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| 12 Theory slides |
| 9 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.
Factoring
Tearrik is watching the clock on the wall just waiting for the school bell to ring so he can prepare for a fun weekend with friends and family. Just before the bell rang, his math teacher assigned the following challenge.
At first glance, he thought that the task was very simple since there is only one number whose cube is 125. Is Tearrik right? Are there no more solutions to the equation? Find all the solutions to the equation.
In Tearrik's courses, he previously learned that some real-life situations such as saving a constant amount of money weekly or shooting a basketball can be modeled by linear and quadratic equations, respectively.
Now, he wonders if there are situations involving other types of equations. Specifically, Tearrik wants to know if an equation could contain a polynomial. Luckily for Tearrik, his teacher is planning on introducing the topic next class.
On the weekend, Tearrik and his friends decided to go to the amusement park to have some fun.
LHS−7=RHS−7
Rearrange equation
LHS⋅5=RHS⋅5
Zero Property of Multiplication
Distribute 5
Calculate quotient
Substitute values
Calculate power and product
-(-a)=a
Subtract term
Calculate root
Factor out 2
ba=b/2a/2
State solutions
(I), (II): Add and subtract terms
LHS⋅5=RHS⋅5
Multiply
Distribute 5
Calculate quotient
LHS−44=RHS−44
Rearrange equation
LHS⋅(-1)=RHS⋅(-1)
Zero Property of Multiplication
Distribute (-1)
Associative Property of Addition
Factor out t2
Factor out (t−6)
After arriving home and feeling excited about solving some polynomial equations at the amusement park, Tearrik sees a note written by his sister. Tearrik gets right to his homework so he can finish in time to watch a movie with his family!
Tearrik's homework asks him to factor a pair of polynomial equations.
48x9=12x7⋅4x2, 108x7=12x7⋅9
Factor out 12x7
Substitute expressions
Factor out 3x5
While checking the factorization methods he knows so far, Tearrik notices that he has a formula for factoring the sum and difference of two squares.
Sum of Squares | Difference of Squares |
---|---|
a2+b2=(a+bi)(a−bi) | a2−b2=(a+b)(a−b) |
However, Tearrik wonders if a similar formula exists for the sum of two cubes. The good news is that such a formula does exist and, along with the Zero Product Property, is useful for solving polynomial equations.
The sum of two cubes can be factored as the product of a binomial by a trinomial.
a3+b3=(a+b)(a2−ab+b2)
Notice that the binomial is the sum of the bases a and b, and the trinomial is the sum of the bases squared minus the product of the bases.
Multiply parentheses
am⋅an=am+n
Commutative Property of Addition
Associative Property of Addition
Add and subtract terms
After reading the formula and having in mind that a cube can also be thought of as a three-dimensional object, Tearrik wondered whether there is a geometric way of deducting the formula. Indeed, there is one way of visualizing the formula geometrically. First, consider two cubes, one of side a and another of side b.
In need of a break from such a fun weekend, Tearrik decided to just relax in his room. Looking around, he sees a die and a Rubik's cube that have been laying around forever. He wonders about the sum of their volumes. He knows that each side of the die measures 2 centimeters but does not know the dimensions of the Rubik's cube.
Substitute values
Calculate power and product
Subtract term
a3+b3=(a+b)(a2−ab+b2)
Calculate power and product
Zero Property of Multiplication
(I): LHS−7=RHS−7
(I): LHS/2=RHS/2
Use the Quadratic Formula: a=4,b=-14,c=49
Calculate power and product
-(-a)=a
Subtract term
-a=a⋅i
Split into factors
a⋅b=a⋅b
Calculate root
Factor out 2
ba=b/2a/2
Write as a sum of fractions
Equation | Solutions |
---|---|
8x3+350=7 | x1x2x3=-27=47+473i=47−473i
|
Substitute values
Commutative Property of Addition
Subtract term
ba=b⋅2a⋅2
Add fractions
Add and subtract terms
a0=0
Substitute expressions
Factor out (a−b)
Commutative Property of Addition
The difference of two cubes can be factored as the product of a binomial by a trinomial.
a3−b3=(a−b)(a2+ab+b2)
Notice that the binomial is the difference of the bases a and b, and the trinomial is the sum of the bases squared plus the product of the bases.
Multiply parentheses
am⋅an=am+n
Commutative Property of Addition
Associative Property of Addition
Subtract terms
Rewrite 64 as 43
a3−b3=(a−b)(a2+ab+b2)
Calculate power
Use the Quadratic Formula: a=1,b=4,c=16
Calculate power and product
Subtract term
-a=a⋅i
Split into factors
a⋅b=a⋅b
Calculate root
Factor out 2
ba=b/2a/2
Equation | Solutions |
---|---|
x3−27=37 | x1=4 x2=-2−23i x3=-2+23i |
Real Solutions | Imaginary Solutions |
---|---|
x=4 | x=-2−23i |
x=-2+23i |
Graph the polynomial function.
