2. Polynomial Root Theorems
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Chapter 2
2.
Polynomial Root Theorems
This lesson delves into the fascinating world of polynomial root theorems, emphasizing the significance of understanding the roots of polynomials. It highlights methods to determine the number of roots using the Fundamental Theorem of Algebra. However, the theorem doesn't specify the nature of these roots, be it integer, rational, real, or imaginary. The lesson further explores techniques to uncover these details. Through engaging examples, readers are introduced to concepts like the Rational Root Theorem, which aids in identifying integer and rational roots. The Factor Theorem is also discussed, offering insights into factoring polynomials of lesser degrees. The lesson is enriched with practical exercises and scenarios, making it an engaging learning experience.
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Lesson Settings & Tools
| | 13 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Polynomial Root Theorems
Slide of 13
When presented with a polynomial, it is possible to determine the number of roots by using the Fundamental Theorem of Algebra. However, this theorem does not give any detailed information about the roots — whether they are integer, rational, real, or imaginary. This lesson will explore methods to find these amazing inner workings of a root.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Polynomial Functions and Their Roots
Recalling the Fundamental Theorem of Algebra, a polynomial function of degree greater than 0 always has a root. These roots can be found using diverse methods like the quadratic formula for quadratic polynomial functions. ax^2 + bx + c = 0 ⇓ x = - b ± sqrt(b^2 - 4ac)/2a However, factoring a polynomial function of greater degrees can be tough, as there are not easy to use formulas to do so. The good news is that looking at the terms of a polynomial function can be useful to find more characteristics of a given function. Consider the following functions and their roots.
| Polynomial Function | Roots |
|---|---|
| p(x) = x^3 - 6 x^2 + 11 x - 6 | 1, 2, 3 |
| g(x) =x^4 - 4 x^3 - 19 x^2 + 106 x - 120 | -5, 2, 3, 4 |
| m(x)=x^3 + 6 x^2 - 55 x - 252 | -9, -4, 7 |
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Rational Root Theorem
Consider a polynomial where every coefficient is an integer. P(x) = a_n x^n + a_(n-1)x^(n-1)+⋯+a_1x+a_0 The following properties are true for the roots of the polynomial.
- Every integer root is a factor of the constant a_0.
- All rational roots must have a numerator that is a factor of a_0 and a denominator that is a factor of the leading coefficient a_n.
Proof
Rational Root TheoremTo prove this theorem, each of the properties will be considered individually. To start with the first property, consider a polynomial P(x) with integer coefficients that has an integer root x_r. This means that P(x_r)=0.
a_n x_r^n + a_(n-1)x_r^(n-1)+⋯+a_1x_r+a_0 = 0
Now a_0 is subtracted from both sides of the equation.
Since the coefficients of P(x) are integers and x_r is an integer, the expression between the parentheses results in an integer. Also, - a_0 is the product of an integer and x_r, which means that x_r is a factor of a_0. This completes the proof of the first property. 
This time the equation will be modified in a different way.
Since the coefficients of P(x), p, and q are all integers, the expression between parentheses results in an integer. Also, since pq is written in its simplest form, p and q do not have a common factor, which means that p and q^n do not have a common factor either. Therefore, p is a factor of a_0.
If the modification is done differently, it is possible to come to the other property.
With the same reasoning as before, it is possible to conclude that q, the denominator of the root, is a factor of a_n, finishing the proof.
a_n x_r^n + a_(n-1)x_r^(n-1)+⋯+a_1x_r+a_0 = 0
SubEqn
LHS-a_0=RHS-a_0
a_n x_r^n + a_(n-1)x_r^(n-1)+⋯+a_1x_r = - a_0
FactorOut
Factor out x_r
x_r(a_n x_r^(n-1) + a_(n-1)x_r^(n-2)+⋯+a_1) = - a_0
For the second property, suppose that the polynomial P(x) has a rational root pq, such that the fraction is written in its simplest form. Again, it is obtained that P( pq)=0.
a_n (p/q)^n + a_(n-1)(p/q)^(n-1)+⋯+a_1(p/q)+a_0 = 0
PowQuot
(a/b)^m=a^m/b^m
a_n (p^n/q^n)+ a_(n-1)(p^(n-1)/q^(n-1))+⋯+a_1(p/q)+a_0 = 0
MultEqn
LHS * q^n=RHS* q^n
a_n(p^n) + a_(n-1)(p^(n-1)q)+⋯ + a_1(pq^(n-1))+a_0(q^n) = 0
SubEqn
LHS-a_0(q^n)=RHS-a_0(q^n)
a_n(p^n) + a_(n-1)(p^(n-1)q)+⋯ + a_1(pq^(n-1)) = - a_0(q^n)
FactorOut
Factor out p
p(a_n p^(n-1) + a_(n-1)p^(n-2)q+⋯ + a_1q^(n-1)) = - a_0 q^n
The numerator of the root, p, is a factor of a_0.
