Solving Polynomial Equations By Factoring and Using Synthetic Division

Summary

This tutorial focuses on solving polynomial equations using various methods, including factoring by grouping, substitution, and the rational zero theorem with synthetic division. The instructor demonstrates how to factor polynomials, identify common factors, and solve for real and imaginary roots. Examples include solving cubic and quartic polynomials, and the tutorial emphasizes recognizing patterns such as the difference of perfect squares. By the end, viewers will understand multiple techniques for solving polynomial equations, equipping them with practical methods for tackling different types of polynomial problems.

Takeaways

• 🔢 The tutorial focuses on solving polynomial equations using various methods.
• 📐 Factoring by grouping is a technique used when the ratios of the coefficients of the first two terms match those of the last two terms.
• 🔄 Factoring involves taking out the greatest common factor (GCF) from each group of terms.
• 🔄 After factoring, if possible, further factor the expression, such as a difference of squares, to completely factor the polynomial.
• 🔍 When factoring by grouping is not possible, consider reducing the polynomial to a quadratic form using substitution.
• 🔄 Substitute xⁿ with a variable (like 'a') to transform the polynomial into a quadratic form that can be factored.
• 🔢 For polynomials that cannot be factored by grouping or reduced to quadratic form, use the rational zero theorem to list possible rational zeros.
• 🔄 Synthetic division is used to find the remaining zeros of a polynomial after identifying one zero.
• 🔄 Imaginary numbers can be used to factor polynomials with a sum of perfect squares.
• 📝 The tutorial provides step-by-step examples for each method, demonstrating how to solve specific polynomial equations.

Q & A

What method can be used to solve the equation x³ + 3x² - 4x - 12 = 0?

The equation can be solved by factoring by grouping because the ratio of the first two coefficients (1 and 3) is the same as the ratio of the last two coefficients (-4 and -12).

How is the equation x³ + 3x² - 4x - 12 factored?

First, group the terms and factor out the greatest common factors. The equation is factored as (x + 3)(x² - 4). Then, factor x² - 4 as a difference of squares to get (x + 3)(x + 2)(x - 2).

What are the solutions to the equation x³ + 3x² - 4x - 12 = 0?

The solutions are x = -3, x = -2, and x = 2.

How can you solve the equation x⁴ - 5x² - 36 = 0?

This equation can be solved by substitution. Let a = x², then solve the resulting quadratic equation a² - 5a - 36 = 0, which factors to (a - 9)(a + 4).

What are the real and imaginary solutions to x⁴ - 5x² - 36 = 0?

The real solutions are x = ±3, and the imaginary solutions are x = ±2i.

How do you solve x³ - 3x² - 6x + 8 = 0 using the Rational Root Theorem?

First, list all possible rational zeros using the factors of the constant term and the leading coefficient. Then, use synthetic division to test each possible zero. After finding one root, use synthetic division to simplify the equation and find the remaining roots.

What are the solutions to the equation x³ - 3x² - 6x + 8 = 0?

The solutions are x = 1, x = 4, and x = -2.

How do you apply synthetic division to find the remaining zeros after identifying one zero in a cubic equation?

After identifying one zero using the Rational Root Theorem, apply synthetic division by dividing the original polynomial by (x - root). The quotient will be a quadratic equation that can be solved using factoring or other methods to find the remaining zeros.

How can you factor the polynomial x⁴ - 7x³ + 7x² + 35x - 60?

Start by using the Rational Root Theorem to find a root. Once a root is identified, use synthetic division to simplify the polynomial and then factor the resulting expression.

What are the solutions to the polynomial equation x⁴ - 7x³ + 7x² + 35x - 60 = 0?

The solutions are x = 3, x = 4, x = ±√5.

Outlines

00:00

🔍 Solving Polynomial Equations with Factor by Grouping

In this paragraph, the focus is on solving a cubic polynomial equation through factor by grouping. The example given is \( x^3 + 3x^2 - 4x - 12 = 0 \), where the ratios of the coefficients suggest using grouping. By extracting the greatest common factor (GCF) in both parts of the polynomial, the expression simplifies, and further factoring leads to the equation's solutions: \( x = -3 \), \( x = -2 \), and \( x = 2 \). This method emphasizes the efficiency of factoring when the ratio of coefficients aligns.

