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out of interest what did you get for English? (adv/std?)

Suppose f(x) = x^3 + bx + c, where b and c are constants. Suppose that the equation f(x) = 0 has three distinct roots p,q,r. i) Find p + q + r ii) find p^2 + q^2 + r^2 iii) Since the roots are real and distinct, the graph of y= f(x) has two turning points, at x = u and x = v, and...

thanks again Trebla

Let w = cos(2pi/7) + isin(2pi/7). Prove that 1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0

For part (ii), could w also possibly be -1 or -3 (since they're also factors of 3)?

consider P(x) = ax^4 + bx^3 + cx^2 + dx + e, where a,b,c,d and e are integers. Suppose w is an integer such that P(w) = 0. (i) Prove that w divides e. (ii) Prove that the polynomial Q(x) = 5x^4 + 2x^3 -3x^2 -x + 3 does not have an integer root.

oooh sorry there was a typo in the question. its meant to be (1 - 3w + w^2)(1 + w - 8w^2). But i understand the method now anyway. I just wanted to double check, is the answer 36?

If w is the complex cube root of unity, evaluate (1 - 3w + w^2)(1 + w + 8w^2)

a) Find the least positive integer k such that cos(4pi/9) + isin(4pi/9) is a solution of z^k =1 b) Show that if the complex number w is a solution of z^n =1, then so is w^m, where m and n are arbitrary integers.

oooh i got it thanks guys

oooh. the diagram shows that the arg(z) = arg(conjugate z). So how does that show that -arg(z) = arg(conjugate of z)? is it because arg(z) = -arg(z)

hmmm.. i understood everything you did, except the last part where you said -arg(z) = arg(conjugate z)

I was just wondering, is the arg(1/z) = arg(conjugate of z)? If so, how can you prove this?

reverse chain rule what?

How'd you know the integral of (Cosec x Cot x) = - Cosec x? Is that just a general rule? (It isn't on the standard integrals :()

integrate cosx/(sinx)^2

Thank you, thank you! +Rep :)

Can someone please show me how to integrate: sin^2 x.Cos^2 x answer: 1/8 (x - 0.25sin4x) Thanks :)

You've already got another thread on this.

oooh. at the start of the question, i multiplied the numerator and denominator by -1 because i wanted the leading terms of each to be positive :S As a result, i got integral from 1 to 0 of (2x + 3) -1/(x+2)