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Fundamental Theorem of Algebra and Factoring | Study Mondo
Topics / Polynomial and Rational Functions / Fundamental Theorem of Algebra and Factoring Fundamental Theorem of Algebra and Factoring Apply the Fundamental Theorem of Algebra to find all roots of polynomials and completely factor polynomial expressions.
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Start Interactive Lesson โ Fundamental Theorem of Algebra and Factoring
The Fundamental Theorem of Algebra
Theorem: Every polynomial equation of degree n โฅ 1 n \geq 1 n โฅ 1 with complex coefficients has exactly n n n complex roots (counting multiplicity).
Implications
A polynomial of degree n n n has exactly n n roots in the complex number system
๐ Practice Problems
1 Problem 1easy โ Question:Find all roots of p ( x ) = x 3 โ 7 x 2 + 14 x โ 8 p(x) = x^3 - 7x^2 + 14x - 8 p ( x ) = x 3 โ and write the complete factorization.
Explain using: ๐ Simple words ๐ Analogy ๐จ Visual desc. ๐ Example ๐ก Explain
โ ๏ธ Common Mistakes: Fundamental Theorem of Algebra and FactoringAvoid these 4 frequent errors
1 Forgetting the constant of integration (+C) on indefinite integrals
โพ 2 Confusing the Power Rule with the Chain Rule
โพ 3 Not checking continuity before applying the Mean Value Theorem
โพ 4 Dropping negative signs when differentiating trig functions
โพ ๐ Real-World Applications: Fundamental Theorem of Algebra and FactoringSee how this math is used in the real world
โ๏ธ Optimizing Package Design
Engineering
โพ ๐ฅ Predicting Drug Dosage Decay
Medicine
โพ ๐ฌ Calculating Distance from Velocity
Physics
โพ ๐ฐ Revenue Optimization
Finance
โพ
๐ Worked Example: Related Rates โ Expanding CircleProblem: A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 2 2 2 cm/s. How fast is the area of the circle increasing when the radius is 10 10 10 cm?
1 Identify the known and unknown rates Click to reveal โ
2 Write the relationship between variables
3 Differentiate both sides with respect to time
๐งช Practice Lab Interactive practice problems for Fundamental Theorem of Algebra and Factoring
โพ ๐ Related Topics in Polynomial and Rational Functionsโ Frequently Asked QuestionsWhat is Fundamental Theorem of Algebra and Factoring?โพ Apply the Fundamental Theorem of Algebra to find all roots of polynomials and completely factor polynomial expressions.
How can I study Fundamental Theorem of Algebra and Factoring effectively?โพ Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Fundamental Theorem of Algebra and Factoring study guide free?โพ Yes โ all study notes, flashcards, and practice problems for Fundamental Theorem of Algebra and Factoring on Study Mondo are 100% free. No account is needed to access the content.
What course covers Fundamental Theorem of Algebra and Factoring?โพ Fundamental Theorem of Algebra and Factoring is part of the AP Precalculus course on Study Mondo, specifically in the Polynomial and Rational Functions section. You can explore the full course for more related topics and practice resources.
๐ก Study Tipsโ Work through examples step-by-step โ Practice with flashcards daily โ Review common mistakes n
Some roots may be repeated (multiplicity > 1)
Some roots may be real, some may be complex
Complex roots with real coefficients come in conjugate pairs
Example The polynomial p ( x ) = x 3 โ 6 x 2 + 11 x โ 6 p(x) = x^3 - 6x^2 + 11x - 6 p ( x ) = x 3 โ 6 x 2 + 11 x โ 6 is degree 3, so it has exactly 3 roots.
If we factor: p ( x ) = ( x โ 1 ) ( x โ 2 ) ( x โ 3 ) p(x) = (x - 1)(x - 2)(x - 3) p ( x ) = ( x โ 1 ) ( x โ 2 ) ( x โ 3 )
The three roots are x = 1 , 2 , 3 x = 1, 2, 3 x = 1 , 2 , 3 (all real).
Relationship Between Roots and Factors If r r r is a root of p ( x ) p(x) p ( x ) , then ( x โ r ) (x - r) ( x โ r ) is a factor of p ( x ) p(x) p ( x ) .
Conversely: If ( x โ r ) (x - r) ( x โ r ) is a factor of p ( x ) p(x) p ( x ) , then r r r is a root of p ( x ) p(x) p ( x ) .
