C2 Chapter 1 question help please!

Hello, I'm stuck on the last 3 questions of my mixed exercise and wondered if anyone could help me!

Q13) The remainder obtained when

x^3 + dx^2 - 5x + 6

is divided by

(x - 1)

is twice the remainder obtained when the same expression is divided by

(x + 1)

I've tried this question a few times and keep getting it wrong, the answer is apparently 18 yet when I sub in 18 the answer isn't twice the remainder of the expression divided by

(x + 1)

my working out must be really wrong somewhere if someone could shed some light please!!

Q14)

a) Show that

(x - 2)

is a factor of

f(x) = x^3 + x^2 - 5x -2

Did this part just fine but part B is where I'm confused, I don't even understand what the question is asking!

b) Hence, or otherwise, find the exact solutions of the equation

f(x) = 0

Q15) Given that

-1

is a root of the equation

2x^3 - 5x^2 - 4x + 3

, find the two positive roots.

Have no idea what I'm meant to do here!!

Thanks a lot

Reply 1

ghostwalker

Original post by popppyss

Hello, I'm stuck on the last 3 questions of my mixed exercise and wondered if anyone could help me!

Q13) The remainder obtained when

x^3 + dx^2 - 5x + 6

is divided by

(x - 1)

is twice the remainder obtained when the same expression is divided by

(x + 1)

I've tried this question a few times and keep getting it wrong, the answer is apparently 18 yet when I sub in 18 the answer isn't twice the remainder of the expression divided by

(x + 1)

my working out must be really wrong somewhere if someone could shed some light please!!

Book's wrong IMHO; d = -18. If you've getting anything else, post some working.

Q14)

a) Show that

(x - 2)

is a factor of

f(x) = x^3 + x^2 - 5x -2

Did this part just fine but part B is where I'm confused, I don't even understand what the question is asking!

b) Hence, or otherwise, find the exact solutions of the equation

f(x) = 0

Q15) Given that

-1

is a root of the equation

2x^3 - 5x^2 - 4x + 3

, find the two positive roots.

Have no idea what I'm meant to do here!!

Thanks a lot

In both of these you are looking to find the other roots (I assume in Q15 it is equal to 0, and you've just missed that part out), so you need to divide by the factor you've been given, and you end up with a quadratic to solve.

Reply 2

JohnnySPal

Q13) Do you know the remainder theorem? The information given tells you that r_1 = f(1), r_2 = f(-1) and r_1 = 2r_2; where r_1 and r_2 are the remainders obtained by dividing by (x-1) and (x+1) respectively. You'll be able to solve these three equations to find the value of d.

Q14) You know that (x-2) is a root. Therefore you can divide f(x) cleanly. Do this to end up with a polynomial of the form (x-2)g(x). g(x) will (obviously?) be a quadratic; and hopefully one that you'll be able to easily factorise (if not, use the quadratic equation). From this expression you can work out the exact solutions of f(x)=0.

Q15) Same principle as Q14. You know that -1 is a root, and therefore that (x+1) is a factor. Rinse, lather, repeat.

(edited 13 years ago)

Reply 3

popppyss

Original post by ghostwalker

...

Original post by JohnnySPal

...

Sorry, I looked back and it is -18 I misread the answer.

I copied out question 15 fully and I haven't been given a factor unless I'm being dence and the factor is -1?

I know the remainder theorem but I think I'm doing it severely wrong on question 13.... here's some of my working, please correct it! I got a different answer 5 times I just can't work out how on earth you get -18?

[br][br]f(1) = (1)^3 + d(1)^2 - 5(1) + 6 [br] 1 + d - 5 + 6 = 0[br] d = 2[br][br]f(-1) = (-1)^3 + d(-1)^2 - 5(-1) + 6[br] -1 + d + 5 + 6 = 0[br] d = - 10

then from there I'd been solving it by simultaneous equations getting - 8 as D which clearly is wrong so could you please put the correct working out for me to look at?

Thanks a lot

Any tips for part B question 14?

(edited 13 years ago)

Reply 4

ghostwalker

Original post by popppyss

[br][br]f(1) = (1)^3 + d(1)^2 - 5(1) + 6 [br] 1 + d - 5 + 6 = 0[br] d = 2[br][br]f(-1) = (-1)^3 + d(-1)^2 - 5(-1) + 6[br] -1 + d + 5 + 6 = 0[br] d = - 10

OK. You're problem is that you've equated each of the values of the function to 0. This would only be valid if (x-1) and (x+1) were factors of f(x), i.e. they gave a remainder of zero when you divided them into f(x), or to put it another way, 1 and -1 were roots of f(x). But that is not the case here.

You're using the REMAINDER THEOREM, not the FACTOR THEOREM.

The remainder obtained when dividing by (x-1) is simply f(1), as you were working out.
And you are told that this is twice the remainder obtained when dividing by (x+1), which is f(-1).

So your initial equation will be f(1) = 2 f(-1).

Just one equation which will involve "d" on both sides and just needs re-arranging to find "d".