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Topic 2/3
15 Flashcards in this deck.
A polynomial of degree $n$ can have up to $n$ roots, which may be real or complex. Real roots are solutions where the polynomial equals zero on the real number line, while complex roots involve imaginary numbers of the form $a + bi$, where $i$ is the square root of $-1$.
The Fundamental Theorem of Algebra states that every non-constant single-variable polynomial with complex coefficients has exactly as many complex roots as its degree, counted with multiplicity. This theorem guarantees that factoring a polynomial into linear factors is always possible within the complex number system.
$$ P(x) = a_n(x - r_1)(x - r_2)\dots(x - r_n) $$When a polynomial has real roots, it can be factored into linear factors corresponding to those roots. For example, if $r$ is a real root of the polynomial $P(x)$, then $(x - r)$ is a factor of $P(x)$.
Example: Factor $P(x) = x^2 - 5x + 6$.
First, find the roots by solving $x^2 - 5x + 6 = 0$: $$ x = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} \Rightarrow x = 3 \text{ or } x = 2 $$ Thus, the factored form is: $$ P(x) = (x - 2)(x - 3) $$
Polynomials with complex roots can be factored into quadratic factors with real coefficients. If a polynomial has a complex root $a + bi$, its conjugate $a - bi$ is also a root. The corresponding quadratic factor is $(x - (a + bi))(x - (a - bi)) = x^2 - 2ax + (a^2 + b^2)$.
Example: Factor $P(x) = x^2 + 4x + 8$.
First, find the roots using the quadratic formula: $$ x = \frac{-4 \pm \sqrt{16 - 32}}{2} = \frac{-4 \pm \sqrt{-16}}{2} = -2 \pm 2i $$ Thus, the quadratic factor is: $$ P(x) = (x - (-2 + 2i))(x - (-2 - 2i)) = (x + 2 - 2i)(x + 2 + 2i) = x^2 + 4x + 8 $$
For polynomials of degree three or higher, factoring may involve multiple steps:
Example: Factor $P(x) = x^3 - 3x^2 + 4x - 12$.
First, apply the Rational Root Theorem. Possible rational roots are $\pm1$, $\pm2$, $\pm3$, $\pm4$, $\pm6$, $\pm12$. Testing $x=2$: $$ 2^3 - 3(2)^2 + 4(2) - 12 = 8 - 12 + 8 - 12 = -8 \neq 0 $$ Testing $x=3$: $$ 3^3 - 3(3)^2 + 4(3) - 12 = 27 - 27 + 12 - 12 = 0 $$ Thus, $(x - 3)$ is a factor. Perform polynomial division: $$ P(x) = (x - 3)(x^2 + 0x + 4) $$ Factor the quadratic: $$ x^2 + 4 = (x + 2i)(x - 2i) $$ Thus, the complete factorization is: $$ P(x) = (x - 3)(x + 2i)(x - 2i) $$
A root's multiplicity indicates how many times it appears as a solution. A root with multiplicity greater than one will appear multiple times in the factored form. For example, if $r$ is a root of multiplicity $k$, then $(x - r)^k$ is a factor.
Example: Factor $P(x) = (x - 1)^2(x + 2)$.
The roots are $x = 1$ (with multiplicity 2) and $x = -2$.
The Complex Conjugate Root Theorem states that if a polynomial with real coefficients has a complex root $a + bi$, then its conjugate $a - bi$ is also a root. This theorem is essential for factoring polynomials with complex roots, ensuring that factors with real coefficients are used.
Example: Factor $P(x) = x^4 + 1$.
First, find the roots: $$ x^4 + 1 = 0 \Rightarrow x^4 = -1 \Rightarrow x = \pm \sqrt{ \pm i } $$ The complex roots are $x = \pm \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$ and $x = \pm \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$. Thus, the factors are: $$ P(x) = \left(x^2 + \sqrt{2}x + 1\right)\left(x^2 - \sqrt{2}x + 1\right) $$
The roots of a polynomial correspond to the x-intercepts (real roots) and do not directly correspond to visible points on the graph for complex roots. However, the presence of complex roots affects the graph's shape. For instance, a polynomial with complex roots of even multiplicity remains entirely above or below the x-axis, depending on the leading coefficient.
Factoring polynomials with complex roots is crucial in various applications, including engineering, physics, and computer science. It aids in solving differential equations, modeling oscillatory systems, and analyzing signal processing algorithms.
Example: In electrical engineering, characteristic equations of circuits often lead to polynomials with complex roots, representing oscillations in the system.
Several techniques enhance the efficiency of factoring polynomials:
Example: Factor $P(x) = x^4 - 5x^2 + 4$.
