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Algebra 2 Solving Polynomial Functions

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Section 6: Polynomial Functions: Lecture 6 | 21:10 min

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Lecture Comments (6)

	0 answers Post by Mulin Li on April 17, 2019 In example one, after you factor the difference of two cubes, isn't the (a^2+ab+b^2) part a perfect square trinomial?
	1 answer Last reply by: Dr Carleen Eaton Fri Jun 8, 2018 11:33 PM Post by John Stedge on June 5, 2018 In the last example you can factor the (x^2+4) into (x+2) (x+2)
	1 answer Last reply by: Dr Carleen Eaton Tue Mar 13, 2018 7:15 PM Post by MICHELLE AYOH on March 3, 2018 very helpful ;)
	0 answers Post by Nsikan Esenowo on July 19, 2010 Thank you for all your effort put together on this site to make my Maths' problems solve. I learned a lot receiving lectures from you. You are the best that ever happen to me in this field. Maths made easy.

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Solving Polynomial Functions

When factoring, always factor completely.
Always begin factoring by factoring the greatest common factor, and then factor the remaining expression.
If the polynomial has 4 terms, try factoring by grouping.
Factor trinomials by trial and error.
Memorize the formulas for the sum and difference of two cubes.

Solving Polynomial Functions

Factor 8x³ + y³

This problem is a Difference Of Cubes
8x³ + y³ = (2x)³ + y³ because 2 to the third power = 8
The Pattern Looks like this:
x³ + y³ = (x + y)(x² − xy + y²); where x = (2x) and y = y
Subsitute x and y into the pattern
(2x)³ + y³ = (2x + y)((2x)² − (2x)y + y²)
Simplify Using Rules of Exponents

(2x)³ + y³ = (2x + y)(4x² − 2xy + y²)

Factor 27a³ − 64b³

This problem is a Difference Of Cubes
27a³ − 64b³ = (3a)³ − (4b)³ because 3 to the third power = 27 and 4 to the third power = 64
The Pattern Looks like this:
x³ − y³ = (x − y)(x² + xy + y²); where x = (3a) and y = 4b
Subsitute x and y into the pattern
(3a)³ − (4b)³ = (3a − 4b)((3a)² + (3a)(4b) + (4b)²)
Simplify Using Rules of Exponents

(3a)³ − (4b)³ = (3a − 4b)(9a² + 123ab + 16b²)

Factor 6x³y³z³ − 8x⁴y³z⁴ − 2x²y³z³

Find the GCF, then cancel out anything you took out form each term.
Step 1: What is the GCF of 6, 8 and 2?
The GCF of those number is 2
Step 2:What is the maximum number of x's can you take out?
You can take out a maximum of 2 x, x²
Step 3: What is the maximum number of y's can you take out?
You can take out a maximum of 2 y′s, y²
Step 4:What is the maximum number of z's can you take out?
You can take out a maximum of 2 z′s, z²
What is the GCF then?
2x²y²z²
Cancel out anything you took out form each term

2x²y²z²(3xyz − 4x²yz² + yz)

Write in quadratic form: 3x⁴ − 2x² − 21

Write the expression in the format ax² + bx + c
3(x²)² − 2(x)² − 21
Let p = x²
Notice that we can factor 3p² − 2p − 21
ax² + bx + c = 0
ax² + bx + c = (x + [m/a])(x + [n/a])
Find two numbers "m" and "n" such that:
a*c = m*n
b = m + n
− 63 = m*n
− 2 = m + n
The only numbers that work is 7 and − 9
3p² − 2p − 21 = (p + [7/3])(p + [( − 9)/3]) = (3p + 7)(p − 3)
Given that p = x², the final answer then becomes

3x⁴ − 2x² − 21 = (3x² + 7)(p² − 3)