Multiplicity If ( x โ r ) k (x - r)^k ( x โ r ) k is a factor but ( x โ r ) k + 1 (x - r)^{k+1} ( x โ r ) k + 1 is not, then r r r is a root of multiplicity k k k .
Effects of multiplicity on graphs:
Multiplicity 1 (simple root): Graph crosses the x-axis
Multiplicity 2 (double root): Graph touches but doesn't cross (turning point)
Multiplicity 3: Graph crosses with a flattening
Even multiplicity: Graph touches x-axis
Odd multiplicity: Graph crosses x-axis
Complete Factorization Every polynomial with real coefficients can be factored into:
Linear factors : ( x โ r ) (x - r) ( x โ r ) where r r r is a real root
Irreducible quadratic factors : a x 2 + b x + c ax^2 + bx + c a x 2 + b x + c where b 2 โ 4 a c < 0 b^2 - 4ac < 0 b 2 โ 4 a c < 0
Linear Factorization Theorem If p ( x ) p(x) p ( x ) is a polynomial of degree n โฅ 1 n \geq 1 n โฅ 1 , then:
p ( x ) = a n ( x โ r 1 ) ( x โ r 2 ) โฏ ( x โ r n ) p(x) = a_n(x - r_1)(x - r_2)\cdots(x - r_n) p ( x ) = a n โ ( x โ r 1 โ ) ( x โ r 2 โ ) โฏ ( x โ r n โ )
where a n a_n a n โ is the leading coefficient and r 1 , r 2 , โฆ , r n r_1, r_2, \ldots, r_n r 1 โ , r 2 โ , โฆ , r n โ are the roots (possibly complex, possibly repeated).
Finding All Roots
Step-by-step process:
Count the roots : Degree n n n means n n n roots total
Find rational roots : Use Rational Root Theorem
Factor out linear factors : Use synthetic division
Reduce the polynomial : Continue until you have a quadratic or simpler
Find remaining roots : Use quadratic formula if needed
Include complex conjugates : If you find a + b i a + bi a + bi , also include a โ b i a - bi a โ bi
Complex Conjugate Root Theorem Theorem: If a polynomial has real coefficients and a + b i a + bi a + bi (where b โ 0 b \neq 0 b ๎ = 0 ) is a root, then the complex conjugate a โ b i a - bi a โ bi is also a root.
Consequence for Factoring If a + b i a + bi a + bi and a โ b i a - bi a โ bi are roots, the corresponding quadratic factor is:
[ x โ ( a + b i ) ] [ x โ ( a โ b i ) ] = x 2 โ 2 a x + ( a 2 + b 2 ) [x - (a + bi)][x - (a - bi)] = x^2 - 2ax + (a^2 + b^2) [ x โ ( a + bi )] [ x โ ( a โ bi )] = x 2 โ 2 a x + ( a 2 + b 2 )
This is a real quadratic (no imaginary coefficients).
Writing Polynomials from Roots Given roots, we can construct the polynomial:
Example: Roots are 2 , 3 , โ 1 2, 3, -1 2 , 3 , โ 1
Factors: ( x โ 2 ) ( x โ 3 ) ( x + 1 ) (x - 2)(x - 3)(x + 1) ( x โ 2 ) ( x โ 3 ) ( x + 1 )
Polynomial: p ( x ) = ( x โ 2 ) ( x โ 3 ) ( x + 1 ) = x 3 โ 4 x 2 + x + 6 p(x) = (x - 2)(x - 3)(x + 1) = x^3 - 4x^2 + x + 6 p ( x ) = ( x โ 2 ) ( x โ 3 ) ( x + 1 ) = x 3 โ 4 x 2 + x + 6
With complex roots: If roots are 1 , 2 + i , 2 โ i 1, 2 + i, 2 - i 1 , 2 + i , 2 โ i :
p ( x ) = ( x โ 1 ) [ x โ ( 2 + i ) ] [ x โ ( 2 โ i ) ] p(x) = (x - 1)[x - (2 + i)][x - (2 - i)] p ( x ) = ( x โ 1 ) [ x โ ( 2 + i )] [ x โ ( 2 โ i )]
= ( x โ 1 ) [ x 2 โ 4 x + 5 ] = (x - 1)[x^2 - 4x + 5] = ( x โ 1 ) [ x 2 โ 4 x + 5 ]
= x 3 โ 5 x 2 + 9 x โ 5 = x^3 - 5x^2 + 9x - 5 = x 3 โ 5 x 2 + 9 x โ 5
Number of Real vs Complex Roots For a polynomial of degree n n n with real coefficients :
Complex roots come in conjugate pairs
If n n n is odd, there is at least one real root
If n n n is even, there may be no real roots
Example Scenarios (degree 4):
4 real roots
2 real, 2 complex (conjugate pair)
0 real, 4 complex (two conjugate pairs)
Descartes' Rule of Signs For positive roots: Count sign changes in p ( x ) p(x) p ( x ) . The number of positive real roots is either equal to the number of sign changes or less by an even number.