05:01

🔧 Using Substitution to Solve Higher-Degree Polynomials

This paragraph explains how to solve a quartic polynomial, \( x^4 - 5x^2 - 36 = 0 \), by reducing it to quadratic form through substitution. By substituting \( a = x^2 \), the equation becomes a trinomial, which is then factored. After factoring and substituting back for \( x \), the real solutions (\( x = -3, 3 \)) and imaginary solutions (\( x = \pm 2i \)) are determined, emphasizing the versatility of substitution for higher-degree polynomials.

10:02

⚖️ Solving Cubic Polynomials Using Rational Zeros and Synthetic Division

Here, a cubic equation \( x^3 - 3x^2 - 6x + 8 = 0 \) is tackled. Since factor by grouping is not viable, the rational zero theorem is used to list possible rational roots, starting with \( x = 1 \), which is verified. Synthetic division simplifies the equation to a quadratic form, allowing it to be factored, revealing the three real solutions: \( x = 4, -2, 1 \). The method showcases rational zeros and synthetic division for non-factorable polynomials.

📐 Factoring a Quartic Polynomial Using Grouping and Synthetic Division

This paragraph demonstrates solving a fourth-degree polynomial \( x^4 - 7x^3 + 7x^2 + 35x - 60 = 0 \). After testing potential rational zeros using the rational zero theorem, \( x = 3 \) is found to be a solution. Synthetic division reduces the equation, and grouping is used to factor the remaining terms. The final solutions include real numbers and irrational roots (\( x = 4, \pm \sqrt{5} \)). The paragraph highlights the effectiveness of multiple factoring techniques.

Keywords

💡Polynomial Equations

A polynomial equation is a mathematical expression consisting of variables and coefficients, where the variables are raised to integer powers. In the video, the speaker solves different polynomial equations, such as x^3 + 3x^2 - 4x - 12 = 0, by factoring and using various algebraic techniques.

💡Factoring by Grouping

Factoring by grouping is a method for simplifying a polynomial by grouping terms that share common factors and factoring them out. This is used in the video when solving the equation x^3 + 3x^2 - 4x - 12 = 0, where the speaker groups terms to factor the expression more efficiently.

💡Greatest Common Factor (GCF)

The greatest common factor (GCF) is the largest factor that divides two or more terms. In the video, the GCF is used to factor out terms from grouped polynomial expressions, such as when factoring x^3 + 3x^2 into x^2(x + 3).

💡Difference of Perfect Squares

This is a factoring technique used when an expression is the difference between two squares. It can be factored into the form (a + b)(a - b). In the video, this is applied to factor x^2 - 4 into (x + 2)(x - 2).

💡Quadratic Form

A quadratic form is a type of polynomial that can be rewritten as a quadratic equation, usually after substitution. In the video, x^4 - 5x^2 - 36 = 0 is reduced to a quadratic form using substitution, where a = x^2, allowing it to be solved more easily.

💡Rational Zero Theorem

The Rational Zero Theorem is used to find possible rational roots of a polynomial equation by considering the factors of the constant term and the leading coefficient. In the video, this theorem is used to identify potential zeros in equations like x^3 - 3x^2 - 6x + 8 = 0.

💡Synthetic Division

Synthetic division is a shortcut method for dividing polynomials when one of the factors is linear. In the video, synthetic division is applied after identifying one of the zeros (e.g., x = 1) to find other solutions for the polynomial.

💡Imaginary Numbers

Imaginary numbers are numbers that involve the square root of negative one, denoted as i. In the video, when solving x^2 + 4, the speaker introduces imaginary numbers to factor it as (x + 2i)(x - 2i), leading to imaginary solutions.

💡Zeros of a Function

The zeros of a function are the values of x that make the function equal to zero. In the video, finding the zeros of a polynomial equation is the goal, and the speaker finds both real and imaginary zeros by factoring and using the Rational Zero Theorem.

💡Trinomial

A trinomial is a polynomial with three terms. In the video, the speaker works with trinomials like x^2 - 5x - 36, factoring them by identifying numbers that multiply to give the constant term and add to give the coefficient of the middle term.

Highlights

Introduction to solving polynomial equations using factoring techniques.

Illustrating how to factor by grouping when the ratio of the first two coefficients is the same as the ratio of the last two.

Step-by-step breakdown of factoring x^3 + 3x^2 - 4x - 12 by grouping.

Explanation of the difference of perfect squares used to factor x^2 - 4 into (x + 2)(x - 2).

Introduction of a second polynomial equation reducible to a quadratic form, x^4 - 5x^2 - 36 = 0, using substitution.

Detailed breakdown of factoring a quadratic form and how substitution simplifies complex polynomials.

Introduction to imaginary numbers when factoring x^2 + 4 using complex numbers (2i and -2i).

Application of the Rational Zero Theorem to find possible rational solutions for higher-degree polynomials.

Using synthetic division to evaluate potential rational zeros and simplify the polynomial.

Demonstrating how synthetic division is applied to the equation x^3 - 3x^2 - 6x + 8 = 0 to find all real solutions.

Further example of polynomial solving using Rational Zero Theorem and synthetic division on a fourth-degree polynomial, x^4 - 7x^3 + 7x^2 + 35x - 60.

Explanation of factoring by grouping to simplify the equation into x^2 terms.

Factoring x^2 - 5 using difference of perfect squares, resulting in complex and real roots.

General rule: for a polynomial of degree n, there should be n solutions, including real and complex numbers.

Summary of techniques: factoring by grouping, factoring by substitution, the Rational Zero Theorem, and synthetic division to solve polynomial equations.

Transcripts

in this tutorial i'm going to focus on

solving polynomial equations

so let's start with this one

x cubed plus

3x squared

minus 4x minus 12

is equal to zero

go ahead and solve this particular

equation

now

notice that

the ratio

of the first two coefficients is the

same as the ratio of the last two

so for example

one and three

has the same ratio

as negative four and negative twelve

negative twelve divided by negative four

is equal to three

and 3 divided by 1 is 3.

whenever you see that

you can factor by grouping

and in fact that's the best way or the

fastest way to get the answer

so let's take out the gcf in the first

two terms

the greatest common factor is x squared

x cubed divided by x squared is x

and three x squared divided by x squared

is three

now in the last two terms let's take out

the gcf as well

the greatest common factor is negative

four

negative four x divided by negative four

is x

negative twelve divided by negative four

is plus three

and so if you see this common factor

then

you're on or by track so let's take out

x plus three

so this whole thing divided by x plus

three

is x squared and this whole thing

divided by x plus three is negative four

now we're not done yet we can still

factor x squared minus four

it's a difference of perfect squares

the square root of x squared is x the

square root of four is two

one is going to be positive and the

other will be negative

so now

the polynomial expression

has been completely factored at this

point we can set each factor equal to

zero

and so we're going to get three answers

so x is equal to negative three

it's equal to negative two

and positive two as well

so that's how you can solve this

particular polynomial equation

now let's move on to our next example

so let's say that

x to the fourth

minus 5x squared

minus 36 is equal to zero

how can we solve this particular

polynomial equation

notice that

this equation is reducible

to a quadratic form so we can

factor it using substitution

let's say that a is equal to x squared

so that means that a squared is equal to

x to the fourth so let's replace x to

the fourth with a squared

and x squared with a so now we have a

trinomial with a leading coefficient of

one

so what two numbers multiply to negative

36 but add to negative five

so this is going to be

nine and four but negative 9 and

positive 4.

negative 9 times 4 is negative 36 but

negative 9 plus 4 is negative 5.

so to factor it's going to be a minus 9

and a plus 4.

so at this point i'm gonna replace a

with x squared so it's the same as x

squared minus nine

times x squared plus four

now you can factor x squared minus nine

it's going to be x plus three

and x minus 3 if you have a difference

of perfect squares

if you have a sum of perfect squares you

can factor using imaginary numbers

so it's going to be x plus 2i

and x minus 2i

so because it's

a polynomial of degree 4 there's gonna

be four answers but in this case

two of them are real numbers and the

other two are imaginary numbers

so x is equal to negative 3

3

negative 2i

and 2i so you can write it this way x is

equal to plus or minus 3 or

plus or minus 2i

so those are the four answers for this

problem

let's try this one x cubed minus three x

squared

minus six x plus eight

is equal to zero

find all the zeros of the function go

ahead and solve for the value of x

now we can't factor by grouping

because

the first two terms do not have the same

ratio as the last two

negative three divided by one

is not the same as eight divided by

negative six

which reduces to negative four thirds

so we can't factor by grouping

and we can't reduce it to quadratic form

so the last method that we can use

is we need to list all the possible

rational zeros using the rational zero

theorem and then use synthetic division

to find out the other zeros

so p

represents the factors of the constant

term eight

so factors of eight are one

two

four and eight

q

represents the factors of one the lean

coefficient

so

let's evaluate the function at x equals

one let's see if that is a possible

answer

one to the third minus three times one

squared minus six times one

plus eight let's see if it's equal to

zero

so one to the third is one three times

one squared is three

and then we have six plus eight

so one minus three is negative two

and negative two minus six is negative

eight negative eight plus eight is equal

to zero

because it's equal to zero

that means that this is one of the

answers x

is equal to one

now let's find the other answers

so let's use synthetic division

the coefficients are one

negative three negative six and eight

so let's bring down the one

one times one

is one

negative three plus one is negative two

and then one times negative two is

negative two

negative 6 plus negative 2 is negative 8

and then 1 times negative 8

is

negative 8 and 8 plus negative 8 is 0.