Let $y = x^2$: $$ y^2 - 5y + 4 = (y - 1)(y - 4) $$ Substitute back: $$ (x^2 - 1)(x^2 - 4) = (x - 1)(x + 1)(x - 2)(x + 2) $$
The Rational Root Theorem helps identify potential rational roots of a polynomial. If a polynomial $P(x) = a_nx^n + \dots + a_0$ has a rational root $\frac{p}{q}$, then $p$ divides $a_0$ and $q$ divides $a_n$. This theorem restricts the possible candidates for rational roots, streamlining the factoring process.
Example: Find rational roots of $P(x) = 2x^3 - 3x^2 - 8x + 12$.
Possible roots: $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12, \pm\frac{1}{2}, \pm\frac{3}{2}, \pm\frac{6}{2}= \pm3$. Testing $x=2$: $$ 2(8) - 3(4) - 8(2) + 12 = 16 - 12 - 16 + 12 = 0 $$ Thus, $(x - 2)$ is a factor. Proceed with polynomial division: $$ P(x) = (x - 2)(2x^2 + x - 6) $$ Factor the quadratic: $$ 2x^2 + x - 6 = (2x - 3)(x + 2) $$ Thus, the complete factorization is: $$ P(x) = (x - 2)(2x - 3)(x + 2) $$
Synthetic division is a simplified method of polynomial division when dividing by a linear factor of the form $(x - c)$. It reduces computational steps and helps identify and verify roots efficiently.
Example: Use synthetic division to factor $P(x) = x^3 - 6x^2 + 11x - 6$ by $x=1$.
Set up the synthetic division:
1 | 1 -6 11 -6 | 1 -5 6 ---------------- 1 -5 6 0The remainder is $0$, so $(x - 1)$ is a factor. The reduced polynomial is $x^2 - 5x + 6$, which factors further: $$ P(x) = (x - 1)(x - 2)(x - 3) $$
Combining the Rational Root Theorem with synthetic division accelerates the factoring process, especially for higher-degree polynomials. It allows for systematic testing of possible roots and efficient reduction of the polynomial's degree.
Example: Factor $P(x) = 3x^4 - 2x^3 - 11x^2 + 6x + 12$.
Possible rational roots: $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{4}{3}$. Testing $x=2$: $$ 3(16) - 2(8) - 11(4) + 6(2) + 12 = 48 - 16 - 44 + 12 + 12 = -(-)0 $$ Thus, $(x - 2)$ is a factor. Perform synthetic division:
2 | 3 -2 -11 6 12 | 6 8 -6 0 ---------------------- 3 4 -3 0 12Since the remainder is not zero, $x=2$ is not a root. Continue testing until a valid root is found. For brevity, suppose $x= -1$ is tested and found to be a root. Eventually, the complete factorization is obtained: $$ P(x) = (x - 2)(3x^3 + 4x^2 - 3x + 6) $$ Further factoring the cubic may involve complex roots or additional methods.
Aspect | Real Roots | Complex Roots |
Definition | Solutions where the polynomial equals zero on the real number line. | Solutions involving imaginary numbers, in the form $a + bi$. |
Factors | Linear factors of the form $(x - r)$. | Quadratic factors of the form $(x^2 - 2ax + (a^2 + b^2))$. |
Graphical Representation | Correspond to x-intercepts on the graph. | Do not produce x-intercepts; affect the graph's curvature. |
Multiplicity Impact | Roots can have multiplicities, leading to repeated factors. | Complex roots always come in conjugate pairs, ensuring real coefficients. |
Factoring Techniques | Rational Root Theorem, synthetic division, polynomial division. | Complex Conjugate Root Theorem, quadratic factorization. |
Applications | Solve real-world problems with real solutions, graphing. | Engineering, physics, signal processing involving oscillations and waves. |
Use the Rational Root Theorem: Quickly identify potential real roots by testing factors of the constant term over factors of the leading coefficient.
Practice Synthetic Division: Mastering synthetic division can save time when factoring higher-degree polynomials.
Check for Multiplicity: Always verify if a root has a higher multiplicity to ensure accurate factoring.
Memorize Key Formulas: Familiarize yourself with formulas for factoring quadratics, especially those involving complex roots, to streamline the process during exams.
Factoring polynomials with complex roots plays a crucial role in electrical engineering, particularly in analyzing alternating current (AC) circuits. Additionally, the discovery of complex numbers stemmed from the need to solve polynomial equations that couldn't be resolved using only real numbers. Interestingly, complex roots are essential in computer graphics for rendering realistic animations and simulations.
Mistake 1: Forgetting to consider the multiplicity of roots.
Incorrect: Factoring $P(x) = (x - 2)(x - 2)$.
Correct: Recognizing the double root and writing $P(x) = (x - 2)^2$.
Mistake 2: Incorrectly applying the Complex Conjugate Root Theorem.
Incorrect: Assuming a single complex root without its conjugate.
Correct: Always pairing complex roots as $(a + bi)$ and $(a - bi)$ to maintain real coefficients.