For negative roots: Count sign changes in p ( โ x ) p(-x) p ( โ x ) .
Example p ( x ) = x 3 โ 4 x 2 + x + 6 p(x) = x^3 - 4x^2 + x + 6 p ( x ) = x 3 โ 4 x 2 + x + 6
Signs: + โ + + + - + + + โ + + โ 2 sign changes
Positive roots: 2 or 0
p ( โ x ) = โ x 3 โ 4 x 2 โ x + 6 p(-x) = -x^3 - 4x^2 - x + 6 p ( โ x ) = โ x 3 โ 4 x 2 โ x + 6
Signs: โ โ โ + - - - + โ โ โ + โ 1 sign change
Negative roots: 1
7 x 2 +
14 x โ
8
๐ก Show Solution Solution:
Given: p ( x ) = x 3 โ 7 x 2 + 14 x โ 8 p(x) = x^3 - 7x^2 + 14x - 8 p ( x ) = x 3 โ 7 x 2 + 14 x โ 8
Step 1: Check for rational roots
By Rational Root Theorem, possible rational roots are: ยฑ 1 , ยฑ 2 , ยฑ 4 , ยฑ 8 \pm 1, \pm 2, \pm 4, \pm 8 ยฑ 1 , ยฑ 2 , ยฑ 4 , ยฑ 8
Test x = 1 x = 1 x = 1 :
p ( 1 ) = 1 โ 7 + 14 โ 8 = 0 p(1) = 1 - 7 + 14 - 8 = 0 p ( 1 ) = 1 โ 7 + 14 โ 8 = 0 โ
So x = 1 x = 1 x = 1 is a root, and ( x โ 1 ) (x - 1) ( x โ 1 ) is a factor.
Step 2: Divide by ( x โ 1 ) (x - 1) ( x โ 1 ) using synthetic division
& 1 & -7 & 14 & -8 \\
1 & & 1 & -6 & 8 \\
\hline
& 1 & -6 & 8 & 0
\end{array}$$
Result: $p(x) = (x - 1)(x^2 - 6x + 8)$
**Step 3: Factor the quadratic**
$$x^2 - 6x + 8 = (x - 2)(x - 4)$$
**Step 4: Complete factorization**
$$p(x) = (x - 1)(x - 2)(x - 4)$$
**All roots:** $x = 1, 2, 4$
**Verification:**
- Degree 3 polynomial โ 3 roots โ
- All roots are real โ
- Expand: $(x - 1)(x - 2)(x - 4)$
- $(x - 1)(x^2 - 6x + 8) = x^3 - 6x^2 + 8x - x^2 + 6x - 8 = x^3 - 7x^2 + 14x - 8$ โ
2 Problem 2medium โ Question:A polynomial of degree 4 has roots at x = 2 x = 2 x = 2 (multiplicity 2) and x = โ 1 ยฑ i x = -1 \pm i x = โ 1 ยฑ i . Find the polynomial in standard form with leading coefficient 1.
๐ก Show Solution Solution:
Given information:
Degree 4
Root: x = 2 x = 2 x = 2 with multiplicity 2
Roots: x = โ 1 + i x = -1 + i x = โ 1 + i and x = โ 1 โ i x = -1 - i (complex conjugate pair)
3 Problem 3hard โ Question:Find all roots of p ( x ) = x 4 + 5 x 2 + 4 p(x) = x^4 + 5x^2 + 4 p ( x ) = x 4 + 5 x 2 + 4 and write the complete factorization over the complex numbers.
๐ก Show Solution Solution:
Given: p ( x ) = x 4 + 5 x 2 + 4 p(x) = x^4 + 5x^2 + 4 p ( x ) = x 4 + 5 x 2 +
4 Problem 4medium โ Question:A polynomial of degree 5 has zeros at x = 2, x = -1, and x = 3 + i. What other zeros must it have?
