, that's what I got.

For 4, you could just try multiplying out (x^2+1)(ax+b) and then compare what you get with the polynomial you want.

#10 Joel

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Posted 29 November 2005 - 12:04 AM

QUOTE(Dave @ Nov 28 2005, 11:34 PM)
if you have one factor

you can find the other by synthetically dividing the equation by the factor

View Post




Thats how I would normally find the second factor but I've only ever done it with first factors like x - 2 = 0, x = 2.

What happens when the x term is squared like in this question?

x^2 + 1 = 0
x^2 = -1
x = 1?

Is the 1st factor x = 1?

#11 John

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Posted 29 November 2005 - 07:55 AM

there will be no factor from that, as at Higher Level you can't find the square root of a negative number.

One way to check if there are any "real" factors is to stick the equation you are synthetically dividing into GraphIt or something similar, and see if it cut's the x axis.

#12 The Wedge Effect

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Posted 29 November 2005 - 08:27 AM

Yeah, square roots of negative numbers is not covered until Advanced Highers, in the complex numbers topic.

#13 dfx

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Posted 29 November 2005 - 06:47 PM

Unless you complete the square?

#14 duncad

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Posted 30 November 2005 - 07:14 AM

Q4
When you use synthetic division on a polynomial the bottom line reveals a quadratic. In this case the quadratic is (x^2 + 1).
Therefore the bottom line of the synthetic division is:- 1 0 1 0.
You should be able to deduce that p = 1 and q = -1.
Now work backwards to reveal that the other factor is (x - 1).




#15 duncad

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Posted 30 November 2005 - 02:58 PM

Q12b
Set out your synthetic division showing zero remainder and with the x coefficient = a.
Now work backwards to get the coefficient to be -17.

ie in the second last column you will have a + 8 = -9

#16 Joel

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Posted 30 November 2005 - 08:34 PM

Thanks for the help everyone.






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