Find the roots of a complex square polynomial

Question

I'm stuck with this exercise and would welcome some help. I've been asked to:

Find the roots in $\mathbb{C}$ and factorize in $\mathbb{C[x]}$ the polynomial $x^2 + (3+2i)x + 5 + i$

I tried many things. The first attempt was to see if I could factorize it just by looking at it. Then, I tried the formula $x=\frac{-b \pm w}{2a}$, where $w^2= b^2-4ac$

I did that and got $w^2= 8i + 15$ and then wrote $w= c+di$, since it's a complex number, and wrote a system of equations:

$a^2-b^2= -15$

$2ab= 8$

$a^2 + b^2= 289$

With that system I get that $a=\pm\sqrt{137}$, with that information I get $b$ and then go back to my previous equation with $x$, but that is already far from the answer, that is: $x_1=-1+i$, $x_2=-2-3i$

There's something obviously wrong in my reasoning. Could someone help me spot it? Thanks a lot.

Answer 1

You should have got $$ b^2-4ac=(3+2i)^2-4(5+i)=9+12i-4-20-4i=8i-15 $$ Now, setting $8i-15=(r+is)^2$ you get $$ \begin{cases} r^2-s^2=-15 \\[4px] 2rs=8 \\[4px] r^2+s^2=\sqrt{8^2+15^2}=\sqrt{289}=17 \end{cases} $$ Thus $2r^2=2$ and $r=\pm1$, together with $s=4/r$, so the square roots are $w_1=1+4i$ and $w_2=-1-4i$.

You can compute the roots as $$ \frac{-b+w_1}{2} \qquad\qquad \frac{-b+w_2}{2} $$ Notice $-b$, not $-b^2$.

Answer 2

after the quadratic formula we have $$x_{1,2}-\frac{3+2i}{2}\pm\sqrt{\left(\frac{3+2i}{2}\right)^2-(5+i)}$$ after simplification we get $$x_1=-2-3i$$ $$x_2=-1+i$$

Answer 3

Hint Note that $b^2-4ac=-15+8i=1^2+2\cdot (4 i)\cdot 1+(4i)^2=(\pm(1+4i))^2$.

Answer 4

Just use the standard formula for solving a quadratic equation.

It works for complex coefficients the same as real.

Answer 5

$-2-3i+(-1+i)=-3-2i$ and $(-2-3i)(-1+1)=2-2i+3i+3=5+i$.

Thus, by the Viete's theorem we obtain the answer: $\{-2-3i,-1+i\}$

and $x^2 + (3+2i)x + 5 + i=1(x-x_1)(x-x_2)=(x+2+3i)(x+1-i)$.

Answer 6

You have a mistake (of sign) in $w^2$: $$ w^2=b^2-4ac=(9-4+12i)-(20+4i)=8i-15=(4i+1)^2 $$