Math 8 | Quarter 1- Week 2 | Solving Problems Involving Factors of Polynomials | Acute Angels TV
Summary
TLDRIn this educational video, Teacher Eliza teaches students how to solve problems involving factors of polynomials. She walks through three word problems, demonstrating step-by-step solutions, including finding two consecutive integers with a product of 272, determining the dimensions of a rectangle with an area of 65 square feet, and identifying a number that satisfies a given polynomial equation. The lesson emphasizes the importance of defining variables, setting up equations, and factoring to find solutions, aiming to equip students with the skills to tackle similar problems.
Takeaways
- 📚 The lesson is focused on solving problems involving factors of polynomials.
- 🔍 The first word problem involves finding two consecutive integers whose product is 272.
- 📝 The teacher introduces a method to define and set up an equation for the problem, leading to the equation n^2 + n = 272.
- 🧩 The equation is factored into (n + 17)(n - 16) = 0, yielding two possible integer solutions.
- 🔑 The solutions to the first problem are the pairs of integers (-16, -17) and (16, 17).
- 🏠 The second problem deals with finding the dimensions of a rectangle given its area and the relationship between length and width.
- 📐 The dimensions are found using the equation 3x^2 - 2x = 65, which is then factored to find the width.
- 📏 The width of the rectangle is determined to be 5, and the length is 13, confirming the area is indeed 65 square feet.
- 🔢 The third problem asks to find a number where three times the number minus 2 equals negative 9 times the square of the number.
- 🤔 The equation is set to 9x^2 + 3x - 2 = 0 and factored to find the possible values of the number.
- 🎯 The final values for the number are x = -2/3 and x = 1/3, which are verified to satisfy the original equation.
- 👩🏫 The lesson is taught by Teacher Eliza Mae Kunan, a grade 8 mathematics teacher, emphasizing the importance of understanding factors in polynomials.
Q & A
What is the main topic of the learning episode presented by Teacher Eliza?
-The main topic of the learning episode is solving problems involving factors of polynomials.
What is the first word problem presented by Teacher Eliza, and what is the mathematical equation derived from it?
-The first word problem is about finding two consecutive integers whose product is 272. The derived equation is n^2 + n = 272.
How does Teacher Eliza suggest setting up the equation for the first word problem?
-Teacher Eliza suggests setting up the equation by defining the first integer as n and the second consecutive integer as n + 1, and then multiplying these two integers to get the equation n(n + 1) = 272, which simplifies to n^2 + n = 272.
What is the next step after setting up the equation for the first word problem?
-The next step is to set the equation equal to zero by subtracting 272 from both sides, resulting in n^2 + n - 272 = 0.
How does Teacher Eliza approach factoring the quadratic equation from the first word problem?
-Teacher Eliza approaches factoring by looking for two numbers that multiply to -272 and add up to 1 (the coefficient of the middle term). The correct pair of factors is -16 and 17.
What are the possible values of the first integer 'n' in the first word problem?
-The possible values of 'n' are -17 and 16, as derived from setting each factor equal to zero: n + 17 = 0 or n - 16 = 0.
What is the second word problem presented by Teacher Eliza, and what is the formula used to solve it?
-The second word problem is about finding the dimensions of a rectangle with a given area of 65 square feet. The formula used is the area of a rectangle, which is length times width, leading to the equation 3x^2 - 2x = 65.
How does the equation for the second word problem get transformed to set it equal to zero?
-The equation is transformed by subtracting 65 from both sides, resulting in 3x^2 - 2x - 65 = 0.
What is the method used by Teacher Eliza to factor the trinomial equation from the second word problem?
-Teacher Eliza uses trial and error with the factors of -65 to find the correct pair that, when multiplied with the respective terms, gives the middle term of the trinomial. The correct pair is -5 and 13.
What are the possible dimensions of the rectangle in the second word problem?
-The possible dimensions of the rectangle are width 5 feet and length 13 feet, as derived from solving x = 5 and substituting it into the length formula 3x - 2.
