One of the roots is $\displaystyle \small x=1$ because if we plug in 1, we get 0. We can factor the polynomial as

$\displaystyle \small \small f(x)=x^3-x^2+2x-2=(x-1)(x^2+2)$

So now we solve the roots of $\displaystyle \small x^2+2$.

$\displaystyle x^2=-2 \rightarrow x= \pm\sqrt{-2}= \pm i\sqrt2$

The root will not be real.

The roots of this polynomial are $\displaystyle \small x=\sqrt{2}i, -\sqrt{2}i$.

So, the roots are $\displaystyle \small x=1, \sqrt{2}i, -\sqrt{2}i$

If this polynomial has a real zero at 1.5, that means that the polynomial has a factor that when set equal to zero has a solution of $\displaystyle x = 1.5$. We can figure out what this is this way:

$\displaystyle x - 1.5 = 0$ multiply both sides by 2

$\displaystyle 2x - 3 = 0$ is the factor

Now that we have one factor, we can divide to find the other two solutions:

$\displaystyle \begin{matrix} x^2 - 2x + 26\\ 2x - 3 \overline{| 2x^3 - 7 x^2 + 58 x - 78} \\ \underline{2x^3 - 3x^2} \enspace \enspace \enspace \enspace \\ \enspace \enspace \enspace \enspace -4x^2 + 58 x \\ \enspace \enspace \enspace \enspace \underline{-4x^2 + 6x} \\ \enspace \enspace \enspace \enspace \enspace \enspace \enspace52x - 78 \\ \enspace \enspace \enspace \enspace \enspace \enspace \enspace \underline{52x - 78} \\ \enspace \enspace \enspace \enspace \enspace \enspace \enspace 0 \end{matrix}$

To finish solving, we can use the quadratic formula with the resulting quadratic, $\displaystyle x^2 - 2x + 26$: 

$\displaystyle x = \frac{2 \pm \sqrt {4 - 4(1)(26)}}{2(1)} = \frac{2 \pm \sqrt{-100}}{2}= \frac{2 \pm 10i }{2 } = 1 \pm 5i$

We know that the real zero of this polynomial is 3, so one of the factors must be $\displaystyle x - 3$. To find the other factors, we can divide the original polynomial by $\displaystyle x - 3$, either by long division or synthetic division:

$\displaystyle \begin{matrix} \underline{3}| \enspace 1 \enspace 5 \enspace 1 \enspace -75 \\\underline{ + \enspace \enspace \enspace 3 \enspace 24 \enspace \enspace 75 }\\ \enspace \enspace 1 \enspace 8 \enspace 25 \enspace \enspace \enspace 0 \end{matrix}$

This gives us a second factor of $\displaystyle x^2 + 8 x + 25$ which we can solve using the quadratic formula:

$\displaystyle x = \frac{-8 \pm \sqrt{64 - 4(1)(25)}}{2} = \frac{-8 \pm \sqrt{-36}}{ 2 } = \frac{-8 \pm 6i }{2} = -4 \pm 3i$

Since we know that one of the zeros of this polynomial is 3, we know that one of the factors is $\displaystyle x - 3$. To find the other two zeros, we can divide the original polynomial by $\displaystyle x - 3$, either with long division or with synthetic division:

$\displaystyle \begin{matrix} \underline{3}| \enspace 1 \enspace -1 \enspace -1 \enspace -15 \\\underline{ + \enspace \enspace \enspace \enspace 3 \enspace \enspace \enspace \enspace 6 \enspace \enspace \enspace 15 }\\ 1 \enspace \enspace \enspace 2 \enspace \enspace \enspace 5 \enspace \enspace \enspace 0 \end{matrix}$

This gives us the second factor of $\displaystyle 1x^2 + 2 x + 5$. We can get our solutions by using the quadratic formula:

$\displaystyle \\x = \frac{-2 \pm \sqrt{4 - 4(1)(5)}}{2} \\ \\= \frac{-2 \pm \sqrt{ 4 - 20 }}{2 } \\ \\= \frac{-2 \pm \sqrt{-16}}{2} \\ \\= \frac{-2 \pm 4i }{2}\\ \\ = -1 \pm 2i$

First, factorize the equation using grouping of common terms:

$\displaystyle 2x^5+x^4-2x-1\rightarrow x^4(2x+1)-1(2x+1)\rightarrow (x^4-1)(2x+1)\rightarrow (x+1)(x-1)(x^2+1)(2x+1)$

Next, setting each expression in parenthesis equal to zero yields the answers.

$\displaystyle (x+1)=0\rightarrow x=-1$ $\displaystyle (x-1)=0\rightarrow x=1$ $\displaystyle (x^2+1)=0\rightarrow x=\pm i$ 

$\displaystyle (2x+1)=0\rightarrow x=\frac{-1}{2}$

First, pull out the common t and then factorize using quadratic factoring rules:

This equation has solutions at two values: when $\displaystyle t=0$ and when 

$\displaystyle (t^2-3)^2=0\rightarrow t=\pm \sqrt{3}$  Therefore, But since the degree on the former equation is one and the degree on the latter equation is two, the multiplicities are 1 and 2 respectively.

This one is a bit of a journey. The expressions for the first two zeroes are easily calculated, $\displaystyle (x+2)$ and $\displaystyle (x-5)$ respectively. The last expression must be broken up into two equations:

 which are then set equal to zero to yield the expressions $\displaystyle (x+2-i\sqrt{2})$ and $\displaystyle (x+2+i\sqrt{2})$

Finally, we multiply together all of the parenthesized expressions, which multiplies out to $\displaystyle x^4+x^3-16x^2-58x-60=0$

First, factor the expression by grouping:

$\displaystyle x^3-x^2+4x-4\rightarrow x^2(x-1)+4(x-1)\rightarrow (x^2+4)\cdot(x-1)$

To find the complex zeroes, set the term $\displaystyle (x^2+4)$ equal to zero:

$\displaystyle x^2=-4\rightarrow x=\pm \sqrt{-2}=\pm 2i$

First, factorize the equation using grouping of common terms:

$\displaystyle 2x^{5}+x^{4}-2x-1\rightarrow x^{4}(2x+1)-1(2x+1)\rightarrow (x^{4}-1)(2x+1)\rightarrow (x+1)(x-1)(x^{2}+1)(2x+1)$

Next, setting each expression in parentheses equal to zero yields the answers.

$\displaystyle (x+1)=0\rightarrow x=-1(x-1)=0\rightarrow x=1(x^{2}+1)=0\rightarrow x=\pm i{(2x+1)=0\rightarrow x=\frac{-1}{2}}$

First, pull out the common t and then factorize using quadratic factoring rules: $\displaystyle t^{5}-6t^{3}+9t\rightarrow t(t^{4}-6t^{2}+9)\rightarrow t(t^{2}-3)^{2}$

This equation has a solution as two values: when $\displaystyle t=0$, and when $\displaystyle \left ( t^{2}-3 \right )^{2}=0\rightarrow t=\pm \sqrt{3}$. Therefore, But since the degree on the former equation is one and the degree on the latter equation is two, the multiplicities are 1 and 2 respectively.