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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

^{3}+ y

^{3}

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

^{3}+ y

^{3}= (2x + y)(4x

^{2}− 2xy + y

^{2})

^{3}− 64b

^{3}

- This problem is a Difference Of Cubes
- 27a
^{3}− 64b^{3}= (3a)^{3}− (4b)^{3}because 3 to the third power = 27 and 4 to the third power = 64 - The Pattern Looks like this:
- x
^{3}− y^{3}= (x − y)(x^{2}+ xy + y^{2}); 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

^{3}− (4b)

^{3}= (3a − 4b)(9a

^{2}+ 123ab + 16b

^{2})

^{3}y

^{3}z

^{3}− 8x

^{4}y

^{3}z

^{4}− 2x

^{2}y

^{3}z

^{3}

- 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, x
^{2} - Step 3: What is the maximum number of y's can you take out?
- You can take out a maximum of 2 y′s, y
^{2} - Step 4:What is the maximum number of z's can you take out?
- You can take out a maximum of 2 z′s, z
^{2} - What is the GCF then?
- 2x
^{2}y^{2}z^{2} - Cancel out anything you took out form each term

^{2}y

^{2}z

^{2}(3xyz − 4x

^{2}yz

^{2}+ yz)

^{4}− 2x

^{2}− 21

- Write the expression in the format ax
^{2}+ bx + c - 3(x
^{2})^{2}− 2(x)^{2}− 21 - Let p = x
^{2} - Notice that we can factor 3p
^{2}− 2p − 21 - ax
^{2}+ bx + c = 0 - ax
^{2}+ 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
- 3p
^{2}− 2p − 21 = (p + [7/3])(p + [( − 9)/3]) = (3p + 7)(p − 3) - Given that p = x
^{2}, the final answer then becomes

^{4}− 2x

^{2}− 21 = (3x

^{2}+ 7)(p

^{2}− 3)

^{4}− 6x

^{2}+ 1

- Write the expression in the format ax
^{2}+ bx + c - 5(x
^{2})^{2}− 6(x)^{2}+ 1 - Let p = x
^{2} - Notice that we can factor 5p
^{2}− 6p + 1 - ax
^{2}+ bx + c = 0 - ax
^{2}+ 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
- 5p
^{2}− 6p + 1 = (p − [5/5])(p − [1/5]) = (p − 1)(5p − 1) - Given that p = x
^{2}, we get

^{4}− 6x

^{2}+ 1 = (x

^{2}− 1)(5x

^{2}− 1)

^{8}− 17x

^{4}+ 1

- Write the expression in the format ax
^{2}+ bx + c - 16(x
^{4})^{2}− 17(x^{4}) + 1 - Let p = x
^{4} - Notice that we can factor 16p
^{2}− 17p + 1 - ax
^{2}+ bx + c = 0 - ax
^{2}+ 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
- 16p
^{2}− 17p + 1 = (p − [16/16])(p − [1/16]) = (p − 1)(16p − 1) - Given that p = x
^{4}, we get - 16x
^{8}− 17x^{4}+ 1 = (x^{4}− 1)(16x^{4}− 1) - We repeat the process to factor out (x
^{4}− 1) and (16x^{4}− 1). Notice how they will will become difference of squares - Let q = x
^{2} - (q
^{2}− 1)(16q^{2}− 1) = (q^{2}− 1)((4q)^{2}− 1) = (q + 1)(q − 1)(4q + 1)(4q − 1) - By plugging back in q = x
^{2}we get

^{8}− 17x

^{4}+ 1 = (x

^{2}+ 1)(x

^{2}− 1)(4x

^{2}+ 1)(4x

^{2}− 1)

^{4}+ 3x

^{2}− 4 = 0

- Remember that working with x to the 4th power is the same as working with x squared.
- x
^{2}+ bx + c = 0 - x
^{2}+ 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
- (x
^{2}+ 4)(x^{2}− 1) = 0 - Second term is difference of squares, so that will give us.
- (x
^{2}+ 4)(x − 1)(x + 1) = 0 - Using the Zero Product Property we get
- x
^{2}+ 4 = 0 , x − 1 = 0 , x + 1 = 0 - Solving the first equation gives us the following imaginary roots
- 0 = x
^{2}+ 4 = - x
^{2}= − 4 - x = ±√{ − 4}
- = ±i√4 =
- = ±2i
- Solution is then

x = − 2i

x = 1

x = − 1

^{4}− 6x

^{2}+ 8 = 0

- Remember that working with x to the 4th power is the same as working with x squared.
- x
^{2}+ bx + c = 0 - x
^{2}+ 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
- (x
^{2}− 4)(x^{2}− 2) = 0 - First term is difference of squares, so that will give us.
- (x − 2)(x + 2)(x
^{2}− 2) = 0 - Using the Zero Product Property we get
- x − 2 = 0 , x + 2 = 0 , x
^{2}− 2 = 0 - Solving the third equation involves square roots.
- x
^{2}− 2 = 0 - x
^{2}= 2 - x = ±√2
- Solution is then

x = − √2

x = 2

x = − 2

^{4}− 2x

^{2}− 63 = 0

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

x = − i√7

x = 3

x = − 3

^{4}+ 8x

^{2}+ 16 = 0

- Remember that working with x to the 4th power is the same as working with x squared.
- x
^{2}+ bx + c = 0 - x
^{2}+ 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
- (x
^{2}+ 4)(x^{2}+ 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
- (x
^{2}+ 4)(x^{2}+ 4) = 0 - Using the Zero Product Property we get
- x
^{2}+ 4 = 0 , x^{2}+ 4 = 0 - Solving the first equation involves square roots and imaginary numbers.
- x
^{2}+ 4 = 0 - x
^{2}= − 4 - x = ±√{ − 4}
- x = ±i√4
- x = ±2i
- Solution is then,

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

### Algebra 2

### 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 4x ^{4} - 6x^{2} - 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 2x ^{3}; 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 a ^{2} - b^{2}, and an example would be x^{2} - 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: x ^{2} - 3^{2}.*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 x ^{2} + 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 cases*0178

*of difference of two squares or perfect square trinomials.*0186

*It might be something such as, say, x ^{2} + 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 x ^{2}; Outer is -2x; Inner is 4x; and then, this is -8.*0253

*So, x ^{2} + 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 3x ^{3} - 4x^{2} + 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 x ^{2}, 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 x ^{2}: 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 x ^{2} + 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: 64x ^{3} + 125.*0375

*Well, if I think about this, the cube root of 64 is 4; the cube root of x ^{3} is x; the cube root of 125 is 5.*0381

*Here, this is in this form, a ^{3} + b^{3}; 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, a ^{3} + b^{3}, which equals*0414

*(a + b) times (a ^{2} - ab + b^{2}), 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 a ^{2}, so that is going to give me (4x)^{2}, minus a times b, plus b^{2}.*0433

*OK, this is 4x + 5, and this is going to give me 16x ^{2} - 20x + (5^{2} 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, 64x ^{3} + 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 actually*0502

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

*And quadratic form is an ^{2} + 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 6x ^{8} + 5x^{4} + 9, and I want to get it into this form, an^{2} + 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 x ^{8}?"*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 (x ^{4})^{2} is x^{8}.*0567

*So, all I did is took my x ^{8} and put it in a form where it is something squared.*0574

*Now, I am going to assign another variable this value, x ^{4}.*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 x ^{4}.*0585

*Therefore, y ^{2} equals x^{8}, since y equals x^{4},*0596

*because if y equals x ^{4}, I could just substitute and say that this is really (x^{4})^{2}.*0608

*OK, so again, what I did is looked at what I had, x ^{8}; 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 x ^{4}.*0626

*Then, I took this and wrote it as a different variable, y = x ^{4}.*0628

*I assigned the variable y the value x ^{4}; so that is what this means--something n^{2}.*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 x ^{4}, and that is going to be y^{2}.*0646

*Plus...my next coefficient is 5.*0654

*OK, here I have an x ^{4}, but I said y = x^{4}, 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 x ^{4}.*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, 27a ^{3} - 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 a ^{3} 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

*a ^{2} + ab + b^{2}; OK, if a is 3a, then that is going to give me 3a - b (which is 2).*0778

*Here, I am going to have a ^{2}; so that is 3a^{2} + a times b + b^{2}.*0788

*This is going to give me 9a ^{2} + 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 x ^{2}; I have an x here; and an x^{3}; so they all have at least an x.*0858

*For y, there is one y here; there is a y ^{3} here; and there is a y^{2} here.*0865

*For z, I have z ^{3}, z^{2}, and z^{4}, so they all have at least z^{2}.*0871

*So, the greatest common factor is 3xyz ^{2}; 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 z ^{2} 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 y ^{2}; z^{2} is gone.*0895

*Here, I pulled out the factor of 3; that leaves behind 8; one x is gone; I have x ^{2}; one y is gone; and z^{2} 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 an ^{2} + bn + c.*0935

*Here, I have x ^{6}; I want to find some variable, assign it a certain value, and then square it.*0946

*So, x to the something, squared, equals x ^{6}.*0958

*Well, if I take 6 and divide it by 2, I am going to get 3; therefore, (x ^{3})^{2} = x^{6}, because 3 times 2 is 6.*0964

*I am going to let y equal x ^{3}; therefore, y^{2} is going to equal x^{6}, because that would be (x^{3})^{2}.*0975

*Now, let's look at what we have: we have our leading coefficient of 10, x ^{6}...*0995

*well, x ^{6} is going to be the same as y^{2}; minus 12; now, I have x^{3} here.*1001

*Well, I said that y equals x ^{3}, 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 x ^{3},*1016

*which allowed me to write that as 10y ^{2} - 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 x ^{4} trinomial, you could think*1050

*that it is similar to working with a quadratic, except you have x ^{2} here in the factors, instead of just x.*1057

*So, x ^{2} times x^{2} is x^{4}.*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; x ^{2} + 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 x ^{2} + 4 = 0, or x^{2} - 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 x ^{2} = -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: x ^{2} = 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, (x ^{2} + 4) times (x^{2} - 6),*1231

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

*In this case, I ended up with x ^{2} 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

0 answers

Post by Nsikan Esenowo on July 19, 2010

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