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Precalculus : Fundamental Theorem of Algebra

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Example Questions

Example Question #11 : Express A Polynomial As A Product Of Linear Factors

Factorize the following expression completely to its linear factors: $f(x)=2x^{2}+3xy-10x-15y$

Possible Answers:

$\left ( x-5 \right )\cdot \left ( x+y \right )$

$(x-5)\cdot (2x-y)$

$(x+5)\cdot (2x+y)$

$\left ( x+5 \right )\cdot \left ( 2x+y \right )$

$(x-5)\cdot (2x+3y)$

Correct answer:

$(x-5)\cdot (2x+3y)$

Explanation:

Use the grouping method to factorize common terms: $2x^{2}+3xy-10x-15y\rightarrow x(2x+3y)-5(2x+3y)\rightarrow (x-5\)\cdot \left ( 2x+3y \right )$

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Example Question #11 : Express A Polynomial As A Product Of Linear Factors

Factorize the following polynomial expression completely to its linear factors: $f(x)=10x^{2}-4x+15xy-6y$

Possible Answers:

$(2x+3)\cdot (5x-2)$

$(2x+3y\cdot (5x+2)$

$(2x+3y)\cdot (5x-2)$

$(2x-3y\cdot (5x+2)$

Correct answer:

$(2x+3y)\cdot (5x-2)$

Explanation:

Use the grouping method to factorize common terms: $10x^{2}-4x+15xy-6y\rightarrow 2x(5x-2)\rightarrow (2x+3y)\cdot (5x-2)$

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Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

What are the roots of

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

including complex roots, if they exist?

Possible Answers:

$\small x=1, \sqrt{2}i, -\sqrt{2}i$

$\small x=1,-1,\sqrt{2}$

$\small x=1$

$\small x=1, \sqrt{2}, -\sqrt{2}$

Correct answer:

$\small x=1, \sqrt{2}i, -\sqrt{2}i$

Explanation:

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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Example Question #2 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

The polynomial y = 2x^3 - 7 x ^ 2 + 58 x - 78 has a real zero at 1.5. Find the other two zeros.

Possible Answers:

$x = 2 \pm 5 i$

x = 6, -4

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

$x = 1 \pm 5i$

$x = 1 \pm 10 i$

Correct answer:

$x = 1 \pm 5i$

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 x = 1.5 . We can figure out what this is this way:

x - 1.5 = 0 multiply both sides by 2

2x - 3 = 0 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^2 - 2x + 26 :

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

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Example Question #3 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

If the real zero of the polynomial y = x^3 + 5 x ^2 + x - 75 is 3, what are the complex zeros?

Possible Answers:

$8 \pm 5i$

$-8 \pm 6i$

$-4 \pm 4i$

$4 \pm 4i$

$-4 \pm 3i$

Correct answer:

$-4 \pm 3i$

Explanation:

We know that the real zero of this polynomial is 3, so one of the factors must be x - 3 . To find the other factors, we can divide the original polynomial by x - 3 , 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 x^2 + 8 x + 25 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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Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

The polynomial y = x^3 - x ^ 2 - x - 15 intersects the x-axis at point (3,0) . Find the other two solutions.

Possible Answers:

$1 \pm 4i$

$-1 \pm 2i$

$- 1 \pm 4i$

$-1 \pm 5 i$

$-2 \pm 4i$

Correct answer:

$-1 \pm 2i$

Explanation:

Since we know that one of the zeros of this polynomial is 3, we know that one of the factors is x - 3 . To find the other two zeros, we can divide the original polynomial by x - 3 , 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 1x^2 + 2 x + 5 . 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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Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Find all the real and complex zeroes of the following equation: 2x^5+x^4-2x-1

Possible Answers:

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

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

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

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

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

Correct answer:

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

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.

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

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

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Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

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

Possible Answers:

$t=0, \pm \sqrt{3i}$ (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 $\pm \sqrt{3i}$

$t=0, \pm \sqrt{3}$ (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 $\pm \sqrt{3}$

Correct answer:

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

Explanation:

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

This equation has solutions at two values: when t=0 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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Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

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

Possible Answers:

x^4-16x^2-58x-60

x^3-16x^2-58x-60

x^4-x^3-16x^2-58x-60

x^4+x^3+16x^2+58x-60

x^4+x^3-16x^2-58x-60

Correct answer:

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, (x+2) and (x-5) 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 x^4+x^3-16x^2-58x-60=0

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Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

The third degree polynomial expression f(x)=x^3-x^2+4x-4 has a real zero at x=1 . Find all of the complex zeroes.

Possible Answers:

$x=\pm \sqrt{2}$

$x=\pm 2i$

$x=2\pm \sqrt{2i}$

$x=\pm 4i$

$x=2\pm 2i$

Correct answer:

$x=\pm 2i$

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:

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

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Precalculus : Fundamental Theorem of Algebra

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #11 : Express A Polynomial As A Product Of Linear Factors

Example Question #11 : Express A Polynomial As A Product Of Linear Factors

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #2 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #3 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

Example Question #1 : Find Complex Zeros Of A Polynomial Using The Fundamental Theorem Of Algebra

All Precalculus Resources

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