What is the third word problem presented by Teacher Eliza, and what is the final equation to solve it?
-The third word problem is about finding a number where the difference of three times the number and 2 is the same as negative 9 times the square of the number. The final equation to solve is 9x^2 + 3x - 2 = 0.
How does Teacher Eliza determine the correct factors for the trinomial equation in the third word problem?
-Teacher Eliza determines the correct factors by trying different pairs of factors of -2 and finding that the pair -1 and 2, when used, results in the middle term of the trinomial, which is 3x.
What are the possible values of the number 'x' in the third word problem?
-The possible values of 'x' are -2/3 and 1/3, as derived from solving 3x + 2 = 0 and 3x - 1 = 0.
How does Teacher Eliza verify the solutions for the third word problem?
-Teacher Eliza verifies the solutions by substituting the values of 'x' back into the original equation and checking if they satisfy the equation.
Outlines
📚 Introduction to Solving Polynomial Factors
In this educational video, Teacher Eliza introduces viewers to the concept of solving problems involving factors of polynomials. She sets the expectation that by the end of the lesson, students will be able to solve such problems. The first word problem involves finding two consecutive integers whose product is 272. The teacher guides the students through defining the integers, setting up the equation, and applying algebraic methods to solve for the integers, ultimately finding the pairs -17 and -16, and 16 and 17.
📏 Solving for Rectangle Dimensions
The second paragraph presents a word problem about finding the dimensions of a rectangle given the area and a relationship between its length and width. The width is represented by x, and the length by 3x - 2. The area is given as 65 square feet. The teacher demonstrates how to set up the equation, factor it, and solve for the possible values of x, which are -4.3 and 5. Since a width cannot be negative, the width is determined to be 5 feet, and the length is calculated to be 13 feet, confirming the area as 65 square feet.
🔍 Finding a Number in a Polynomial Equation
The final paragraph discusses a problem where the difference between three times a number and 2 is equal to negative 9 times the square of the number. The teacher formulates this into a quadratic equation, 9x^2 + 3x - 2 = 0, and factors it to find the possible values of x, which are -2/3 and 1/3. These solutions are then checked by substituting them back into the original equation to confirm their validity, concluding the lesson on solving polynomial factor problems.
Mindmap
Keywords
💡Polynomials
💡Factors
💡Consecutive Integers
💡Equation
💡Factoring
💡Rectangle
💡Area
💡Trinomial
💡Coefficient
💡Word Problems
💡Solving Equations
Highlights
Introduction to solving problems involving factors of polynomials.
The first word problem involves finding two consecutive integers whose product is 272.
Defining the first integer as 'n' and the second as 'n+1' to set up the equation.
Formulating the equation n^2 + n = 272 to find the consecutive integers.
Applying the subtraction property of equality to set the equation to zero.
Factoring the quadratic equation n^2 + n - 272 = 0.
Identifying the correct pair of factors for -272 that sum to 1 (n+17 and n-16).
Solving for n to find the two sets of consecutive integers: -17 and -16 or 16 and 17.
Introduction to the second word problem about finding the dimensions of a rectangle with a given area.
Defining the width as 'x' and the length as '3x-2' with an area of 65 square feet.
Setting up the equation 3x^2 - 2x = 65 to find the rectangle's dimensions.
Solving the quadratic equation 3x^2 - 2x - 65 = 0 by factoring.
Finding the correct factors of -65 that lead to the middle term -2x (3x+13 and x-5).
Determining the width and length of the rectangle to be 5 feet and 13 feet, respectively.
Introduction to the third word problem involving a number and its square.
Defining the equation 3x - 2 = -9x^2 and setting it to zero.
Factoring the trinomial 9x^2 + 3x - 2 = 0 to find the number.
Identifying the correct pair of factors for -2 that result in the middle term +3x (3x+2 and 3x-1).
Solving for x to find the possible values of -2/3 and 1/3.