To make the graph, a graphing calculator could be used.
Notice that the y-coordinates of the x-intercepts are equal to zero. That is, if the graph of f cuts the x-axis at x0, then f(x0)=0. Consequently, the x-intercepts are the solutions to the equation defined in the first step. Therefore, identify them.
There are three x-intercepts, which means that there are three solutions. In this case, the two left-hand side solutions will be approximated.
x=-2.5
Calculate power and product
Multiply
Add and subtract terms
Value | Substitution | Solution? |
---|---|---|
x≈0.7 | f(0.7)=20(0.7)3−44(0.7)2−179(0.7)+140=0 | ✓ |
x=4 | f(4)=20(4)3−44(4)2−179(4)+140=0 | ✓ |
As verified, the three values obtained in the fourth step are solutions to the polynomial equation. Note that these three values could also be obtained by graphing y=P(x) and y=Q(x) on the same coordinate plane and determining the x-coordinates of their points of intersection.
Using this method, Tearrik noticed that a polynomial equation of the form x3+C=0 has only one x-intercept. Therefore, it has only one real solution.
The previous conclusion makes perfect sense with the two factoring formulas Tearrik studied before.
Method | Formula |
---|---|
Sum of Two Cubes | x3+a3=(x+a)(x2−ax+a2) |
Difference of Two Cubes | x3−a3=(x−a)(x2+ax+a2) |
Tearrik and tío Angelito are now taking a break from woodwork. Tío Angelito tells Tearrik to pull up a chair — he has a story to tell. Way back in the day, he used to go diving off the coast. Being that he loves math, when he dives he tries to follow the trajectory of a polynomial.
By following the given polynomial, he was able to model the trajectory of one of his favorite dives. In this polynomial, D(t) represents the depth, in meters, at which tío Angelito was t minutes after he started diving. Negative values of D(t) mean that he was underwater.
D(t)=-90
LHS+90=RHS+90
Rearrange equation
t | 1.5 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 11.2 |
---|---|---|---|---|---|---|---|---|---|---|---|---|
f(t) | ≈77 | 0 | -40 | 0 | 54 | 80 | 60 | 0 | -70 | -96 | 0 | ≈42 |
Given those four intercepts, it can be determined that there were four moments in which tío Angelito was exactly 90 meters underwater — namely, at 2, 4, 8, and 11 minutes after he began his dive.
LHS+200=RHS+200
Rearrange equation
t | 1.5 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 11.2 |
---|---|---|---|---|---|---|---|---|---|---|---|---|
g(t) | ≈187 | 110 | 70 | 110 | 164 | 190 | 170 | 110 | 40 | 14 | 110 | ≈152 |
LHS+150=RHS+150
Rearrange equation
t | 1.5 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 11.2 |
---|---|---|---|---|---|---|---|---|---|---|---|---|
h(t) | ≈137 | 60 | 20 | 60 | 114 | 140 | 120 | 60 | -10 | -36 | 60 | ≈102 |
Each of the questions can also be solved by analyzing the graph of y=D(t). As done in the previous parts, begin the process of graphing the equation by making a table of values.
t | 1.5 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 11.2 |
---|---|---|---|---|---|---|---|---|---|---|---|---|
D(t) | ≈-13 | -90 | -130 | -90 | -36 | -10 | -30 | -90 | -160 | -186 | -90 | ≈-48 |
Based on the drawn graph, some conclusions can be made.
x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
---|---|---|---|---|---|---|---|---|---|---|---|---|
y | ≈658 | ≈507 | ≈333 | ≈183 | ≈78 | ≈21 | ≈1 | ≈0.5 | ≈0 | ≈-15 | ≈-47.5 | ≈-85 |
According to the table, there is only one sign change, which means that between -4 and 7 there is only one x-intercept — at x=4. Therefore, the graph of the polynomial should look as follows.
Furthermore, from -3 to the left the graph goes up, so it is expected to continue going up to the left of x=-4 as well. Similarly, from 6 onwards the graph decreases, so it is expected to continue descending to the right of x=7.
This is all that can be deduced from the table of values. However, some things are mistaken. According to a graphing calculator, the graph of the polynomial is the following.
As seen, the graph intersects the x-axis twice between 2 and 3. Also, the behavior on both ends is contrary to the one previously found. Therefore, the table of values should include more values and also consider decimal numbers.
A table of values might not precisely reflect all the characteristics of a graph.