๐ก Show Solution Step 1: Apply Fundamental Theorem of Algebra:
A polynomial of degree 5 has exactly 5 zeros (counting multiplicities)
Step 2: List known zeros:
โข x = 2
โข x = -1
โข x = 3 + i
Step 3: Apply Complex Conjugate Root Theorem:
If a polynomial has real coefficients and 3 + i is a zero,
then 3 - i must also be a zero
Step 4: Count zeros so far:
2, -1, 3 + i, 3 - i = 4 zeros
Step 5: Find remaining zero:
5 - 4 = 1 more zero needed
This could be any real number or a repeated root
Step 6: Most complete answer:
Must have: 3 - i
May have: one additional real zero or a repeated root
Answer: Must have 3 - i; needs one more zero (real or repeated)
5 Problem 5hard โ Question:Factor completely over the complex numbers: f(x) = xโด - 16
๐ก Show Solution Step 1: Recognize difference of squares:
xโด - 16 = (xยฒ)ยฒ - 4ยฒ
= (xยฒ - 4)(xยฒ + 4)
Step 2: Factor xยฒ - 4 (difference of squares):
xยฒ - 4 = (x - 2)(x + 2)
Step 3: Factor xยฒ + 4 (sum of squares):
Over complex numbers: xยฒ + 4 = xยฒ - (-4)
xยฒ = -4
x = ยฑ2i
So: xยฒ + 4 = (x - 2i)(x + 2i)
Step 4: Write complete factorization:
f(x) = (x - 2)(x + 2)(x - 2i)(x + 2i)
Step 5: Verify (optional):
4 zeros: 2, -2, 2i, -2i
Degree 4 polynomial has 4 zeros โ
Answer: f(x) = (x - 2)(x + 2)(x - 2i)(x + 2i)
Rational Functions and Asymptotes
Are there practice problems for Fundamental Theorem of Algebra and Factoring?โพ
Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
x
=
โ 1 โ
i
From x = 2 x = 2 x = 2 (multiplicity 2): ( x โ 2 ) 2 (x - 2)^2 ( x โ 2 ) 2
From x = โ 1 + i x = -1 + i x = โ 1 + i : [ x โ ( โ 1 + i ) ] = ( x + 1 โ i ) [x - (-1 + i)] = (x + 1 - i) [ x โ ( โ 1 + i )] = ( x + 1 โ i )
From x = โ 1 โ i x = -1 - i x = โ 1 โ i : [ x โ ( โ 1 โ i ) ] = ( x + 1 + i ) [x - (-1 - i)] = (x + 1 + i) [ x โ ( โ 1 โ i )] = ( x + 1 + i )
Step 2: Combine complex conjugate factors
[ x โ ( โ 1 + i ) ] [ x โ ( โ 1 โ i ) ] = [ ( x + 1 ) โ i ] [ ( x + 1 ) + i ] [x - (-1 + i)][x - (-1 - i)] = [(x + 1) - i][(x + 1) + i] [ x โ ( โ 1 + i )] [ x โ ( โ 1 โ i )] = [( x + 1 ) โ i ] [( x + 1 ) + i ]
This is a difference of squares pattern: ( a โ b ) ( a + b ) = a 2 โ b 2 (a - b)(a + b) = a^2 - b^2 ( a โ b ) ( a + b ) = a 2 โ b 2
= ( x + 1 ) 2 โ i 2 = ( x + 1 ) 2 โ ( โ 1 ) = ( x + 1 ) 2 + 1 = (x + 1)^2 - i^2 = (x + 1)^2 - (-1) = (x + 1)^2 + 1 = ( x + 1 ) 2 โ i 2 = ( x + 1 ) 2 โ ( โ 1 ) = ( x + 1 ) 2 + 1
= x 2 + 2 x + 1 + 1 = x 2 + 2 x + 2 = x^2 + 2x + 1 + 1 = x^2 + 2x + 2 = x 2 + 2 x + 1 + 1 = x 2 + 2 x + 2