Verification of the solutions by substituting back into the original equation.
Conclusion of the lesson with a summary and acknowledgment of the learners.
Transcripts
[Music]
good day acute angels welcome to a new
learning episode
this is teacher eliza your graded
mathematics teacher
today you will learn about solving
problems involving factors of
polynomials
at the end of this lesson you are
expected to
solve problems involving factors of
polynomials
[Music]
let us have our word problem number one
the product of two consecutive integers
is two hundred seventy-two
find the value of each integer
the first thing that you need to do is
to define the integers
let n
be the first integer and let n plus 1 be
the second integer
the word product here means to multiply
so we need to multiply the two integers
together
the first integer which is n multiplied
by the second integer which is n plus
one
equals two hundred seventy-two
now we multiply everything out
n times n is equal to n squared
and n times 1 is equal to n so the
equation will become n squared plus n is
equal to 272
now set the equation equal to zero
in order to do that
we need to apply the subtraction
property of equality to subtract 272
from both sides of the equation
so n squared plus n minus 272
is equal to zero
once the equation was equated to zero
then it is time to factor and solve
since the first term is n squared we
know that the factoring must take the
form
quantity n plus blank
multiplied by quantity n minus blank
equals to zero
it will take this form because when we
multiply the two linear terms the first
term must be n squared and the only way
to get that is to multiply n by n
therefore the first term in each factor
must be n
to finish this we need to determine the
two numbers that need to go in the blank
spots
we can narrow down the
possibilities by listing all the factors
of the third term which is negative 272.
here are the factors of negative 272
and on the second column
is the sum of each pair
take note that the sum of the correct
pair of numbers or factors
must be equal to the coefficient of the
second term in our example our second
term is n
and its coefficient is 1.
[Music]
so in this case the pair of factors
negative 16 and 17
has a sum of one so this is the pair
that we are looking for
now let us substitute or put these two
factors in the blank spots
so n
plus 17
our quantity n plus 17 multiplied by
quantity n minus sixteen is equal to
zero
this means n plus seventeen is equal to
zero or n minus sixteen is equal to zero
for our first value of n we have n
equals negative 17 or n equals 16.
we have solved two values of n
let us find out what will be the other
value of the second integer if n is
equal to negative 17 or 16.
if n is equal to negative 17
we will just substitute negative 17 to n
plus one
then n plus one or negative seventeen
plus one is equal to negative sixteen
hence the two integers are negative
seventeen and negative sixteen
[Music]
next what about if n is equal to
positive 16
if n is equal to 16
then n plus 1 or 16 plus 1 is equal to
17.
hence the two integers are 16 and 17.
[Music]
both of these pairs are correct
the answer to this problem or the value
of the two integers are
negative 16 and negative seventeen
or sixteen and seventeen
let's move on with word problem number
two
the length of a rectangle is two feet
less than 3 times the width
if the area is 65 feet squared
find its dimensions
let x be the width of the rectangle
and 3x minus 2 its length
and its given area is 65 bit squared
the formula to get the area of the
rectangle is length times width
so we have to substitute the given
expressions for these terms which are
quantity three x minus two and x
equals sixty-five
now multiply everything out
three x multiplied by x is three x
squared and negative two multiplied by x
is negative two x
so this equation will become three x
squared minus two x equals sixty-five
now it's time to set the equation equal
to 0.
