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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.

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

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

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 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

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