Fundamental Theorem of Algebra


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Obstacles Resolvedβ€”A Surprising Result

Student Outcomes

  • Students understand the Fundamental Theorem of Algebra; that all polynomial expressions factor into linear terms in the realm of complex numbers. Consequences, in particular, for quadratic and cubic equations are understood.

New York State Common Core Math Algebra II, Module 1, Lesson 40

Algebra 2 Engage NY Lesson 40, Part 1 of 2

Algebra 2 Engage NY Lesson 40, Part 2 of 2

Classwork

Opening Exercise
Write each of the following quadratic expressions as a product of linear factors. Verify that the factored form is equivalent.
a. π‘₯2 + 12π‘₯ + 27
b. π‘₯2 βˆ’ 16
c. π‘₯2 + 16
d. π‘₯2 + 4π‘₯ + 5

Example 1
Consider the polynomial 𝑃(π‘₯) = π‘₯3 + 3π‘₯2 + π‘₯ βˆ’ 5 whose graph is shown to the right.
a. Looking at the graph, how do we know that there is only one real solution?
b. Is it possible for a cubic polynomial function to have no zeros?
c. From the graph, what appears to be one solution to the equation π‘₯3 + 3π‘₯2 + π‘₯ βˆ’ 5 = 0?
d. How can we verify that this value is a solution?
e. According to the remainder theorem, what is one factor of the cubic expression π‘₯3 + 3π‘₯2 + π‘₯ βˆ’ 5?
f. Factor out the expression you found in part (e) from π‘₯3 + 3π‘₯2 + π‘₯ βˆ’ 5.
g. What are all of the solutions to π‘₯3 + 3π‘₯2 + π‘₯ βˆ’ 5 = 0?
h. Write the expression π‘₯3 + 3π‘₯2 + π‘₯ βˆ’ 5 in terms of linear factors.




Exercises 1–2
Write each polynomial in terms of linear factors. The graph of 𝑦 = π‘₯3 βˆ’ 3π‘₯2 + 4π‘₯ βˆ’ 12 is provided for Exercise 2.

  1. 𝑓(π‘₯) = π‘₯3 + 5π‘₯
  2. 𝑔(π‘₯) = π‘₯3 βˆ’ 3π‘₯2 + 4π‘₯ βˆ’12

Example 2
Consider the polynomial function 𝑃(π‘₯) = π‘₯4 βˆ’ 3π‘₯3 + 6π‘₯2 βˆ’ 12π‘₯ + 8, whose corresponding graph 𝑦 = π‘₯4 βˆ’ 3π‘₯3 + 6π‘₯2 βˆ’ 12π‘₯ + 8 is shown to the right. How many zeros does 𝑃 have?
a. Part 1 of the fundamental theorem of algebra says that this equation will have at least one solution in the complex numbers. How does this align with what we can see in the graph to the right?
b. Identify one zero from the graph.
c. Use polynomial division to factor out one linear term from the expression π‘₯4 βˆ’ 3π‘₯3 + 6π‘₯2 βˆ’ 12π‘₯ + 8.
d. Now we have a cubic polynomial to factor. We know by part 1 of the fundamental theorem of algebra that a polynomial function will have at least one real zero. What is that zero in this case?
e. Use polynomial division to factor out another linear term of π‘₯4 βˆ’ 3π‘₯3 + 6π‘₯2 βˆ’ 12π‘₯ + 8.
f. Are we done? Can we factor this polynomial any further?
g. Now that the polynomial is in factored form, we can quickly see how many solutions there are to the original equation π‘₯4 βˆ’ 3π‘₯3 + 6π‘₯2 βˆ’ 12π‘₯ + 8 = 0.
h. What if we had started with a polynomial function of degree 8?

Lesson Summary
Every polynomial function of degree 𝑛, for 𝑛 β‰₯ 1, has 𝑛 roots over the complex numbers, counted with multiplicity.
Therefore, such polynomials can always be factored into 𝑛 linear factors, and the obstacles to factoring we saw before have all disappeared in the larger context of allowing solutions to be complex numbers.

The Fundamental Theorem of Algebra:

  1. If 𝑃 is a polynomial function of degree 𝑛 β‰₯ 1, with real or complex coefficients, then there exists at least one number π‘Ÿ (real or complex) such that 𝑃(π‘Ÿ) = 0.
  2. If 𝑃 is a polynomial function of degree 𝑛 β‰₯ 1, given by 𝑃(π‘₯) = π‘Žπ‘›π‘₯𝑛 + π‘Žπ‘›-1π‘₯π‘›βˆ’1 + … + π‘Ž1π‘₯ + π‘Ž0 with real or complex coefficients π‘Žπ‘–, then 𝑃 has exactly 𝑛 zeros π‘Ÿ1, π‘Ÿ2, … , π‘Ÿπ‘› (not all necessarily distinct), such that 𝑃(π‘₯) = π‘Žπ‘›(π‘₯ βˆ’ π‘Ÿ1)(π‘₯ βˆ’ π‘Ÿ2) … (π‘₯ βˆ’ π‘Ÿπ‘›).

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