to subtract 65 from both sides of the
equation we need to apply the
subtraction property of equality so 3x
squared minus two x minus sixty-five is
equal to zero
[Music]
since the equation was already equated
to zero
then it is time to solve and factor
for this given trinomial the factoring
must take the form quantity three x plus
blank multiplied by quantity x minus
blank equals to zero
since three x and x are the factors of
the first term 3x squared
or x may come first and 3x may be
written as the first term
of the second factor
[Music]
now we need to complete or determine the
two numbers that need to go in the blank
spots in order to do that we have to get
the factors of the third term of the
given trinomial negative sixty-five here
are the factors of negative sixty-five
negative one and sixty-five
one and negative sixty-five
negative five and thirteen
and five and negative thirteen
to find the correct pair of numbers or
factors
we will plug in each of these factors
and multiply out until we get the
correct pair
let us start with negative 1 and 65
now we multiply the either terms 65 and
x
which is equal to 65x and the outer
terms 3x times negative 1 is negative 3x
now get the sum of these two products
65x plus negative 3x is equal to 62x
this is not the product we are looking
for since this is not equal to the
second term of the given trinomial which
is negative 2x
now let us try another pair of factors
let us have 1 and 6 negative 65
the product of the inner terms is x
and the product of the outer terms is
negative 195 x
get the product or sum rather of these
two products we have
negative 194x
which is still not the product we are
after
now what about the third pair of factors
negative 5 and 13
let us plug these two numbers or these
two factors
13 times x is equal to 13x and the outer
terms 3x times negative 5 is equal to
negative 15x
add these two products 13x plus negative
15x is equal to negative 2x
now negative 2x is equal to the second
term of our trinomial
therefore
this is the pair of factors that we are
looking for
we can now proceed and solve
this equation
this means 3x plus 13 is equal to 0 or x
minus 5 is equal to 0.
for our first value of x we have
x is equal to negative 13 over 3 or x is
equal to negative 4.3
and the second value of x is equal to
5 or x is equal to 5.
the width of the rectangle is
x is equal to five or five
its length is three x minus two
so we have to substitute the value of x
which is five
so three times five minus two is equal
to thirteen
and its area is length times width or
five times three is equal to sixty-five
therefore the dimensions of the
rectangle are
its width is five
its length is 13
and its area is 65 ft squared
we are down with our last word problem
the difference of three times a number
and 2 is the same as negative 9 times
the square of the number
find the number
let x be the number and 3x minus 2
equals negative 9x squared be the given
equation we will set this given equation
equal to zero
we will apply addition property of
equality and add 9x squared on both
sides of the equation
so this will become nine x squared plus
three x minus two equals zero
now it's time to factor this trinomial
the factoring will take the form
quantity three x plus blank multiplied
by quantity 3x minus blank equals zero
3x and 3x are the first terms for each
factor since these are the factors
of 9x squared
now to get the second terms we have to
determine the factors of the third term
of the trinomial which is negative two
and these are
negative one and two
one and negative two
let us try first the first pair of
factors which are negative 1 and 2.
multiply the inner terms 2 and negat 2
and 3x we have 6x
and multiply 3x and negative 1 we have
negative 3x
6x plus negative 3x is equal to
positive 3x since 3x is the second term
of the trinomial therefore the pair of
factors negative 1 and 2 are what we are
looking for
we can now proceed in solving this
equation
so this becomes 3x plus 2 equals 0
or three x minus one equals zero
our first value of x is
x is equal to negative two-thirds
and the second value of x is one-thirds
to check these two values of x that we
have obtained let us substitute them to
the given equation three x minus two
equals negative nine x squared
let us start first with negative
two-thirds
if x is negative two-thirds
then three multiplied by negative
two-thirds minus two equals negative
nine multiplied by negative two-thirds
squared
multiply
three and negative two-thirds
and also negative nine and negative
two-thirds so this becomes negative two
minus two equals negative four
negative two minus two is equal to
negative four
which is equal to negative four
now if x is equal to one third
then three x minus two equals negative
nine x squared will become
three multiply by one third minus two
equals negative nine multiplied by one
third squared
multiply three by one third and negative
nine by one third squared
and the equation will become
one minus two equals negative one
one minus two is equal to negative one
which is equal to negative one
therefore the possible values of x are
negative two-thirds and one-third
and that ends our discussion on solving
problems involving factors of
polynomials
great job great learners
thank you for your time and effort
i hope you have learned a lot from this
discussion
again this is teacher eliza mae kunan
your grade 8 mathematics teacher
have a good day and god bless
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