+870 Let P(x)=2x3+(2i-1)x2-(1+i)x-i
(i being the imaginary unit, i²=-1)
N.B.: This exercise was given in the baccalaureat, which is the French equivalent of an A level.
+33654 Solutions as follows:
P(x)=2x3+(2i−1)x2−(1+i)x−i
a). To see if -i is a root, substitute x = -i into P(x)
P(−i)=2(−i)3+(2i−1)(−i)2−(1+i)(−i)−iP(−i)=2i−2i+1+i−1−i=0
P(-i) = 0, so, yes, x = -i is a root.
b). (x+i)(ax2+bx+c)→ax3+(b+ai)x2+(c+bi)x+ci
For this to match P(x) we compare powers of x:
x3a=2x2b+ai=2i−1b+2i=2i−1b=−1xc+bi=−(1+i)c−i=−1−ic=−1
c). We already know that x = -i is one solution, so we just need to solve the quadratic ax^2 + bx + c =0 to find the others.
2x2−x−1=0
This factorises as
(x−1)(2x+1)=0
so the other two solutions are x = 1 and x = -1/2
+33654 Solutions as follows:
P(x)=2x3+(2i−1)x2−(1+i)x−i
a). To see if -i is a root, substitute x = -i into P(x)
P(−i)=2(−i)3+(2i−1)(−i)2−(1+i)(−i)−iP(−i)=2i−2i+1+i−1−i=0
P(-i) = 0, so, yes, x = -i is a root.
b). (x+i)(ax2+bx+c)→ax3+(b+ai)x2+(c+bi)x+ci
For this to match P(x) we compare powers of x:
x3a=2x2b+ai=2i−1b+2i=2i−1b=−1xc+bi=−(1+i)c−i=−1−ic=−1
c). We already know that x = -i is one solution, so we just need to solve the quadratic ax^2 + bx + c =0 to find the others.
2x2−x−1=0
This factorises as
(x−1)(2x+1)=0
so the other two solutions are x = 1 and x = -1/2
+870 Perfect Alan; your answers are totally exact; you deserved 20/20 and a brownie for your answer.
