Suppose $p(x)$ Is a real polynomial, find $a$ and $b$ and factorize

Question

having some troubles with this question:

Suppose $p(x) = x^3 + ax^2 + x + b$ is a real polynomial with $2 + i$ as a zero, find $a$ and $b$ and factorize $p(x)$ into a real linear and irriductable quadratic factors.

Any pointers of where to go from here would be a great help

Answer 1

By the fundamental theorem of Algebra (in this context, polynomials with real coefficients have complex roots appear in conjugate), trivially $2-i$ is also a zero. And by Vieta's theorem, taking the root other than the conjugate pair (must be real) be $k$, one must get: $$k(2-i)+(2+i)(2-i)+k(2+i)=1$$ $$4k+5=1$$ $$k=-1$$ Thus, we could answer the second question first: $$P(x)=(x+1)(x-2+i)(x-2-i)$$ $$=(x+1)(x^2-4x+5)$$ And again by comparing the coefficient (or straightly Vieta's theorem), we could get $$-a=4+k=3\Rightarrow a=-3$$ $$-b=5k=-5\Rightarrow b=5$$

Remarks:

Vieta's theorem of polynomials of degree $3$:

For the three roots of $P(x)=x^3+ax^2+bx+c$, $i, j, k\in\mathbb C$, notice that the polynomials could be expressed in the form $$P(x)=(x-i)(x-j)(x-k)$$ Expanding it, $$P(x)=x^3+(-(i+j+k))x^2+(ij+ik+jk)x+(-ijk)$$ Comparing the coefficient, $$-a=i+j+k, b=ij+ik+jk, -c=ijk$$

Answer 2

Hint:

If $2+i$ is a root, $2-i$ is also a root.

$$p(x)=(x-k)(x-2-i)(x-2+i)$$

$(x-2-i)(x-2+i)=(x-2)^2+1$.

comparing coefficients and obtain the answer.

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\begin{align} x^3+ax^2+x+b&=(x-k)(x^2-4x+5)\\ &=x^3-(k+4)x^2+(4k+5)x-5k \end{align}

Therefore, $-k-4=a$, $4k+5=1$ and $-5k=b$

Answer 3

HINT: if $$2+i$$ is a solution of $$p(x)=0$$ then $$x=2-i$$ is also a solution of $$p(x)=0$$