Write in quadratic form: 5x⁴ − 6x² + 1

Write the expression in the format ax² + bx + c
5(x²)² − 6(x)² + 1
Let p = x²
Notice that we can factor 5p² − 6p + 1
ax² + bx + c = 0
ax² + bx + c = (x + [m/a])(x + [n/a])
Find two numbers "m" and "n" such that:
a*c = m*n b = m + n
5 = m*n
− 6 = m + n
The only numbers that work is − 5 and − 1
5p² − 6p + 1 = (p − [5/5])(p − [1/5]) = (p − 1)(5p − 1)
Given that p = x², we get

5x⁴ − 6x² + 1 = (x² − 1)(5x² − 1)

Write in quadratic form: 16x⁸ − 17x⁴ + 1

Write the expression in the format ax² + bx + c
16(x⁴)² − 17(x⁴) + 1
Let p = x⁴
Notice that we can factor 16p² − 17p + 1
ax² + bx + c = 0
ax² + bx + c = (x + [m/a])(x + [n/a])
Find two numbers "m" and "n" such that:
a*c = m*n
b = m + n
5 = m*n
− 6 = m + n
The only numbers that work is − 16 and − 1
16p² − 17p + 1 = (p − [16/16])(p − [1/16]) = (p − 1)(16p − 1)
Given that p = x⁴, we get
16x⁸ − 17x⁴ + 1 = (x⁴ − 1)(16x⁴ − 1)
We repeat the process to factor out (x⁴ − 1) and (16x⁴ − 1). Notice how they will will become difference of squares
Let q = x²
(q² − 1)(16q² − 1) = (q² − 1)((4q)² − 1) = (q + 1)(q − 1)(4q + 1)(4q − 1)
By plugging back in q = x² we get

16x⁸ − 17x⁴ + 1 = (x² + 1)(x² − 1)(4x² + 1)(4x² − 1)

Solve x⁴ + 3x² − 4 = 0

Remember that working with x to the 4th power is the same as working with x squared.
x² + bx + c = 0
x² + bx + c = (x + m)(x + n)
Find two numbers "m" and "n" such that:
c = m*n
b = m + n
−4 = m*n
3 = m +n
The only two numbers that work is 4 and − 1
(x² + 4)(x² − 1) = 0
Second term is difference of squares, so that will give us.
(x² + 4)(x − 1)(x + 1) = 0
Using the Zero Product Property we get
x² + 4 = 0 , x − 1 = 0 , x + 1 = 0
Solving the first equation gives us the following imaginary roots
0 = x² + 4 =
x² = − 4
x = ±√{ − 4}
= ±i√4 =
= ±2i
Solution is then

x = 2i
x = − 2i
x = 1
x = − 1

Solve x⁴ − 6x² + 8 = 0

Remember that working with x to the 4th power is the same as working with x squared.
x² + bx + c = 0
x² + bx + c = (x + m)(x + n)
Find two numbers m and n such that:
c = m*n
b = m + n
8 = m*n
−6 = m + n
The only two numbers that work is − 4 and − 2
(x² − 4)(x² − 2) = 0
First term is difference of squares, so that will give us.
(x − 2)(x + 2)(x² − 2) = 0
Using the Zero Product Property we get
x − 2 = 0 , x + 2 = 0 , x² − 2 = 0
Solving the third equation involves square roots.
x² − 2 = 0
x² = 2
x = ±√2
Solution is then

x = √2
x = − √2
x = 2
x = − 2

Solve x⁴ − 2x² − 63 = 0

Remember that working with x to the 4th power is the same as working with x squared.
x² + bx + c = 0
x² + bx + c = (x + m)(x + n)
Find two numbers m and n such that:
c = m*n
b = m + n
−63 = m*n
−2 = m + n
The only two numbers that work is 7 and − 9
(x² + 7)(x² − 9) = 0
Second term is difference of squares, so that will give us.
(x² + 7)(x − 3)(x + 3) = 0
Using the Zero Product Property we get
x² + 7 = 0 , x − 3 = 0 , x + 3 = 0
Solving the first equation involves square roots and imaginary numbers.
x² + 7 = 0
x² = − 7
x = ±√{ − 7}
x = ±i√7
Solution is then, solving the remaining two equations

x = i√7
x = − i√7
x = 3
x = − 3

Solve x⁴ + 8x² + 16 = 0

Remember that working with x to the 4th power is the same as working with x squared.
x² + bx + c = 0
x² + bx + c = (x + m)(x + n)
Find two numbers m and n such that:
c = m*n
b = m + n
16 = m*n
8 = m + n
The only two numbers that work is 4 and 4
(x² + 4)(x² + 4) = 0
This is a special case, whenever the same term repeats, it is called multiplicity 2, if it were to repeat once more,
it would be called roots with multiplicity 3
(x² + 4)(x² + 4) = 0
Using the Zero Product Property we get
x² + 4 = 0 , x² + 4 = 0
Solving the first equation involves square roots and imaginary numbers.
x² + 4 = 0
x² = − 4
x = ±√{ − 4}
x = ±i√4
x = ±2i
Solution is then,

x = 2i, x = − 2i, x = 2i, and x = − 2i
You can also say:
x = 2i multiplicity 2 and x = − 2i multiplicity 2

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Answer

Solving Polynomial Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Factoring Polynomials

Sum and Difference of Two Cubes

Examples: Two Cubes

Quadratic Form

Example: Quadratic Form

Example 1: Factor Polynomial

Example 2: Factor Polynomial

Example 3: Quadratic Form

Example 4: Solve Polynomial Function

Intro 0:00
Factoring Polynomials 0:06

Greatest Common Factor (GCF)
Difference of Two Squares
Perfect Square Trinomials
General Trinomials
Grouping

Sum and Difference of Two Cubes 6:03

Examples: Two Cubes

Quadratic Form 8:22

Example: Quadratic Form

Example 1: Factor Polynomial 12:03
Example 2: Factor Polynomial 13:54
Example 3: Quadratic Form 15:33
Example 4: Solve Polynomial Function 17:24

Algebra 2

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

Transcription: Solving Polynomial Functions

Welcome to Educator.com.0000

In today's lesson, we will be working on solving polynomial equations.0002

And we are going to start out by reviewing some factoring techniques.0006

And the techniques that you will use for factoring polynomials are familiar from earlier work.0010

And these include greatest common factor, difference of two squares, perfect square trinomials, general trinomials, and factoring by grouping.0015

So, just as we did with quadratic equations, if you are going to be working with a polynomial,0025

the first thing you want to do is: if there is a greatest common factor, factor that out first.0031

For example, if I had something like 4x⁴ - 6x² - 12x, I have a greatest common factor of 2x.0039

So, I have pulled that out first; and that would give me 2x, leaving behind 2x³; here that would leave behind 3x; and here, -6.0055

And then, I would work on factoring that farther, if it is possible.0069

OK, the difference of two squares: this is the greatest common factor, and now difference of two squares.0074

Recall that these are in the form a² - b², and an example would be x² - 9.0085

This could be factored out into (x + 3) (x - 3); this factors into (a - b) times (a + b).0094

Here, in this case, a equaled x and b equaled 3: x² - 3².0107

In a few minutes, we will be talking about the difference and sum of two cubes; we will go on a little farther than just working with squares.0118

Next, perfect square trinomials: you know that, if you recognize these, they are easy to work with.0128

So, you want to be on the lookout for these perfect square trinomials.0136

For example, one that we have seen earlier, working with quadratic equations, is x² + 8x + 16.0140

And this factors as (x + 4)², or (x + 4) (x + 4)--a perfect square trinomial.0151

And all of these are covered in detail in earlier lectures, so go ahead back and review these, if necessary,0166

to make sure you have them down before you work on factoring polynomials.0173

General trinomials: you would recognize these in a form that they don't fit into the special cases0178

of difference of two squares or perfect square trinomials.0186

It might be something such as, say, x² + 2x -8; and you need to just use some trial and error on this.0189

For example, for this one, you would say, "OK, the first term in each factor has to be x."0198

And since I have a negative in front of the constant here, I have to have a positive here and a negative here,0203

because a positive times a negative is going to give me a negative.0209

Then, I am going to look at the factors of 8, and I am going to say, "OK, I have 1 and 8, and 2 and 4."0212

And then, I want to find factors of 8 that, when one is positive and one is negative, sum up to 2.0221

And I can see that 1 and 8 are too far apart; they are never going to give me 2.0227

2 and 4 are perfect; and I see that, if I make the 4 positive and the 2 negative,0231

I am going to get 2, which is going to be the coefficient for my middle term.0240

This is going to factor out to (x + 4) (x - 2); and I can always check this by using FOIL,0246

which will tell me that the First term gives me x²; Outer is -2x; Inner is 4x; and then, this is -8.0253

So, x² + 2x - 8--it gave me this trinomial back.0262

Grouping: recall that we use grouping when we are factoring polynomials with 4 terms--factoring by grouping.0272

Let's say you have some things such as 3x³ - 4x² + 6x - 8.0282

I am going to handle this by grouping the first two terms, and then grouping the second two terms.0290

Now, I am going to pull out any common factors I see; and I do have a common factor of x², to leave behind 3x - 1.0299

Over here, I have a common factor of 2; I am going to pull that out, and that is going to leave me with 3x - 4.0311

I am pulling out the x²: that is going to leave me with 3x - 4 here; here I am going to pull out the 2; that is going to leave me with 3x - 4.0321

This is what I wanted to happen; and what I have here is a common binomial factor.0329

I am going to factor that out (pull that out in front), and that leaves behind x² + 2.0334

So, factoring by grouping: group the first two terms; group the second two; pull out a common factor.0343

If you have a common binomial factor left behind, then you pull that out in front.0349

Make sure you are familiar with all of these; review the ones that you need to.0355

And we are going to be using these to solve polynomial equations.0358

Now, we are going on to a new concept with factoring: and that is the sum and difference of two cubes.0363

You are familiar with working with squares; and now we are talking about the sum and difference of two cubes.0369

Let's look at an example: 64x³ + 125.0375

Well, if I think about this, the cube root of 64 is 4; the cube root of x³ is x; the cube root of 125 is 5.0381

Here, this is in this form, a³ + b³; here, a equals 4x, and b equals 5.0399

Now, by memorizing this formula, I can just factor this out.0409

So, if I know that this is in this form, a³ + b³, which equals0414

(a + b) times (a² - ab + b²), then I know that I have that a is 4x and b is 5.0420

So, I just have to substitute that in.0430

Here, I am supposed to have a², so that is going to give me (4x)², minus a times b, plus b².0433

OK, this is 4x + 5, and this is going to give me 16x² - 20x + (5² is) 25.0446

So again, just recognize that this is actually the sum of two cubes, so it is going to factor out to this form.0467

For the difference of two cubes, it is a very similar idea--only you are going to end up with a negative sign here and a positive here.0475

It is the same idea, just different signs.0483

So, 64x³ + 125 factors out to this.0485

And then, from there, in some cases, you may be able to go on with your factoring.0493

Let's talk about quadratic form: we worked with quadratic equations earlier on, and you can actually0502

put higher-order polynomials into a form called quadratic form.0507

And quadratic form is an² + bn + c, where n is an expression of x.0512

Well, what does that mean--"n is an expression of x?"0520

Let's use an example: let's say I have 6x⁸ + 5x⁴ + 9, and I want to get it into this form, an² + bn + c.0524

Well, let's look at this more closely: I want to get this into some form where I have a number squared, so I have a square here.0540

Now, I am going to think about: "x to the what, squared, equals x⁸?"0553

Thinking about raising a power to a power, I know that I have to multiply 2 times something to get 8.0558

So, 8 divided by 2 is 4; so that tells me that (x⁴)² is x⁸.0567

So, all I did is took my x⁸ and put it in a form where it is something squared.0574

Now, I am going to assign another variable this value, x⁴.0580

I could use y; I could use z; I am going to go ahead and use y--I am going to let y equal x⁴.0585

Therefore, y² equals x⁸, since y equals x⁴,0596

because if y equals x⁴, I could just substitute and say that this is really (x⁴)².0608

OK, so again, what I did is looked at what I had, x⁸; and I said,0616

"What would have to be the exponent here, so that I could just square it and get this back?"0621

Well, the exponent would have to be x⁴.0626

Then, I took this and wrote it as a different variable, y = x⁴.0628

I assigned the variable y the value x⁴; so that is what this means--something n².0634

Here, I am just going to say it is y; so this gives me...my leading coefficient is 6.0640

Now, instead of putting n, I just decided I am going to use y to represent x⁴, and that is going to be y².0646

Plus...my next coefficient is 5.0654

OK, here I have an x⁴, but I said y = x⁴, so I am just going to write it as y.0657

And then, my constant...so now, I have taken this and rewritten it as this equation, where y is equal to x⁴.0664

Now that this is in quadratic form, I can use techniques such as the quadratic formula and factoring, learned earlier on,0672

in order to (if this were an equation, let's say, then I could) solve it using these techniques for working with quadratics.0679

Once I find the value of y, whatever y turns out to be, if I know y, I can find x.0688

If y is 16, then I would just have to say, "OK, this equals 2"; then x would equal 2.0696

So, by putting this equation into quadratic form, I can use quadratic techniques to solve for y.0707

Once I have y, I can solve for x, which makes quadratic form very useful when working with certain polynomials.0715

Looking at this first example, 27a³ - 8, I recognize this as the difference of two cubes.0725

And this is in this form; here, the cube root of 27 is 3; the cube root of a³ is a; so this is what I have.0731

The cube root of 8 is 2; therefore, what I have in here is a, and what I have right here is b.0751

And sometimes it helps to actually write these out instead of doing it in your head, just to keep everything straight.0764

Since I know that a is 3a, and b equals 2, I just have to substitute here to factor: a - b--I am going to rewrite it down here:0768

a² + ab + b²; OK, if a is 3a, then that is going to give me 3a - b (which is 2).0778

Here, I am going to have a²; so that is 3a² + a times b + b².0788

This is going to give me 9a² + 6a + 4.0802

And one important thing is just to watch the signs, working with the difference of two cubes or the sum of two cubes.0810

And then, figure out what your a and b values are by finding the cube root of these terms.0818

And then, just substitute in a and b into this formula to get factorization.0823

In the next example, we are asked to factor this trinomial; and the first step is to factor out a greatest common factor, if there is one.0836

And I am looking here for a greatest common factor.0846

And I see that I have a 3 that can be pulled out of all of these terms.0852

For x's, I have x²; I have an x here; and an x³; so they all have at least an x.0858

For y, there is one y here; there is a y³ here; and there is a y² here.0865

For z, I have z³, z², and z⁴, so they all have at least z².0871

So, the greatest common factor is 3xyz²; so I am going to pull that out and see what is left behind.0877

In my first term, the 3 is gone; I am left with an x.0885

The y is gone, and the z² is gone; that leaves me just xz.0889

Here, I have a negative sign; I pulled a 3 out, leaving behind a 4; the x is gone; one y is gone, so I have y²; z² is gone.0895

Here, I pulled out the factor of 3; that leaves behind 8; one x is gone; I have x²; one y is gone; and z² is gone.0904

OK, so I am left with the greatest common factor, times what was left behind in here.0915

And I am looking, and there are no common factors left.0921

I don't recognize this as the difference of two squares, or anything that I can factor out.0924

So, this is as far as I can go with the factorization.0929

Now, I am asked to write this polynomial in quadratic form; recall that quadratic form is an² + bn + c.0935

Here, I have x⁶; I want to find some variable, assign it a certain value, and then square it.0946

So, x to the something, squared, equals x⁶.0958

Well, if I take 6 and divide it by 2, I am going to get 3; therefore, (x³)² = x⁶, because 3 times 2 is 6.0964

I am going to let y equal x³; therefore, y² is going to equal x⁶, because that would be (x³)².0975

Now, let's look at what we have: we have our leading coefficient of 10, x⁶...0995

well, x⁶ is going to be the same as y²; minus 12; now, I have x³ here.1001

Well, I said that y equals x³, so I am just going to put y here; plus 14.1010

This is my given equation (this polynomial) in quadratic form; and again, I did that by assigning y the value of x³,1016

which allowed me to write that as 10y² - 12y + 14.1024

I could then use methods for solving quadratic equations, if this were an equation.1029

Then, I could solve this and find y; once I have y, I could go back in and just find the cube root of that, which would give me x.1034

In this example, I am actually asked to solve the equation; I am going to start out by factoring.1045

And this is a trinomial; and again, when you see that x⁴ trinomial, you could think1050

that it is similar to working with a quadratic, except you have x² here in the factors, instead of just x.1057

So, x² times x² is x⁴.1063

I have a negative sign here, which means that one term is going to be positive, and one is going to be negative (for the second terms).1068

Let's think about the factors of 24, and let's find the factors that are going to add up to -2.1075

Factors of 24: 1 and 24, 2 and 12, 3 and 8, 4 and 6.1084

Now, this is a lot, but I don't have to work with all of these; I want to find ones that are close together, because their sum is just going to be -2.1096

So, the closest together are 4 and 6; and I see that one is positive and one is negative.1103

Well, if I make 4 positive and 6 negative, their sum will be -2.1108

So, I know that this is what I want: I want for 4 to be positive, and I want the 6 to be negative (and this all equals 0).1116

OK, so I have actually factored this as far out as I can; x² + 4 doesn't factor any farther, nor does this.1126

So, I am just going to go ahead and use the zero product property to solve.1134

And the zero product property tells me that x² + 4 = 0, or x² - 6 = 0,1139

because if either of these terms is 0, the product of the two will be 0, and the equation will be solved.1146

Here, I am going to get x² = -4.1152

Now, when I take the square root of that, I am going to end up with this.1156

Recall from working with imaginary and complex numbers that the square root of -1 equals i.1161

So, this is the same as saying, "The square root of -1 times 4," which equals the square root of -1, times the square root of 4.1171

So, since this equals i, I am just going to write this as this; and that is actually plus or minus;1184

we have to take the positive and the negative, according to the square root property.1191

OK, well, this is a perfect square, so this gives me ± 2i--that is some review of complex numbers.1195

Here, it is a little bit simpler: x² = 6.1206

According to the square root property, if I take the square root of both sides, then I will get x = ±√6.1209

All right, so I was asked to solve this; and I came up with solutions.1217

And the solutions are that x equals 2i, -2i, √6, or -√6.1220

And I figured that out by factoring this into these two factors, (x² + 4) times (x² - 6),1231

and then continuing on to use the zero product property.1243

In this case, I ended up with x² equaling a negative number, so I ended up having to use a complex number for my result.1250

And then, these are the four solutions to this polynomial equation.1256

That concludes this session of Educator.com on polynomial equations.1262

And I will see you next lesson for more work with polynomials.1267

Related Books

Algebra and Trigonometry: Structure and Method, Book 2

Author: McDougal Littell

ISBN: 395977258

Publisher: McDougal Littell

Year: 1999

This classic textbook does a great job of explaining concepts and includes lots of in-depth, worked out examples in every section as well as practice problems.

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Dr. Carleen Eaton

Solving Polynomial Functions

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro