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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.

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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