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"When factorising something with a multiple root, e.g. P(x)=(x-@)^3 Q(x), is there a fast way to do it (e.g. by inspection)?" Not really... you can put x=1 and x=2 into P(x)=(x-@)^3(ax^3+bx^2+cx+d), but you still need to solve simultaneous equations.

product of roots =A*1\A =1 =7\p when p=7, 7/p = 1 so p=7 satisfies the condition

1. say z = r cis@ then 1/z = 1/r cis(-@) If it's either a square or a rhombus (i.e. all four sides are equal), r=1/r so r=1 In addition, Rhombus: argument of z can be anything except pi/2 or -pi/2 Square: z has argument pi/4, 3pi/4, -pi/4 or -3pi/4 Also I think the quadrilateral is...

1. Draw y=f(x) on the right-half of the coordinate axis (quadrants 1 and 4) 2. Flip everything below x-axis to above (like when you draw y=|(x-1)^2 - 1|), giving you a curve in the first quadrant only 3. Mirror it to the bottom 4. Mirror the new curve to the left (As a check... the curve...

"Find p if the quadractic px^2 + 2x +7 = 0 has roots that are reciprocals" let roots be A and 1\A product of roots: A*1\A=1=7\p .: p=7

(z+1)^4 + 16 = 0 (z+1)^4 = -16 (z+1)^4 = 16 cis(pi+2npi), n integer (z+1) = 16^(1/4) cis[1/4(pi+2npi)] = 2 cis[1/4(pi+2npi)] then put n=0,1,2,3 to get 4 roots in total (1-i)z - 2i = 2 (1-i)z = 2(1+i) z=2(1+i)/(1-i) z=2(1+i)/(1-i) * (1+i)/(1+i) (z-1)^3 + 8(z+i)^3=0...

sum of all complex roots of unity (and of anything actually) = 0 so sum of all complex cube roots of unity = 0 these roots are 1, cos2pi/3+isin2pi/3, cos2pi/3-isin2pi/3 in the special case of roots of unity (1), all of the roots can be expressed as powers of one of the roots with an...

(a+b)(aw+bw2)(aw2+bw) = a3+b3 a3+b3 = (a+b)(a2-ab+b2) so need to prove (aw+bw2)(aw2+bw) = (a2-ab+b2) LHS = a2w3 + abw4 + abw2 + b2w3 = a2w3 + abw(w3) + abw2 + b2w3 then on using w3=1, = a2 + abw + abw2 + b2 = a2 + ab(w + w2) + b2 but w+w2+w3=0 (sum of roots = 0) so w+w2=-1...