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The Fundamental Theorem of Algebra: If P(x) is a polynomial of degree n ≥ 1, then P(x) = 0 has exactly n roots, including multiple and complex roots. |
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In plain English, this theorem says that the degree of a polynomial equation tells you how many roots the equation will have.
A linear equation (degree 1) will have one root.
A quadratic equation (degree 2) will have two roots.
A cubic equation (degree 3) will have three roots.
An nth degree polynomial equation will have n roots. |
 Read carefully to fully understand what this theorem is saying.
• "including multiple... roots" - if a polynomial has a repeated root, each repetition of the root is counted.
• "including... complex roots" - the term "complex" is referring to "complex roots with a non-zero imaginary part" (a + bi with b ≠ 0) which implies the conjugate of the complex root will also be counted since such complex roots come in conjugate pairs.
Yes, the Complex Numbers include the Real Numbers, but in this reference, "complex" includes the imaginary component. 
Under Quadratics, we saw how the Fundamental Theorem of Algebra applied to quadratic equations.
Let's extend that usage to polynomial equations in general.
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Determine the Number of Roots |
According to the Fundamental Theorem of Algebra, how many roots does the equation
3x5 + 2x3 - x + 6 = 0 possess? |
Solution: Since the degree of the equation is 5, there will be 5 roots. |
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Determine the Zeros and Degree |
The factored form of a polynomial is given to be P(x) = (x + 2)(x - 3)(x + 1)(x - 2). What are the zeros of the polynomial function, and what is the degree of the polynomial function? |
Solution: Set the factors equal to zero to find the roots (zeros). The roots are -2, 3, -1, and 2. The degree of the polynomial will be degree four. |
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Explain |
Explain why a polynomial of even degree could have no real roots. |
Solution: Remember that the Fundamental Theorem of Algebra includes complex roots with non-zero imaginary parts (which come in conjugate pairs). Since these complex roots come in pairs, there will always be an even number of such roots in total. It is possible that a polynomial of even degree may have only pairs of complex roots, and no real roots. |
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Linear Factors |
How many linear factors does the polynomial P(x) = x5 - 3x4 + 2x2 + x - 1 possess? |
Solution: A factor of the form (x - a) is said to be a linear factor since it is of degree 1. Since the degree of this polynomial is five, we will have 5 roots (zeros) from five factors of the form (x - a). Thus we have five linear factors, allowing for the possibility of a being a complex number. |
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Finding Roots |
Find the roots (zeros) of the polynomial function P(x) = x4 - 16. |
Set the function equal to zero. Since the degree of this polynomial is 4, we will be looking for four zeros. |
x4 - 16 = 0 |
Factor. This problem is dealing with the difference of squares. |
(x2 - 4)(x2 + 4) = 0 |
Factor again. |
(x - 2)(x + 2)(x2 + 4) = 0 |
Set factors equal to zero and solve for x.
Remember that complex roots (with non-zero imaginary parts) come in conjugate pairs. |
x - 2 = 0 → x = 2
x + 2 = 0
→ x = -2
x2 + 4 = 0 → x2 = -4 → x = ±2i
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Solution: |
Roots: 2, -2, 2i, -2i |
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Creating the Equation |
Write an equation in standard form whose zeros are 4 and -2, has a degree of 3, and the root -2 has a multiplicity of 2. |
If the zeros are 4 and -2, then the polynomial will have the factors of (x - 4) and (x + 2). |
Factors: (x - 4)(x + 2) |
If the degree is 3, there will be three factors by the Fundamental Theorem of Algebra. |
Factors: (x - 4)(x + 2)(x - ?) |
If the multiplicity of the root -2 is 2, then there will be two factors of (x + 2). |
Factors: (x - 4)(x + 2)(x + 2) |
One Equation Solution: |
x3 - 12x - 16 = 0 |
| Note: As we saw in Algebra 1, when creating an equation, given certain information, you may not be creating the ONLY possible equation. In this example, an equation such as 2(x3 - 12x - 16) = 0 would also possess these characteristics. |
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