Exam questions on solving cubics and quartics and equations

x^3-3x^2+7x-5 , root 1 = 1-2i

• a). Factorise and then state the roots of this cubic

x^3-3x^2+7x-5 , root 1 = 1-2i

• Root 1 = 1-2i , so root 2 = 1+2i (as conjugate).
• (z-(1-2i))(z-(1+2i)) = z^2 - (sum) + (product) = 0 , z^2-(2)z+(1-4i^2) = 0 , z^2-2z+5 , (z^2-2z+10)(z-a)
• a x 5 = -5 , so a=-1 , and therefore (z+1)(1-2i)(1+2i) , and the roots are: -1 , 1+/-2i

x^4-2x^3+14x^2-8x+40 , root 1 = 1+3i

• a). Factorise and state the roots of this quartic

x^4-2x^3+14x^2-8x+40 , root 1 = 1+3i

• Root 1 = 1+3i , so root 2 = 1-3i (as conjugate)
• (z-(1+3i))(z-(1-3i)) , z^2-(sum)+(product)=0 , z^2-2z+10=0 , (z^2-2z+10)(z^2+bz+c) = 0
• 10 x c = 40 , so c=4 , z^2 x z^2 = z^4 , so a=1
• Now use box method to find other values (z^4/z^2 = z^2, so a=1) , (-2z^3=-2z^3, so b=0) , (10x4=40, so c =4)
• (z^2-2z+10)(z^2+4) = 0 , (z^2-2z+10) = (1-3i)(1+3i) , (z^2+4) = (x+2i)(x-2i)
• Factorised = (1-3i)(1+3i)(x+2i)(x-2i)
• Roots = 2i , -2i , 1+/-3i

Question:

• Find the roots of 5-12i
• Lets say root(5-12i) = a+bi , so (a+bi)^2 = 5-12i
• a^2+2bi+b^2i^2 , a^2+2bi^2-b^2 = 5-12i
• Equate real: a^2-b^2 = 5
• Equate Imaginary: 2b = -12 , so b=-6
• a^2 = 5+ (-6)^2 , a^2= 41 , a= root(41)