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# Maths Polynomial Questions

### #1

Posted 28 November 2005 - 05:28 PM

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 -, and 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

Posted 28 November 2005 - 10:33 PM

"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

T in the Park 2008!

YAS!

### #3

Posted 28 November 2005 - 10:52 PM

### #4

Posted 28 November 2005 - 10:57 PM

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

i.e. you get

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

Hope that helps a little bit

### #5

Posted 28 November 2005 - 11:03 PM

You want

Now, that something must be of the form , so that the ax term will multiply with the term to give you the .

From that, you should be able to deduce what the 'a' is ( ), 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 in so that (x-1) is a factor - there's an example of a question like this in our notes for Polynomials and Quadratics .

### #6

Posted 28 November 2005 - 11:15 PM

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

Posted 28 November 2005 - 11:20 PM

Not sure about question 4 though...

### #8

Posted 28 November 2005 - 11:34 PM

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

If i am not here i am somewhere else

### #9

Posted 28 November 2005 - 11:42 PM

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?

Yep, that's what I got.

For 4, you could just try multiplying out and then compare what you get with the polynomial you want.

### #10

Posted 29 November 2005 - 12:04 AM

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

Posted 29 November 2005 - 07:55 AM

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

Posted 29 November 2005 - 08:27 AM

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### #13

Posted 29 November 2005 - 06:47 PM

### #14

Posted 30 November 2005 - 07:14 AM

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

Posted 30 November 2005 - 02:58 PM

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

Posted 30 November 2005 - 08:34 PM

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