Although the given polynomial has only two terms, we can factor it by noting that 64 is equal to 4^3. Making this change converts the polynomial into the sum of two cubes. P(x) &= x^3 + 64 &⇓ P(x) &= x^3 + 4^3 The sum of two cubes can be factored using the following formula. a^3 + b^3 = (a+b)(a^2 - ab + b^2) In our case, a= x and b= 4. Thus, let's factor the given polynomial.
Now, let's factor the quadratic polynomial. First, we will check whether it has real solutions by calculating the discriminant. To do so, we will substitute a= 1, b= -4, and c= 16 into the expression b^2-4ac.
Since the discriminant is negative, the quadratic polynomial does not have real solutions. Therefore, it cannot be factored using integer coefficients. Consequently, the factored form of P(x) looks as follows. P(x) = (x+4)(x^2-4x+16)
To find the solutions to the given polynomial equation, instead of expanding the cubic power, notice that we can rewrite 216 as 6^3. Making this change converts the given equation into the sum of two cubes.
(x-3)^3 + 216 = 0
⇓
(x-3)^3 + 6^3 = 0
Next, let's factor the expression on the left-hand side by using the same formula we used in Part A. This time a=x-3 and b=6.
Applying the Zero Product Property gives us two equations, one linear and another quadratic. (x+3)(x^2 - 12x + 63) = 0 ↙ ↘ x+3=0 x^2 - 12x + 63 = 0 The solution to the linear equation is x=-3. Let's use the Quadratic Formula to find the solutions to the quadratic equation.
From the above, we can conclude that the quadratic equation has two imaginary solutions. Now, we are ready to list all the solutions to the given equation. x_1 &= -3 x_2 &= 6+3sqrt(3)i x_3 &= 6-3sqrt(3)i
At the intersection of Euler Avenue and Complex Street there is a bank that is shaped like a cube. To build the entrance, a smaller cube was removed from one corner of the building.
According to the given information, the volume of the bank is equal to the volume of the bigger cube minus the volume of the removed cube. The volume of a cube is equal to its side length cubed. Thus, let's start by writing the side length of each cube.
Cube | Side Length (ft) | Volume |
---|---|---|
Bigger Cube (building) | x | x^3 |
Removed Cube (entrance) | 9 | 9^3 |
With this information, we can write a polynomial modeling the volume of the bank. V(x) = x^3 - 9^3 Our next mission is to factor V(x) so that each factor has integer coefficients. To start, we can factor it as a difference of cubes.
To continue factoring V(x), we have to factor the quadratic polynomial. Before doing that, let's find its discriminant. To do so, let's substitute a= 1, b= 9, and c= 81 into the expression b^2-4ac.
Since the discriminant is negative, the quadratic polynomial does not have real solutions. Therefore, it cannot be factored using integer coefficients. Consequently, the factored form of V(x) looks as follows. V(x) = (x-9)(x^2 + 9x + 81)
To find the solutions to the given polynomial equation, notice that we can rewrite 512 as 8^3. Making this change converts the given equation into the difference of two cubes.
(x+1)^3 - 512 = 0
⇓
(x+1)^3 - 8^3 = 0
Next, let's factor the expression on the left-hand side by using the same formula we used in Part A. This time a=x+1 and b=8.
Applying the Zero Product Property gives us two equations, one linear and another quadratic. (x-7) (x^2 + 10x + 73) = 0 ↙ ↘ x-7=0 x^2 + 10x + 73 = 0 The solution to the linear equation is x=7. To find the solutions to the quadratic equation, let's use the Quadratic Formula.
From those solutions, we can conclude that the quadratic equation has two imaginary solutions. Now, we are ready to list all the solutions to the given equation. x_1 &= 7 x_2 &= -5+4sqrt(3)i x_3 &= -5-4sqrt(3)i
The following graph corresponds to a fifth-degree polynomial P(x).
Since we are given the graph of P(x), we can find the real solutions to the equation P(x)=0 graphically. To do so, let's identify the x-intercepts of the graph.
As we can see, the graph has three x-intercepts — one at -9, the second at -4, and a third at 1. Therefore, evaluating the polynomial at any of these values gives 0 as the output. P(-9) &= 0 P(-4) &= 0 P(1) &= 0 Consequently, these three values are solutions to the equation P(x)=0. Moreover, we can see that the graph does not have more x-intercepts. This means that the equation P(x)=0 has only 3 real solutions.
Be aware that the graph of a polynomial does not give us information about the imaginary solutions — only about the real ones. However, since we know the degree of P(x), we can determine the number of imaginary solutions to P(x)=0.
Since the degree of P(x) is 5, it has a total of 5 roots. This implies that the equation P(x)=0 has 5 solutions. In Part A, we concluded that P(x)=0 has 3 real solutions. Consequently, the remaining 2 solutions are imaginary numbers.