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.

This one is a bit of a journey. The expressions for the first two zeroes are easily calculated, $\displaystyle \left ( x+2 \right )$ and $\displaystyle \left ( x-5 \right )$ respectively. The last expression must be broken up into two equations: $\displaystyle x=-2-i\sqrt{2},x=-2+i\sqrt{2}$ which are then set equal to zero to yield the expressions $\displaystyle \left ( x+2-i\sqrt{2} \right )$ and $\displaystyle \left ( x+2+i\sqrt{2} \right )$

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

To determine zeros of $\displaystyle y=5x^2+14x-3$, factorize the polynomial.

$\displaystyle 5x^2+14x-3=(5x-1)(x+3)$

Set each of the factorized components equal to zero and solve for $\displaystyle x$.

$\displaystyle 5x-1=0$

$\displaystyle x=\frac{1}{5}$

$\displaystyle x+3=0$

$\displaystyle x=-3$

The sum of the roots:

$\displaystyle \frac{1}{5}+(-3)= \frac{1}{5}-\frac{15}{5}= -\frac{14}{5}$

The product of the roots:

$\displaystyle (\frac{1}{5})(-3)= -\frac{3}{5}$