Math 8 | Quarter 1- Week 2 | Solving Problems Involving Factors of Polynomials | Acute Angels TV

Summary

In 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

00:00

📚 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.

05:01

📏 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.

10:03

🔍 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.

Keywords

💡Polynomials

Polynomials are algebraic expressions consisting of variables and coefficients, that involve only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. In the video, polynomials are the central theme as they are used to formulate and solve various word problems involving factors, demonstrating their importance in mathematical problem-solving.

💡Factors

Factors are numbers or expressions that divide into another number or polynomial without leaving a remainder. The video discusses solving problems where identifying the factors of polynomials is crucial, such as finding consecutive integers whose product is 272 or determining the dimensions of a rectangle given its area.

💡Consecutive Integers

Consecutive integers are numbers that follow one after the other in order, such as 4 and 5. The script uses the concept of consecutive integers in the first word problem, where the product of two such integers equals 272, to illustrate how to set up and solve an equation based on this relationship.

💡Equation

An equation is a statement that asserts the equality of two expressions, typically denoted by the symbol '='. The video script involves setting up and solving equations derived from word problems, such as 'n squared plus n minus 272 equals zero', to find unknown values.

💡Factoring

Factoring is the process of breaking down a polynomial into a product of its factors. The script demonstrates factoring as a method to solve polynomial equations, such as expressing 'n squared plus n minus 272' as '(n + 17)(n - 16)' to find the values of n.

💡Rectangle

A rectangle is a quadrilateral with four right angles. The video uses the properties of a rectangle, specifically its length and width, to solve for its dimensions when given the area, showcasing the application of algebra in geometry.

💡Area

Area is a measure of the two-dimensional space enclosed within a shape, typically expressed in square units. The script applies the formula for the area of a rectangle (length times width) to set up an equation that is solved to find the rectangle's dimensions.

💡Trinomial

A trinomial is a polynomial equation that has three terms. The video script includes solving trinomial equations, such as '3x squared minus 2x minus 65 equals zero', which requires factoring to find the values of x.

💡Coefficient

A coefficient is a numerical factor in a term of an algebraic expression. In the video, coefficients are used to describe the relationship between terms in polynomials, such as the coefficient '1' in 'n squared plus n minus 272' indicating the linear term.

💡Word Problems

Word problems are mathematical problems that are stated in the form of a sentence or a story. The video script presents several word problems that require the application of algebraic concepts, such as factoring and solving equations, to find practical solutions.

💡Solving Equations

Solving equations involves finding the values of the variables that make the equation true. The script demonstrates the process of solving equations by setting them to zero, factoring, and finding the roots, which is essential for understanding the application of algebra in problem-solving.

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