Step 3: Form the complete polynomial
p ( x ) = ( x โ 2 ) 2 ( x 2 + 2 x + 2 ) p(x) = (x - 2)^2(x^2 + 2x + 2) p ( x ) = ( x โ 2 ) 2 ( x 2 + 2 x + 2 )
First: ( x โ 2 ) 2 = x 2 โ 4 x + 4 (x - 2)^2 = x^2 - 4x + 4 ( x โ 2 ) 2 = x 2 โ 4 x + 4
Then multiply:
( x 2 โ 4 x + 4 ) ( x 2 + 2 x + 2 ) (x^2 - 4x + 4)(x^2 + 2x + 2) ( x 2 โ 4 x + 4 ) ( x 2 + 2 x + 2 )
= x 4 + 2 x 3 + 2 x 2 โ 4 x 3 โ 8 x 2 โ 8 x + 4 x 2 + 8 x + 8 = x^4 + 2x^3 + 2x^2 - 4x^3 - 8x^2 - 8x + 4x^2 + 8x + 8 = x 4 + 2 x 3 + 2 x 2 โ 4 x 3 โ 8 x 2 โ 8 x + 4 x 2 + 8 x + 8
= x 4 โ 2 x 3 โ 2 x 2 + 8 = x^4 - 2x^3 - 2x^2 + 8 = x 4 โ 2 x 3 โ 2 x 2 + 8
Answer: p ( x ) = x 4 โ 2 x 3 โ 2 x 2 + 8 p(x) = x^4 - 2x^3 - 2x^2 + 8 p ( x ) = x 4 โ 2 x 3 โ 2 x 2 + 8
Degree 4 โ
Leading coefficient 1 โ
Has 4 roots total (counting multiplicity) โ
4
Step 1: Recognize this as quadratic in form
Let u = x 2 u = x^2 u = x 2 , then:
u 2 + 5 u + 4 = 0 u^2 + 5u + 4 = 0 u 2 + 5 u + 4 = 0
Step 2: Factor the quadratic
u 2 + 5 u + 4 = ( u + 1 ) ( u + 4 ) = 0 u^2 + 5u + 4 = (u + 1)(u + 4) = 0 u 2 + 5 u + 4 = ( u + 1 ) ( u + 4 ) = 0
So u = โ 1 u = -1 u = โ 1 or u = โ 4 u = -4 u = โ 4
Case 1: x 2 = โ 1 x^2 = -1 x 2 = โ 1
x = ยฑ i x = \pm i x = ยฑ i
Case 2: x 2 = โ 4 x^2 = -4 x 2 = โ 4
x = ยฑ 2 i x = \pm 2i x = ยฑ 2 i
All roots: x = i , โ i , 2 i , โ 2 i x = i, -i, 2i, -2i x = i , โ i , 2 i , โ 2 i
Step 4: Write complete factorization
p ( x ) = ( x โ i ) ( x + i ) ( x โ 2 i ) ( x + 2 i ) p(x) = (x - i)(x + i)(x - 2i)(x + 2i) p ( x ) = ( x โ i ) ( x + i ) ( x โ 2 i ) ( x + 2 i )
Or in terms of real quadratics:
( x โ i ) ( x + i ) = x 2 + 1 (x - i)(x + i) = x^2 + 1 ( x โ i ) ( x + i ) = x 2 + 1
( x โ 2 i ) ( x + 2 i ) = x 2 + 4 (x - 2i)(x + 2i) = x^2 + 4 ( x โ 2 i ) ( x + 2 i ) = x 2 + 4
p ( x ) = ( x 2 + 1 ) ( x 2 + 4 ) p(x) = (x^2 + 1)(x^2 + 4) p ( x ) = ( x 2 + 1 ) ( x 2 + 4 )
Complex factorization: p ( x ) = ( x โ i ) ( x + i ) ( x โ 2 i ) ( x + 2 i ) p(x) = (x - i)(x + i)(x - 2i)(x + 2i) p ( x ) = ( x โ i ) ( x + i ) ( x โ 2 i ) ( x + 2 i )
Real factorization: p ( x ) = ( x 2 + 1 ) ( x 2 + 4 ) p(x) = (x^2 + 1)(x^2 + 4) p ( x ) = ( x 2 + 1 ) ( x 2 + 4 )
Roots: x = ยฑ i , ยฑ 2 i x = \pm i, \pm 2i x = ยฑ i , ยฑ 2 i
Degree 4 โ 4 roots โ
All roots are purely imaginary (no real roots) โ
Expand: ( x 2 + 1 ) ( x 2 + 4 ) = x 4 + 4 x 2 + x 2 + 4 = x 4 + 5 x 2 + 4 (x^2 + 1)(x^2 + 4) = x^4 + 4x^2 + x^2 + 4 = x^4 + 5x^2 + 4 ( x 2 + 1 ) ( x 2 + 4 ) = x 4 + 4 x 2 + x 2 + 4 = x 4 + 5 x 2 + 4 โ