| {{ 'ml-lesson-number-slides' | message : article.intro.bblockCount}} |
| {{ 'ml-lesson-number-exercises' | message : article.intro.exerciseCount}} |
| {{ 'ml-lesson-time-estimation' | message }} |
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.
LHSā7=RHSā7
Rearrange equation
LHSā 5=RHSā 5
Zero Property of Multiplication
Distribute 5
Calculate quotient
Substitute values
Calculate power and product
-(-a)=a
Subtract term
Calculate root
Factor out 2
baā=b/2a/2ā
State solutions
(I),Ā (II):Ā Add and subtract terms
LHSā 5=RHSā 5
Multiply
Distribute 5
Calculate quotient
LHSā44=RHSā44
Rearrange equation
LHSā (-1)=RHSā (-1)
Zero Property of Multiplication
Distribute (-1)
Associative Property of Addition
Factor out t2
Factor out (tā6)
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=12x7ā 4x2, 108x7=12x7ā 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)(aābi) | a2āb2=(a+b)(aāb) |
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)(a2āab+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
amā 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.
Next, place the small cube on top of the bigger one and complete the missing parts to form a prism.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)(a2āab+b2)
Calculate power and product
Zero Property of Multiplication
(I):Ā LHSā7=RHSā7
(I):Ā LHS/2=RHS/2
Use the Quadratic Formula: a=4,b=-14,c=49
Calculate power and product
-(-a)=a
Subtract term
-aā=aāā i
Split into factors
aā bā=aāā bā
Calculate root
Factor out 2
baā=b/2a/2ā
Write as a sum of fractions
Equation | Solutions |
---|---|
8x3+350=7 | x1āx2āx3āā=-27ā=47ā+473āāi=47āā473āāiā
|
Substitute values
Commutative Property of Addition
Subtract term
baā=bā 2aā 2ā
Add fractions
Add and subtract terms
a0ā=0
Substitute expressions
Factor out (aāb)
Commutative Property of Addition
The difference of two cubes can be factored as the product of a binomial by a trinomial.
a3āb3=(aāb)(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
amā an=am+n
Commutative Property of Addition
Associative Property of Addition
Subtract terms
Rewrite 64 as 43
a3āb3=(aāb)(a2+ab+b2)
Calculate power
Use the Quadratic Formula: a=1,b=4,c=16
Calculate power and product
Subtract term
-aā=aāā i
Split into factors
aā bā=aāā bā
Calculate root
Factor out 2
baā=b/2a/2ā
Equation | Solutions |
---|---|
x3ā27=37 | x1ā=4 x2ā=-2ā23āi x3ā=-2+23āi |
Real Solutions | Imaginary Solutions |
---|---|
x=4 | x=-2ā23āi |
x=-2+23āi |
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 x0ā, then f(x0ā)=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? |
---|---|---|
xā0.7 | f(0.7)=20(0.7)3ā44(0.7)2ā179(0.7)+140=0 | ā |
x=4 | f(4)=20(4)3ā44(4)2ā179(4)+140=0 | ā |
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)(x2āax+a2) |
Difference of Two Cubes | x3āa3=(xāa)(x2+ax+a2) |
Tearrik and tĆo Angelito are now taking a break from woodwork. TĆo 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 tĆo 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) | ā77 | 0 | -40 | 0 | 54 | 80 | 60 | 0 | -70 | -96 | 0 | ā42 |
Given those four intercepts, it can be determined that there were four moments in which tĆo 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) | ā187 | 110 | 70 | 110 | 164 | 190 | 170 | 110 | 40 | 14 | 110 | ā152 |
LHS+150=RHS+150
Rearrange equation
t | 1.5 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 11.2 |
---|---|---|---|---|---|---|---|---|---|---|---|---|
h(t) | ā137 | 60 | 20 | 60 | 114 | 140 | 120 | 60 | -10 | -36 | 60 | ā102 |
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) | ā-13 | -90 | -130 | -90 | -36 | -10 | -30 | -90 | -160 | -186 | -90 | ā-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 | ā658 | ā507 | ā333 | ā183 | ā78 | ā21 | ā1 | ā0.5 | ā0 | ā-15 | ā-47.5 | ā-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.