if this is not a 0 then

a mistake occurred

so this is going to be 1x squared which

is 1 degree lower than this one

minus 2x minus 8.

and let's set that equal to 0.

now let's see if we can factor this

expression two numbers that multiply to

8 but add to the middle coefficient of

negative two

are going to be negative four and

positive two

so to factor it's going to be x minus

four

times x plus two

so x is equal to four

and negative two

so we have three real solutions

four

negative two and one

here's another problem that we could try

x to the fourth power

minus seven x cubed

plus seven x squared

plus thirty five x

minus sixty

go ahead and solve this particular

polynomial equation

so using the rational zero theorem

the factors of 60 are going to be 1

2

3

4

5

6

seven and eight doesn't go into 60 but

10 goes into it

uh 12

15

30 and 60 and 20 as well

and then the leading coefficient is one

so there's a lot of factors

but let's start with one

let's see if f of one is equal to zero

so that's gonna be one to the fourth

power

minus seven times one cubed

plus seven times 1 squared

plus 35 times 1 minus 60.

so this is going to be 1 minus 7 plus 7

plus 35 minus 60.

negative 7 plus 7 is 0. and 35 minus 60

is negative 25

so negative 24 is not equal to zero so

this

doesn't work

let's try the next number two

so it's going to be 2 to the fourth

minus 7 times 2 to the third

plus 7

times 2 squared plus

35 times 2 minus sixty

so two to the fourth that's two times

two times two times two

two times two is four times two is eight

times two is sixteen

and then two to the third is eight

two squared is

four

thirty five times two is seventy

and then

seven times eight is fifty six

seven times four is twenty eight

and seventy minus sixty is ten

now sixteen minus fifty six

that's negative forty

and twenty eight plus 10

is 38

negative 40 plus 38 is negative two

which is not zero

so that doesn't work either

now let's try three

so three to the fourth power

minus seven times three to the third

plus seven

times three

squared plus thirty five times three

minus sixty

let's see if this gives us a zero three

to the fourth power is eighty one

three to the third is twenty seven

and three squared

is nine

now thirty five times three thirty times

three is ninety three times five is

fifteen ninety and fifteen is one of

five

now seven times twenty seven

seven times twenty is one forty

times 7 is 49 so 140 and 49 that's going

to be 189

7 times 9 is 63 and 105 minus 60 is 45.

so 81 minus 189 that's going to be

negative

108

and 63 plus 45 is positive 108. so this

adds up to zero

which means

x is equal to three

so that's one of the answers

now let's use synthetic division to find

the other ones

so the coefficients are 1

negative 7

7

35 and negative 60.

so let's bring down the one three times

one is three

negative seven plus three is negative

four

three times negative four is negative

twelve

seven plus negative twelve is negative

five

three times negative 5 is

negative 15

35 minus 15 is 20 and 3 times 20 is 60.

so if you get a 0 that's a good sign

so we started with degree 4 now it's

going to be a degree 3 function so it's

x cubed minus four x squared

minus five x plus twenty

so notice that we can factor by grouping

negative four divided by one is negative

four

twenty divided by negative five is also

negative 4.

so let's take out the gcf which is going

to be x squared

so x cubed divided by x squared is x

negative 4x squared divided by x squared

is negative 4.

and then let's take out a negative 5.

negative 5x divided by negative 5 is x

20 divided by negative 5 is

negative 4.

so it's going to be x minus 4

times x squared minus 5

and all of that

is equal to 0.

now let's factor x squared minus 5

using the difference of perfect squares

technique

so the square root of x squared is x the

square root of 5 is

the square root of 5.

and so there's going to be 4 answers in

total

so we could set x minus 4 equal to 0

x plus the square root of 5 equal to 0

and x minus the square root of 5 equal

to 0.

so x is equal to 4

negative square root 5

and positive square root five

so here's the first answer

the second one

the third one and the fourth

so you should have four answers for

a fourth degree polynomial and that's it

so now you know how to solve polynomial

equations by factoring by grouping

factoring by substitution

and using the rational zero theorem

along with synthetic division thanks

again for watching

you