As an example of how to apply this theorem, consider the following polynomial. p(x) = 6x^4 - 59x^3 + 94x^2 -49x + 8 The integer roots of the equation must be factors of the value of a_0, which in this case is 8. The factors of 8 are 1, 2, 4, and 8. Each of these values is substituted for x into the equation p(x) = 0 to determine which is a root.
| Equation | Result |
|---|---|
| 6* 1^4 - 59* 1^3 + 94* 1^2 -49* 1 + 8 = 0 | 0 = 0 ✓ |
| 6* 2^4 - 59* 2^3 + 94* 2^2 -49* 2 + 8 = 0 | -90 = 0 * |
| 6* 4^4 - 59* 4^3 + 94* 4^2 -49* 4 + 8 = 0 | -924 = 0 * |
| 6* 8^4 - 59* 8^3 + 94* 8^2 -49* 8 + 8 = 0 | 0 = 0 ✓ |
As determined, the integer roots are 1 and 8. The rational roots have a numerator that is a factor of a_0 and a denominator that is a factor of a_n. In this case,a_n is 6 with factors of 1, 2, 3, and 6. The following table displays the possible fractions that can be formed with the factors. The columns of the table are the factors of 8, and the rows are the factors of 6.
| 1 | 2 | 4 | 8 | |
|---|---|---|---|---|
| 2 | 1/2 | 2/2=1 | 4/2=2 | 8/2=4 |
| 3 | 1/3 | 2/3 | 4/3 | 8/3 |
| 6 | 1/6 | 2/6=1/3 | 4/6=2/3 | 8/6=4/3 |
It is important to understand that the integers do not need to be tested because the integer factors have already been found. Now, each fraction will be tested to see if they are roots of the equation.
| Equation | Result |
|---|---|
| 6( 1/2)^4 - 59( 1/2)^3 + 94( 1/2)^2 -49( 1/2) + 8 = 0 | 0 = 0 ✓ |
| 6( 1/3)^4 - 59( 1/3)^3 + 94( 1/3)^2 -49( 1/3) + 8 = 0 | 0 = 0 ✓ |
| 6( 2/3)^4 - 59( 2/3)^3 + 94( 2/3)^2 -49( 2/3) + 8 = 0 | 22/27= 0 * |
| 6( 4/3)^4 - 59( 4/3)^3 + 94( 4/3)^2 -49( 4/3) + 8 = 0 | -100/9= 0 * |
| 6( 8/3)^4 - 59( 8/3)^3 + 94( 8/3)^2 -49( 8/3) + 8 = 0 | -7280/27= 0 * |
| 6( 1/6)^4 - 59( 1/6)^3 + 94( 1/6)^2 -49( 1/6) + 8 = 0 | 235/108 = 0 * |
Example
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Finding the Integer Roots of a Polynomial Function
Izabella is tutoring some of her friends for an upcoming math exam.
To study, they decided to go over some exercises that the math professor assigned for homework in the past. One polynomial function, in particular, caught Izabella's attention. p(x) =x^4 - x^3 - 6x^2 + 14x - 12 Solve following exercises.
a Find the integer roots of the given polynomial function.
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b Find the remaining roots of the function.
{"type":"text","form":{"type":"roots","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":["i"]},"rendered":false,"hideNoSolution":false,"variable":"x"},"text":" "},"formTextBefore":null,"formTextAfter":null,"answer":{"texts":["1+i","1-i"]}}
Hint
a Use the Rational Root Theorem.
b Use synthetic division.
Solution
a The integer roots of a polynomial function can be found using the Rational Root Theorem. This theorem indicates that the integer roots of a polynomial are factors of the constant term a_0, which in this case is -12. Below are the factors of -12 listed.
Positive:& 1,2,3,4,6,12 Negative:& -1,-2,-3,-4,-6,-12 To determine which of these is a root of the given polynomial function, each of the factors will be substituted into the equation p(x) = 0 to check which results in a true statement. First, the positive factors will be substituted.
| Equation | Solution |
|---|---|
| ( 1)^4 - ( 1)^3 - 6( 1)^2 + 14( 1) - 12 = 0 | -4 ≠ 0 * |
| ( 2)^4 - ( 2)^3 - 6( 2)^2 + 14( 2) - 12 = 0 | 0 = 0 ✓ |
| ( 3)^4 - ( 3)^3 - 6( 3)^2 + 14( 3) - 12 = 0 | 30 ≠ 0 * |
| ( 4)^4 - ( 4)^3 - 6( 4)^2 + 14( 4) - 12 = 0 | 140 ≠ 0 * |
| ( 6)^4 - ( 6)^3 - 6( 6)^2 + 14( 6) - 12 = 0 | 936 ≠ 0 * |
| ( 12)^4 - ( 12)^3 - 6( 12)^2 + 14( 12) - 12 = 0 | 18 300 ≠ 0 * |
The value of x=2 is a root of the function. Since the polynomial function is of degree 4, it is possible that one of the negative factors is also a root, so these values will be substituted next.
| Equation | Solution |
|---|---|
| ( - 1)^4 - ( - 1)^3 - 6( - 1)^2 + 14( - 1) - 12 = 0 | -30 ≠ 0 * |
| ( - 2)^4 - ( - 2)^3 - 6( - 2)^2 + 14( - 2) - 12 = 0 | -40 ≠ 0 * |
| ( -3)^4 - ( -3)^3 - 6( -3)^2 + 14( -3) - 12 = 0 | 0 = 0 ✓ |
| ( - 4)^4 - ( - 4)^3 - 6( - 4)^2 + 14( - 4) - 12 = 0 | 156 ≠ 0 * |
| ( -6)^4 - ( - 6)^3 - 6( - 6)^2 + 14( - 6) - 12 = 0 | 1200 ≠ 0 * |
| ( - 12)^4 - ( - 12)^3 - 6( - 12)^2 + 14( - 12) - 12 = 0 | 21 420 ≠ 0 * |
As can be seen in the table, -3 is also a root of p(x). Since every integer root has to be a factor of -12 and every factor was tested, the only integer roots of the given function are -3 and 2.
b The Factor Theorem can be used to factor the given polynomial into a polynomials of a lesser degree. Since 2 is a root of p(x), (x-2) is a factor of p(x). This means that multiplying a polynomial q(x) by (x-2) results in p(x).
rl IR-0.15cm r 2 & |rr 1 & -1 & -6 & 14 & -12
Bring down 1
rl IR-0.15cm r 2 & |rr 1 & -1 & -6 & 14 & -12 & & & & & c 1 & & & &
rl IR-0.15cm r 2 & |rr 1 & -1 & -6 & 14 & -12 & & & & & c 1 & & & &
▼
Evaluating the division
2* 1=2
rl IR-0.15cm r 2 & |rr 1 & -1 & -6 & 14 & -12 & 2 & & & & c 1 & & & &
-1 + 2 = 1
rl IR-0.15cm r 2 & |rr 1 & -1 & -6 & 14 & -12 & 2 & & & & c 1 & 1 & & &
2* 1= 2
rl IR-0.15cm r 2 & |rr 1 & -1 & -6 & 14 & -12 & 2 & 2 & & & c 1 & 1 & & &
- 6 + 2 = - 4
rl IR-0.15cm r 2 & |rr 1 & -1 & -6 & 14 & -12 & 2 & 2 & & & c 1 & 1 & -4 & &
2* -4 = -8
rl IR-0.15cm r 2 & |rr 1 & -1 & -6 & 14 & -12 & 2 & 2 & -8 & & c 1 & 1 & -4 & &
14 - 8 = 6
rl IR-0.15cm r 2 & |rr 1 & -1 & -6 & 14 & -12 & 2 & 2 & -8 & & c 1 & 1 & -4 & 6 &
2 * 6 = 12
rl IR-0.15cm r 2 & |rr 1 & -1 & -6 & 14 & -12 & 2 & 2 & -8 &12 & c 1 & 1 & -4 & 6 &
-12 + 12 = 0
rl IR-0.15cm r 2 & |rr 1 & -1 & -6 & 14 & -12 & 2 & 2 & -8 & 12 & c 1 & 1 & -4 & 6 & 0
rl IR-0.15cm r -3 & |rr 1 & 1 & -4 & 6
▼
Evaluate the division
Bring down 1
rl IR-0.15cm r -3 & |rr 1 & 1 & -4 & 6 & & & & c 1 & & &
-3 * 1 = -3
rl IR-0.15cm r -3 & |rr 1 & 1 & -4 & 6 & -3 & & & c 1 & & &
1 - 3 = -2
rl IR-0.15cm r -3 & |rr 1 & 1 & -4 & 6 & -3 & & & c 1 & -2 & &
-3 * - 2 = 6
rl IR-0.15cm r -3 & |rr 1 & 1 & -4 & 6 & -3 & 6 & & c 1 & -2 & &
-4 + 6 = 2
rl IR-0.15cm r -3 & |rr 1 & 1 & -4 & 6 & -3 & 6 & & c 1 & -2 & 2 &
-3 * 2 = -6
rl IR-0.15cm r -3 & |rr 1 & 1 & -4 & 6 & -3 & 6 & -6 & c 1 & -2 & 2 &
6 - 6 = 0
rl IR-0.15cm r -3 & |rr 1 & 1 & -4 & 6 & -3 & 6 & -6 & c 1 & -2 & 2 & 0
x=- b±sqrt(b^2 - 4ac)/2a
SubstituteValues
Substitute values
x=- ( -2)±sqrt(( 2)^2 - 4( 1)( 2))/2( 1)
x=2±sqrt(-4)/2
| x = 2 + sqrt(-4)/2 | x = 2 - sqrt(-4)/2 |
|---|---|
| x = 2 + 2i/2 | x = 2 - 2i/2 |
| x = 1 + i | x = 1 - i |
Therefore, the remaining roots of the given polynomial function are the complex numbers 1+i and 1-i.
Example
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Finding the Rational Roots of a Polynomial Function
Izabella now wants to challenge her study group to try a problem all on their own. Kevin is ready for it!
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Hint
Remember the Rational Root Theorem.
Solution
Considering the Rational Root Theorem, it is possible to find the integer and the rational roots. According to the theorem, the integer roots of the polynomial must be factors of the constant term of the polynomial, which is 2. Factors of $2$: -2, -1, 1, 2 Each of these factors is substituted into the equation g(x)=0 to determine which, if any, is a root of the function.
| Substitution | Simplify |
|---|---|
| 3( -2)^3 - ( -2)^2 - 6( -2) + 2 = 0 | -14 ≠ 0 * |
| 3( -1)^3 - ( -1)^2 - 6( -1) + 2 = 0 | 4 ≠ 0 * |
| 3( 1)^3 - ( 1)^2 - 6( 1) + 2 = 0 | -2 ≠ 0 * |
| 3( 2)^3 - ( 2)^2 - 6( 2) + 2 = 0 | 10 ≠ 0 * |
No value resulted in a true statement. This means there is no integer root of the function. However, the Rational Root Theorem also allows to look for the rational roots of a polynomial. These roots can be written as follows. x = p/q In this expression, p, the numerator, is a factor of the constant term q, the denominator, is a factor of the leading coefficient, which in this case is 3. Previously the factors of 2 were presented. Below are the factors of 3. Factors of $3$: -3, -1, 1, 3 Since these factors must be denominators and the integer roots were already considered, the only factors to be considered are -3 and 3, which result in the following rational numbers. - 2/3, - 1/3, 1/3, 2/3 Again, each of these numbers is substituted into the equation as before to verity which, if any, is a root of the function.
| Substitution | Simplify |
|---|---|
| 3( - 2/3)^3 - ( - 2/3)^2 - 6( - 2/3) + 2 = 0 | 143 ≠ 0 * |
| 3( -1/3)^3 - ( -1/3)^2 - 6( -1/3) + 2 = 0 | 343 ≠ 0 * |
| 3( 1/3)^3 - ( 1/3)^2 - 6( 1/3) + 2 = 0 | 0 = 0 ✓ |
| 3( 2/3)^3 - ( 2/3)^2 - 6( 2/3) + 2 = 0 | - 149 ≠ 0 * |
rl IR-0.15cm r 13 & |rr 3 & -1 & -6 & 2 &&& & c 3 & & &
▼
Evaluate division
1/3 * 3 = 1
rl IR-0.15cm r 13 & |rr 3 & -1 & -6 & 2 &1&& & c 3 & & &
-1 + 1 =0
rl IR-0.15cm r 13 & |rr 3 & -1 & -6 & 2 &1&& & c 3 & 0 & &
1/3 * 0 = 0
rl IR-0.15cm r 13 & |rr 3 & -1 & -6 & 2 &1&0& & c 3 & 0 & &
-6 + 0 = -6
rl IR-0.15cm r 13 & |rr 3 & -1 & -6 & 2 &1&0& & c 3 & 0 & -6 &
1/3 * (-6) = -2
rl IR-0.15cm r 13 & |rr 3 & -1 & -6 & 2 &1&0&-2 & c 3 & 0 & -6 &
2 -2 = 0
rl IR-0.15cm r 13 & |rr 3 & -1 & -6 & 2 &1&0&-2 & c 3 & 0 & -6 & 0
3x^2 - 6
FactorOut
Factor out 3
3(x^2 - 2)
a = ( sqrt(a) )^2
3(x^2 - (sqrt(2))^2)
FacDiffSquares
a^2-b^2=(a+b)(a-b)
3(x-sqrt(2))(x+sqrt(2))
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The Irrational Roots of a Polynomial
The Rational Root Theorem helps determine the integer and rational roots of a polynomial. One might expect a similar theorem exists used to find irrational roots. Unfortunately, there is not. But wait, it is not all bad news. A theorem exists that does involve the irrational roots of a polynomial.
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Irrational Conjugate Root Theorem
Let P(x) be a polynomial function whose coefficients are rational. If a and b are rational such that sqrt(b) is an irrational number and a+sqrt(b) is a root of P(x), then its irrational conjugate a-sqrt(b) is also a root of P(x).
P(a+sqrt(b)) = 0 ⇒ P(a-sqrt(b)) = 0
Proof
Let a, b be rational numbers and sqrt(b) be an irrational number. Suppose that the irrational number a+sqrt(b) is a root of the polynomial P(x) with rational coefficients c_i. P(a+sqrt(b))= ∑_(i=0)^n c_i ( a+sqrt(b) ) ^i = 0 Since the conjugate of zero is zero, the expression's conjugate on the left-hand side is also zero. P(a+sqrt(b))=∑_(i=0)^n c_i * ( a+sqrt(b) ) ^i = 0 Next, using the properties of conjugation, the polynomial can be rewritten.
| ∑_(i=0)^n c_i * ( a+sqrt(b) ) ^i | Polynomial |
| ∑_(i=0)^n c_i * ( a+sqrt(b) ) ^i | Conjugate of a sum is the sum of the conjugates |
| ∑_(i=0)^n c_i ( a+sqrt(b) ) ^i | Conjugate of a product is the product of the conjugates |
| ∑_(i=0)^n c_i ( a+sqrt(b) ) ^i | Conjugate of a power is the power of the conjugate |
| ∑_(i=0)^n c_i ( a+sqrt(b) ) ^i | Conjugate of a rational number is itself |
This final expression is equal to 0. P(a+sqrt(b)) = ∑_(i=0)^n c_i ( a+sqrt(b) ) ^i =0 Notice that the middle expression is equal to the value of the polynomial at a+sqrt(b). Using this and the fact that the conjugate of a+sqrt(b) is a-sqrt(b), this equation can be rewritten as follows. ∑_(i=0)^n c_i ( a+sqrt(b) )^i &= 0 &⇓ P( a+sqrt(b) ) &= 0 &⇓ P( a-sqrt(b)) &= 0 In conclusion, the irrational number a-sqrt(b) makes the polynomial p(x) zero, and, therefore, it is also a root of P(x).
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Rational and Irrational Roots
Other than being really good at math, Izabella is also known for making excellent cakes. To motivate her friends, she decided to organize a little contest and give a cake to the winner.
{"type":"choice","form":{"alts":["Ali","Vincenzo","Kevin","Zain"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
Hint
Remember the Irrational Conjugate Root Theorem
Solution
It is mentioned that the polynomial is of degree 3. Considering the Fundamental Theorem of Algebra, the given polynomial has three roots. Two of those roots are already given.
x_1 &= - 5 x_2 &= 1 - sqrt(3)
All the roots of the polynomial are needed to determine the correct polynomial. Since every coefficient of the polynomial is an integer and the solution x_2 = 1 - sqrt(3) is an irrational number, by the Irrational Conjugate Root Theorem, the irrational conjugate of 1 - sqrt(3) has to be a root of the function.
x_3 = 1 + sqrt(3)
Using the Factor Theorem, it is possible to write the polynomial by multiplying the binomials produced by the roots.
(x + 5)(x - (1-sqrt(3)))(x - (1+sqrt(3)))_(p(x))
To expand this expression clearly, first the two rightmost binomials will be multiplied.
Now, this result will be multiplied by the binomial (x+5).
With this result, it is possible to write the correct answer.
p(x) = x^3 + 3x^2 - 12x - 10
Therefore, Ali will be the winner. Additionally, it can be checked that the polynomials written by the other friends did not meet the conditions Izabella requested. Zain, in particular, feels sad to have not gotten their answer correct.
(x-(1-sqrt(3)))(x-(1+sqrt(3)))
▼
Simplify
Distr
Distribute (x-(1-sqrt(3)))
x(x-(1-sqrt(3))) - (1+sqrt(3))(x-(1-sqrt(3)))
Distr
Distribute x
x^2-x(1-sqrt(3)) - (1+sqrt(3))(x-(1-sqrt(3)))
Distr
Distribute (1+sqrt(3))
x^2-x(1-sqrt(3)) - x(1+sqrt(3))+(1-sqrt(3))(1+sqrt(3))
Distr
Distribute x
x^2-x + xsqrt(3) - x - xsqrt(3) + (1-sqrt(3))(1+sqrt(3))
ExpandDiffSquares
(a+b)(a-b)=a^2-b^2
x^2-x + xsqrt(3) - x - xsqrt(3) + 1 - 3
SubTerms
Subtract terms
x^2 - 2x - 2
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All Irrational Roots
After Ali won the prized cake, Zain promised to study more efficiently. They thought this could be accomplished by putting on some shades and a fedora while studying at home. Nobody knows if it will work.
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Hint
Consider the Irrational Conjugate Root Theorem.
Solution
It is given that the polynomial function p(x) has a degree of 4. Then, by the Fundamental Theorem of Algebra, the function has four roots, counting repeating roots as multiple roots. Two roots are already given. x_1 &= 2 - sqrt(5) x_2 &= sqrt(7) Since it is known that the coefficients are integers, they can be used to find the other two roots. Recall that the Irrational Conjugate Root Theorem states that if an irrational number a + sqrt(b) is a root of a polynomial function with rational coefficients, then the irrational conjugate a - sqrt(b) is also a root. The given roots can be written as indicated. x &= a + sqrt(b) [0.5em] x_1 &= 2 + (-sqrt(5)) x_2 &= 0 + sqrt(7) Considering the Irrational Conjugate Theorem, the remaining two roots are the irrational conjugates of the given roots. lcl x_3 = 2 - (-sqrt(5)) &⇒& x_3 = 2+sqrt(5) x_4 = 0 - sqrt(7) &⇒& x_4 = -sqrt(7) Zain could be on to something with rocking this new fashion while studying!
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The Complex Roots of a Polynomial
The Rational and Irrational Roots Theorems provide information about the integer, rational, and irrational roots of a polynomial. Therefore, in some way, the real roots of a polynomial are covered by these theorems. Next, a theorem involving the complex roots of a polynomial is presented.
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Complex Conjugate Root Theorem
Let P(x) be a polynomial function whose coefficients are real. If a complex number a+bi is a root of P(x), then the root's complex conjugate, a-bi, is also a root of P(x).
P(a+bi) = 0 ⇒ P(a-bi) = 0
Proof
Let p(x) be a polynomial with real coefficients c_k. P(x)= ∑_(k=0)^n c_kx^k Using the properties of conjugation, the conjugate of the polynomial evaluated at a+bi can be rewritten as expressed in the following table.
| P(a+bi)=∑_(k=0)^n c_k (a+bi)^k | The conjugate of the polynomial evaluated at a+bi |
| P(a+bi)= ∑_(k=0)^n c_k (a+bi)^k | The conjugate of a sum is the sum of the conjugates. |
| P(a+bi)= ∑_(k=0)^n c_k (a+bi)^k | The conjugate of a product is the product of the conjugates. |
| P(a+bi)= ∑_(k=0)^n c_k (a+bi)^k | The conjugate of a power is the power of the conjugate. |
| P(a+bi)= ∑_(k=0)^n c_k(a+bi) ^k | The conjugate of a real number is itself. |
| P(a+bi)= ∑_(k=0)^n c_k(a-bi)^k | The conjugate of a+bi is a-bi. |
| P(a+bi)=P(a-bi) | The polynomial evaluated at a-bi |
It is given that a+bi is a root of the polynomial p(x). P(a+bi)=0 The conjugate of the real number 0 is 0. Therefore, the conjugate of the polynomial evaluated at a+bi is 0. P(a+bi)=0 This equation, along with the last equality in the table, shows that a-bi, the conjugate of a+bi, is also a root of the polynomial. P(a+bi) = P(a-bi) [0.1cm] P(a+bi) = 0 ⇓ P(a-bi)=0
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Integer and Complex Roots
As Zain progresses with their studies, they build up the courage to ask Izabella for extra help with polynomials that have complex roots.
They were given the numbers x_1=4 and x_2=3+i which are roots of the following equation. Consider that b, c, and d are integers. x^3 + bx^2 + cx + d = 0 Zain needs to answer the following questions.
a What are the remaining roots of the equation?
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b Find the values of b, c, and d.
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Hint
a Remember the Complex Conjugate Root Theorem.
b Multiply the binomials produced by the roots of the equation.
Solution
a By the Fundamental Theorem of Algebra, since the given polynomial equation is of degree 3, the equation has three roots. Two were already given.
x_1 &= 4 x_2 &= 3 + i It can be noted that the root x_2 is a complex root. Since all the coefficients are integers, the conjugate of x_2 is also a root of the equation, by the Complex Conjugate Root Theorem. x_2 = 3 + i ⇒ x_3 = 3 - i Since this is the third root, it is the only root remaining aside from the given two.
b The given polynomial equation is the multiplication of the binomials formed with each of its solutions.
(x-(3+i))(x-(3-i))
▼
Simplify
AssociativePropAdd
Associative Property of Addition
((x-3)-i))((x-3)+i)
(a+bi)(a-bi)=a^2+b^2
(x-3)^2 +1
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
x^2 - 6x + 9 + 1
AddTerms
Add terms
x^2-6x+10
(x^2-6x+10)(x-4)
▼
Simplify
Distr
Distribute (x-4)
x^2(x-4)-6x(x-4)+10(x-4)
Distr
Distribute x^2
x^3 - 4x^2 - 6x(x-4)+10(x-4)
Distr
Distribute 6x
x^3 - 4x^2 - 6x^2+ 24x + 10(x-4)
Distr
Distribute 10
x^3 - 4x^2 - 6x^2+ 24x + 10x-40
AddSubTerms
Add and subtract terms
x^3 - 10x^2+ 34x -40
Example
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Only Complex Roots
After studying polynomial functions with complex roots, Zain feels confident enough to finish an assignment all on their own.
The final exercise of the assignment displays a polynomial with real coefficients and it has the following roots. x_1 = 5 + 3i x_2 = 2 - 6i Along with Zain, answer the following questions.
a What is the least degree of the polynomial?
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b Assuming the polynomial's degree is the least possible one, find the remaining roots.
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Hint
a What does the Fundamental Theorem of Algebra say?
b Remember the Complex Conjugate Root Theorem.
Solution
a By the Complex Conjugate Root Theorem, since the polynomial's coefficients are real, the conjugates of any complex root must also be a root. Two complex roots were given.
x_1 = 5 + 3i x_2 = 2 - 6i
Therefore, the polynomial has at least another two different roots. Then, by the Fundamental Theorem of Algebra, the least degree of the polynomial is 4. b As mentioned in Part A, the conjugates of the given roots are also roots of the polynomial by the Complex Conjugate Root Theorem.
5 + 3i ⇒ 5 - 3i 2 - 6i ⇒ 2 + 6i Since the least degree of the polynomial is 4, these are the remaining roots of the polynomial.
Example
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Irrational and Complex Roots
Feeling so proud of Zain, Izabella feels really confident that all of her friends will do well when dealing with their math assignments. To complete this study session, she proposes a problem that has a polynomial with both irrational and complex roots.
Consider a polynomial function g(x) of degree 4 with integer coefficients has the following roots. x_1 &= 7 - 3i x_2 &= 4 + sqrt(5) Help answering the following.
a What are the remaining roots of the equation g(x)=0?
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b Write the expression of g(x) with a leading coefficient of 1.
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Hint
a Recall the Complex Conjugate and the Irrational Conjugate Root Theorems
Solution
a Considering the Fundamental Theorem of Algebra, it is known that a polynomial of degree 4 has 4 solutions. Therefore, there are two remaining solutions. The given solutions can be used to find the remaining solutions. The solution x_1 is a complex number.
x_1 = 7 - 3i Also, it is stated that the polynomial has integer coefficients, which are real coefficients. If a complex number is a root of a polynomial with real coefficients, the Complex Conjugate Root Theorem states that the complex conjugate of such root must also be a solution. x_1 = 7 - 3i ⇒ x_3 = 7 + 3i On the other hand, the other solution is a real number. Since 5 is a prime number, its square root is an irrational number. Therefore, the given root is an irrational number. x_2 = 4 + sqrt(5) This time the Irrational Conjugate Root Theorem is useful. This theorem says that if an irrational number is a root of a polynomial with rational coefficients, the corresponding irrational conjugate is also a root of the polynomial. x_2 = 4 + sqrt(5) ⇒ x_4 = 4 - sqrt(5) Consequently, the two remaining roots are the following. x_3 &= 7 + 3i x_4 &= 4 - sqrt(5)
b The expression of the polynomial g(x) can be obtained by multiplying the binomials generated by each root.
| Root | Binomial |
|---|---|
| 7+3i | x-(7+3i) |
| 7-3i | x-(7-3i) |
| 4+sqrt(5) | x-(4+sqrt(5)) |
| 4-sqrt(5) | x-(4-sqrt(5)) |
(x-(7+3i))(x-(7-3i))
▼
Simplify
AssociativePropAdd
Associative Property of Addition
((x-7)-3i)((x-7)+3i)
(a+bi)(a-bi)=a^2+b^2
(x-7)^2 + 9
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
x^2-14x + 49 + 9
AddTerms
Add terms
x^2-14x + 58
(x-(4+sqrt(5)))(x-(4-sqrt(5)))
▼
Simplify
AssociativePropAdd
Associative Property of Addition
((x-4)-sqrt(5))((x-4)+sqrt(5))
ExpandDiffSquares
(a+b)(a-b)=a^2-b^2
(x-4)^2 - (sqrt(5))^2
CalcPow
Calculate power
(x-4)^2 - 5
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
x^2 - 8x + 16 - 5
SubTerms
Subtract terms
x^2-8x+ 11
(x^2-14x + 58)(x^2-8x+ 11)
▼
Simplify
Distr
Distribute (x^2-8x+ 11)
x^2(x^2-8x+ 11)-14x(x^2-8x+ 11) + 58(x^2-8x+ 11)
Distr
Distribute x^2
x^4-8x^3+ 11x^2-14x(x^2-8x+ 11) + 58(x^2-8x+ 11)
Distr
Distribute 14x
x^4-8x^3+ 11x^2-14x^3+112x^2 - 154x + 58(x^2-8x+ 11)
Distr
Distribute 58
x^4-8x^3+ 11x^2-14x^3+112x^2 - 154x + 58x^2-464x+ 638
SubTerms
Subtract terms
x^4 -22x^3 + 11x^2+112x^2+ 58x^2 - 618x + 638
AddTerms
Add terms
x^4 -22x^3 + 181x^2 - 618x + 638
Closure
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About the Roots
Previously, it was explained how the coefficients of a polynomial function can give information about its roots, including the irrational and complex roots. This can be especially helpful considering that the graph of a function also gives some information. For example, consider a polynomial function p(x). p(x) = (x-2)^2(x-5) ⇓ p(x) = x^3 - 9 x^2 + 24 x - 20 By the Factor Theorem, the roots of the polynomial are 2 and 5, with a multiplicity of 2 and 1, respectively. Now consider the graph of p(x).
Looking at the graph, it can be noted that the graph does not cross the x-axis when touching the root with multiplicity of 2, but the graph crosses the axis on the next root, with multiplicity of 1. Now consider a different function. It will be written factored because the coefficients are big. p_2(x) = (x-2)^3(x-5)^4 The roots of the polynomial are 2 and 5 again, but this time the roots have a multiplicity of 3 and 4 respectively. Consider the graph of p_2(x).
This time, the function's graph crosses the x-axis on root 2 and it does not on root 5. From this, it is possible to draw two conclusions.
- The graph of a function crosses the x-axis on a root if the root's multiplicity is odd.
- The graph of a function does not cross the x-axis on a root if the root's multiplicity is even.
But what happens if the root is a complex number? Think about the following function. p_3(x) = (x-2)^2(x-5)(x-2i)(x+2i) The graph of this function is presented next.
Polynomial Root Theorems
Exercise 3.1
Consider the following rectangular prism.
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To find the value of x, we first need to find an expression for the height. We are told that the height is two times the length minus the width. Let's write this as an algebraic expression. h = 2l -w We can write an expression for h in terms of x by substituting the expressions for the width and length into the this formula.
Now that we have an expression for h in terms of x, we can write an expression for the volume of the rectangular prism.
We found an expression for the volume. Since we know that the volume of the prism is 6 cubic inches, we can substitute this value into the expression for the volume to get a polynomial equation.
Since we are told that x is an integer, we can use the Rational Root Theorem to find this value. This theorem indicates that the factors of the constant term are the possible solutions to the equation. Let's look at the factors of the constant term, 6. Positive Factors:& 1,2,3,6 Negative Factors:& -1,-2,-3,-6 Since x is a number of inches, it has to be a positive number. This means we can disregard the negative factors. Let's substitute each of the positive factors into the equation to check if any result in a true statement.
| Factor | Substitute | Result |
|---|---|---|
| 1 | 15( 1)^3-11( 1)^2+2( 1) - 6 = 0 | 0 = 0 ✓ |
| 2 | 15( 2)^3-11( 2)^2+2( 2) - 6 = 0 | 74 ≠ 0 * |
| 3 | 15( 3)^3-11( 3)^2+2( 3) - 6 = 0 | 306 ≠ 0 * |
| 6 | 15( 6)^3-11( 6)^2+2( 6) - 6 = 0 | 2850 ≠ 0 * |
Since only 1 resulted in a true statement, the value of x has to be 1 inch.
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Polynomial Root Theorems
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