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

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Question

What are the roots of

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

including complex roots, if they exist?

Answer

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

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

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

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

The root will not be real.

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

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

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Question

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

Answer

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, :

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

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Question

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

Answer

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:

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

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Question

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

Answer

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:

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

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Question

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

Answer

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.

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

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

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Question

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

Answer

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.

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Question

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

Answer

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:

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

Finally, we multiply together all of the parenthesized expressions, which multiplies out to

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Question

The third degree polynomial expression has a real zero at . Find all of the complex zeroes.

Answer

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 equal to zero:

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

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Question

Find all the real and complex zeros of the following equation: $2x^{5}+x^{4}-2x-1$

Answer

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 parentheses equal to zero yields the answers.

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

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Question

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

Answer

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.

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Question

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

Answer

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 $x^{4}+x^{3}-16x^{2}-58x-60=0$

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Question

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

Answer

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

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Question

What are the roots of

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

including complex roots, if they exist?

Answer

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

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

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

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

The root will not be real.

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

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

Compare your answer with the correct one above

Question

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

Answer

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, :

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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.

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

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

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

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

Answer

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:

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

Finally, we multiply together all of the parenthesized expressions, which multiplies out to

Compare your answer with the correct one above

Question

The third degree polynomial expression has a real zero at . Find all of the complex zeroes.

Answer

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 equal to zero:

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

Compare your answer with the correct one above

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