Riemann surface
matlab

Problem 2. Find the point of intersection, if there is one, between the following lines:
(a) $y=-x+5$ and $y-2=3(x+1)$
(b) The line passing through $(-1,-2)$ and the origin, and the line $y=2 x-2$.

Solution: (a) First, we write both lines in slope-intercept form,
$$y=-x+5 \quad y=3 x+5$$
If $(x, y)$ is a point of intersection of the lines, it must satisfy both equations. Assuming $(x, y)$ is as such, we have that
$$\begin{gathered} -x+5=3 x+5 \ -x=3 x \ x=0 . \end{gathered}$$
Thus, $x=0$. To find $y$, we can plug $x=0$ into either one of the original equations, and get that $y=5$. Thus, $(0,5)$ is the (unique) point of intersection.
(b) The line passing through $(-1,-2)$ and the origin has slope
$$m=\frac{0-(-2)}{0-(-1)}=2,$$
and can be expressed by the equation $y=2 x$. But, this line is parallel to (and distinct from) the line $y=2 x-2$, so they cannot have any points of intersection.

Problem 3. Find all real roots $x$ of the following polynomials, and factor into irreducible polynomials.
(a) $6 x^2+5 x+1$
(b) $-x^2+x+1$
(c) $2 x^2-3 x+5$
(d) $x^3+6 x^2-7 x$
(e) $x^3-x^2+x-1$
(f) $x^4-2 x^2+1$

Solution: Note that a polynomial is irreducible if it cannot be factored into non-constant polynomials with real coefficients.
(a)
\begin{aligned} 6 x^2+5 x+1 & =6 x^2+3 x+2 x+1 \ & =3 x(2 x+1)+1(2 x+1) \ & =(3 x+1)(2 x+1) . \end{aligned}
This factors the polynomial into irreducibles, and shows that its roots are $x=-\frac{1}{3}$ and $x=-\frac{1}{2}$.
(b) We use the quadratic formula:
\begin{aligned} x & =\frac{-1 \pm \sqrt{1^2-4(-1)(1)}}{2(-1)} \ & =\frac{-1 \pm \sqrt{1+4}}{-2} \ & =\frac{1 \pm \sqrt{5}}{2}, \end{aligned}

Thus, $x=\frac{1 \pm \sqrt{5}}{2}$ are the two real roots of the polynomial. It follows that the polynomial factors as
$$-x^2+x+1=-\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)$$
(c) We use the quadratic formula:
\begin{aligned} x & =\frac{-(-3) \pm \sqrt{(-3)^2-4(2)(5)}}{2(2)} \ & =\frac{3 \pm \sqrt{-31}}{4}, \end{aligned}
which cannot be real. Thus, the polynomial has no real roots, and cannot be factored further (a polynomial of degree 2 or 3 is irreducible if and only if it has no roots).
(d)
\begin{aligned} x^3+6 x^2-7 x & =x\left(x^2+6 x-7\right) \ & =x\left(x^2-x+7 x-7\right) \ & =x(x(x-1)+7(x-1)) \ & =x(x+7)(x-1) \end{aligned}
This factors the polynomial into irreducibles, and shows that the its roots are $x=0, x=-7$ and $x=1$.
(e) It is easy to see that $x^3-x^2+x-1$ has root $x=1$, since
$$(1)^3-(1)^2+1-1=1-1=0$$
So, we can factor out an $(x-1)$. Using polynomial long division,
\begin{aligned} & x-1) \frac{x^2+1}{x^3-x^2+x-1} \ & -x^3+x^2 \ & x-1 \ & \frac{-x+1}{0} \ & \end{aligned}

we get that
$$x^3-x^2+x-1=(x-1)\left(x^2+1\right)$$
and $x^2+1$ has no real roots since $x^2+1>0$ for all $x \in \mathbb{R}$. Thus, this factors the polynomial into irreducibles, and the only real root is $x=1$.
(f) Let $z=x^2$, then
\begin{aligned} x^4-2 x^2+1 & =z^2-2 z+1 \ & =(z-1)(z-1) \ & =\left(x^2-1\right)\left(x^2-1\right) \ & =(x-1)(x+1)(x-1)(x+1) . \end{aligned}
This factors the polynomial into irreducibles, and shows that its roots are $x= \pm 1$.
Problem 4. Solve the following equations for $x$.

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