Pre-Calculus - Find Complex Zeros of a Polynomial Using the Fundamental Theorem of Algebra

Find a fourth-degree polynomial whose zeroes are , and $-2\pm \sqrt{2i}$

$x^{4}+x^{3}-16x^{2}-58x-60$

$x^{4}-x^{3}-16x^{2}-58x-60$

Explanation

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

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

If the real zero of the polynomial is 3, what are the complex zeros?

$-4 \pm 3i$

$-8 \pm 6i$

$4 \pm 4i$

$-4 \pm 4i$

$8 \pm 5i$

Explanation

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

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

This gives us a second factor of which we can solve using the quadratic formula:

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

The third-degree polynomial expression $f(x)=x^{3}-x^{2}+4x-4$ has a real zero at . Find all of the complex zeroes.

Explanation

First, factor the expression by grouping: $x^{3}-x^{2}+4x-4\rightarrow x^{2}(x-1)+4(x-1)\rightarrow (x^{2}+4)\cdot (x-1)$

To find the complex zeroes, set the term $(x^{2}+4)$ equal to zero: $x^{2}=-4\rightarrow x=\pm \sqrt{-2}=\pm 2i$

Find all the zeroes of the following equation and their multiplicity: $g(t)=t^{5}-6t^{3}+9t$

$t=0,\pm \sqrt{3}$ (multiplicity of 1 on 0, multiplicity of 2 on $\pm \sqrt{3}$ )

$t=0,\pm \sqrt{3}$ (multiplicity of 2 on 0, multiplicity of 1 on $\pm \sqrt{3}$ )

$t=0,\pm \sqrt{3i}$ (multiplicity of 2 on 0, multiplicity of 1 on $\pm \sqrt{3i}$ )

$t=0,\pm \sqrt{3i}$ (multiplicity of 1 on 0, multiplicity of 2 on $\pm \sqrt{3i}$ )

Explanation

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

This equation has a solution as two values: when , and when $\left ( t^{2}-3 \right )^{2}=0\rightarrow t=\pm \sqrt{3}$ . 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.

The polynomial has a real zero at 1.5. Find the other two zeros.

$x = 1 \pm 5i$

$x = 1 \pm \frac{\sqrt{108}}{2}$

$x = 1 \pm 10 i$

$x = 2 \pm 5 i$

Explanation

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 . We can figure out what this is this way:

multiply both sides by 2

is the factor

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

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

To finish solving, we can use the quadratic formula with the resulting quadratic, :

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

Find all the zeroes of the following equation and their multiplicity:

$t=0, \pm \sqrt{3}$ (multiplicity of 1 on 0, multiplicity of 2 on $\pm \sqrt{3}$

$t=0, \pm \sqrt{3}$ (multiplicity of 2 on 0, multiplicity of 1 on $\pm \sqrt{3}$

$t=0, \pm \sqrt{3i}$ (multiplicity of 2 on 0, multiplicity of 1 on $\pm \sqrt{3i}$

$t=0, \pm \sqrt{3i}$ (multiplicity of 1 on 0, multiplicity of 2 on $\pm \sqrt{3i}$

Explanation

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

This equation has solutions at two values: when and when

$(t^2-3)^2=0\rightarrow t=\pm \sqrt{3}$ 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.

Find all the real and complex zeroes of the following equation:

$x=1, \frac{-1}{2}$

$x=1,\frac{-1}{2}, i$

$x=1, \frac{-1}{2}, \pm i$

$x=1, -1, \frac{-1}{2}, \pm i$

$x=1, -1, \frac{-1}{2}, i$

Explanation

First, factorize the equation using grouping of common terms:

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

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

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

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

The polynomial intersects the x-axis at point . Find the other two solutions.

$-1 \pm 2i$

$-1 \pm 5 i$

$-2 \pm 4i$

$- 1 \pm 4i$

$1 \pm 4i$

Explanation

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

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

This gives us the second factor of . We can get our solutions by using the quadratic formula:

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

Find a fourth degree polynomial whose zeroes are -2, 5, and $-2\pm \sqrt{2i}$

Explanation

This one is a bit of a journey. The expressions for the first two zeroes are easily calculated, and respectively. The last expression must be broken up into two equations: