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| 13 Theory slides |
| 11 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Polynomial Function | Roots |
---|---|
p(x)=x3−6x2+11x−6 | 1, 2, 3 |
g(x)=x4−4x3−19x2+106x−120 | -5, 2, 3, 4 |
m(x)=x3+6x2−55x−252 | -9, -4, 7 |
LHS−a0=RHS−a0
Factor out xr
For the second property, suppose that the polynomial P(x) has a rational root qp, such that the fraction is written in its simplest form. Again, it is obtained that P(qp)=0.
(ba)m=bmam
LHS⋅qn=RHS⋅qn
LHS−a0(qn)=RHS−a0(qn)
Factor out p
The numerator of the root, p, is a factor of a0.
LHS−anpn=RHS−anpn
Factor out q
Equation | Result |
---|---|
6⋅14−59⋅13+94⋅12−49⋅1+8=0 | 0=0✓ |
6⋅24−59⋅23+94⋅22−49⋅2+8=0 | -90=0× |
6⋅44−59⋅43+94⋅42−49⋅4+8=0 | -924=0× |
6⋅84−59⋅83+94⋅82−49⋅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 a0 and a denominator that is a factor of an. In this case,an 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 | 21 | 22=1 | 24=2 | 28=4 |
3 | 31 | 32 | 34 | 38 |
6 | 61 | 62=31 | 64=32 | 68=34 |
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(21)4−59(21)3+94(21)2−49(21)+8=0 | 0=0✓ |
6(31)4−59(31)3+94(31)2−49(31)+8=0 | 0=0✓ |
6(32)4−59(32)3+94(32)2−49(32)+8=0 | 2722=0× |
6(34)4−59(34)3+94(34)2−49(34)+8=0 | -9100=0× |
6(38)4−59(38)3+94(38)2−49(38)+8=0 | -277280=0× |
6(61)4−59(61)3+94(61)2−49(61)+8=0 | 108235=0× |
Izabella is tutoring some of her friends for an upcoming math exam.
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 | 18300=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 | 21420=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.
Bring down 1
2⋅1=2
-1+2=1
2⋅1=2
-6+2=-4
2⋅-4=-8
14−8=6
2⋅6=12
-12+12=0
Bring down 1
-3⋅1=-3
1−3=-2
-3⋅-2=6
-4+6=2
-3⋅2=-6
6−6=0
x=22+-4 | x=22−-4 |
---|---|
x=22+2i | x=22−2i |
x=1+i | x=1−i |
Therefore, the remaining roots of the given polynomial function are the complex numbers 1+i and 1−i.
Izabella now wants to challenge her study group to try a problem all on their own. Kevin is ready for it!
Remember the Rational Root Theorem.
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× |
Substitution | Simplify |
---|---|
3(-32)3−(-32)2−6(-32)+2=0 | 314=0× |
3(-31)3−(-31)2−6(-31)+2=0 | 334=0× |
3(31)3−(31)2−6(31)+2=0 | 0=0✓ |
3(32)3−(32)2−6(32)+2=0 | -914=0× |
31⋅3=1
-1+1=0
31⋅0=0
-6+0=-6
31⋅(-6)=-2
2−2=0
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.
Let P(x) be a polynomial function whose coefficients are rational. If a and b are rational such that b is an irrational number and a+b is a root of P(x), then its irrational conjugate a−b is also a root of P(x).
P(a+b)=0 ⇒ P(a−b)=0
i=0∑nci⋅(a+b)i | Polynomial |
i=0∑nci⋅(a+b)i | Conjugate of a sum is the sum of the conjugates |
i=0∑nci (a+b)i | Conjugate of a product is the product of the conjugates |
i=0∑nci (a+b)i | Conjugate of a power is the power of the conjugate |
i=0∑nci (a+b)i | Conjugate of a rational number is itself |
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.
Remember the Irrational Conjugate Root Theorem
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.
Consider the Irrational Conjugate Root Theorem.
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.
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
P(a+bi)=k=0∑nck(a+bi)k | The conjugate of the polynomial evaluated at a+bi |
P(a+bi)=k=0∑nck(a+bi)k | The conjugate of a sum is the sum of the conjugates. |
P(a+bi)=k=0∑nck (a+bi)k | The conjugate of a product is the product of the conjugates. |
P(a+bi)=k=0∑nck(a+bi)k | The conjugate of a power is the power of the conjugate. |
P(a+bi)=k=0∑nck(a+bi)k | The conjugate of a real number is itself. |
P(a+bi)=k=0∑nck(a−bi)k | The conjugate of a+bi is a−bi. |
P(a+bi)=P(a−bi) | The polynomial evaluated at a−bi |
As Zain progresses with their studies, they build up the courage to ask Izabella for extra help with polynomials that have complex roots.
Associative Property of Addition
(a+bi)(a−bi)=a2+b2
(a−b)2=a2−2ab+b2
Add terms
Distribute (x−4)
Distribute x2
Distribute 6x
Distribute 10
Add and subtract terms
After studying polynomial functions with complex roots, Zain feels confident enough to finish an assignment all on their own.
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.
Root | Binomial |
---|---|
7+3i | x−(7+3i) |
7−3i | x−(7−3i) |
4+5 | x−(4+5) |
4−5 | x−(4−5) |
Associative Property of Addition
(a+bi)(a−bi)=a2+b2
(a−b)2=a2−2ab+b2
Add terms
Associative Property of Addition
(a+b)(a−b)=a2−b2
Calculate power
(a−b)2=a2−2ab+b2
Subtract terms
Distribute (x2−8x+11)
Distribute x2
Distribute 14x
Distribute 58
Subtract terms
Add terms
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.
Consider the following rectangular prism.
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.