+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.
+33616 Solutions as follows:
\(P(x)=2x^3+(2i-1)x^2-(1+i)x-i\)
a). To see if -i is a root, substitute x = -i into P(x)
\(P(-i)=2(-i)^3+(2i-1)(-i)^2-(1+i)(-i)-i\\ P(-i)=2i-2i+1+i-1-i=0\)
P(-i) = 0, so, yes, x = -i is a root.
b). \((x+i)(ax^2+bx+c)\rightarrow ax^3+(b+ai)x^2+(c+bi)x+ci\)
For this to match P(x) we compare powers of x:
\(x^3\quad a=2\\x^2\quad b+ai=2i-1\quad b+2i=2i-1\quad b=-1\\x\quad c+bi=-(1+i)\quad c-i=-1-i\quad c=-1\)
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.
\(2x^2-x-1=0\)
This factorises as
\((x-1)(2x+1)=0\)
so the other two solutions are x = 1 and x = -1/2
+33616 Solutions as follows:
\(P(x)=2x^3+(2i-1)x^2-(1+i)x-i\)
a). To see if -i is a root, substitute x = -i into P(x)
\(P(-i)=2(-i)^3+(2i-1)(-i)^2-(1+i)(-i)-i\\ P(-i)=2i-2i+1+i-1-i=0\)
P(-i) = 0, so, yes, x = -i is a root.
b). \((x+i)(ax^2+bx+c)\rightarrow ax^3+(b+ai)x^2+(c+bi)x+ci\)
For this to match P(x) we compare powers of x:
\(x^3\quad a=2\\x^2\quad b+ai=2i-1\quad b+2i=2i-1\quad b=-1\\x\quad c+bi=-(1+i)\quad c-i=-1-i\quad c=-1\)
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.
\(2x^2-x-1=0\)
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.
