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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (1)

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Post by julius mogyorossy on December 16, 2013

x=+-2i, so what that is really saying is, x=2, after you square it, make it negative, is that it?

Roots and Zeros

  • A polynomial of degree n has n roots.
  • If the coefficients of the polynomial are real, then complex roots occur as conjugate pairs.
  • Learn and understand Descartes’ rule of signs for the roots of a polynomial.
  • Use synthetic substitution to find the roots of a polynomial of degree 3 or more.

Roots and Zeros

Solve x3 − 5x2 + 6x = 0
  • Find the GCF
  • GCF = x
  • x(x2 − 5x + 6) = 0
  • Find two numbers such that when multiplied = 6; and when added = − 5
  • Those two numbers are − 3 and − 2
  • x(x − 3)(x − 2) = 0
  • Using Zero Product Property we have
  • x = 0, x − 3 = 0, and x − 2 = 0
  • Solve
x = 0, x = 3, and x = 2
Solve x3 − 3x2 − 4x = 0
  • Find the GCF
  • GCF = x
  • x(x2 − 3x − 4) = 0
  • Find two numbers such that when multiplied = − 4; and when added = − 3
  • Those two numbers are − 4 and 1
  • x(x − 4)(x + 1) = 0
  • Using Zero Product Property we have
  • x = 0, x − 4 = 0, and x + 1 = 0
  • Solve
x = 0, x = 4, and x = − 1
Solve 2x3 − 2x2 − 40x = 0
  • Find the GCF
  • GCF = 2x
  • 2x(x2 − x − 20) = 0
  • Find two numbers such that when multiplied = − 20; and when added = − 1
  • Those two numbers are 4 and − 5
  • 2x(x + 4)(x − 5) = 0
  • Using Zero Product Property we have
  • 2x = 0, x + 4 = 0, and x − 5 = 0
  • Solve
x = 0, x = − 4, and x = 5
Solve x4 − 17x2 + 16 = 0
  • Recall that factoring a 4th power polynomial is the same as factoring a 2nd degree polynomial.
  • Find two numbers m and n such that when multiplied, m*n = 16 and when added m + n = − 17
  • Those numbers are − 16 and − 1
  • (x2 − 16)(x2 − 1) = 0
  • Notice that nowe have two difference of squares.
  • (x − 4)(x + 4)(x − 1)(x + 1) = 0
  • Solve using Zero Product Property
  • x − 4 = 0, x + 4 = 0, x − 1 = 0, and x + 1 = 0
  • Solve
x = 4, x = − 4, x = 1, and x = − 1
Solve x4 − 17x2 + 16 = 0
  • Recall that factoring a 4th power polynomial is the same as factoring a 2nd degree polynomial.
  • Find two numbers m and n such that when multiplied, m*n = 16 and when added m + n = − 17
  • Those numbers are − 16 and − 1
  • (x2 − 16)(x2 − 1) = 0
  • Notice that nowe have two difference of squares.
  • (x − 4)(x + 4)(x − 1)(x + 1) = 0
  • Solve using Zero Product Property
  • x − 4 = 0, x + 4 = 0, x − 1 = 0, and x + 1 = 0
  • Solve
x = 4, x = − 4, x = 1, and x = − 1
Solve x4 − 3x2 − 54 = 0
  • Recall that factoring a 4th power polynomial is the same as factoring a 2nd degree polynomial.
  • Find two numbers m and n such that when multiplied, m*n = − 54 and when added m + n = − 3
  • Those numbers are 6 and − 9
  • (x2 + 6)(x2 − 9) = 0
  • Notice that we have one difference of squares.
  • (x2 + 6)(x − 3)(x + 3) = 0
  • Solve using Zero Product Property
  • x2 + 6 = 0, x − 3 = 0, and x + 3 = 0
  • Solve
x = ±i√6 , x = 3, x = − 3
Determine the possible combinations of possible real roots, negative real roots and complex roots.
3x5 − 15x4 − 29x3 + 145x2 + 40x − 200 = 0
  • Type of Root: Total Roots
    How to Find It: You will find the total number of roots by the Degree of the Polynomial
  • Type of Root: Total Positive Real Roots
    How to Find It: Find the number of sign changes or is less than this by an even number. If number of sign changes is 5, then there could be, 5, 3, 1 positive real roots.
  • Type of Root: Total Negative Real Roots
    How to Find It: Compute f(-x). Count the number of sign changes, or is less than this by an even number.
  • Type of root: Complex Roots
    How to Find It: ComplexRoots = Total Roots - Positve Real Roots - Negative Real Roots
  • What is the total possible number of roots?
  • Total Possible roots is 5 because Degree is 5.
  • How Many Positive Real Root? How many Sign Changes are there?
  • According to Descarte's Change of Signs Rule there could be 3 Positive Real Roots, or less than an even number
  • Positive Real Roots = 3 or 1
  • How many Negative Real Roots are there? Compute f( − x) then count the number of sign changes
  • f( − x) = 3( − x)5 − 15( − x)4 − 29( − x)3 + 145( − x)2 + 40( − x) − 200
  • f( − x) = − 3(x)5 − 15x4 + 29x3 + 145x2 − 40x − 200
  • There couuld be 2 Negative Real Roots because there were only 2 sign changes, or less than that by an even number.
  • Negative Real Roots, 2 or 0
  • Complex Roots can be found by adding the different combinations of + Real and - Real Roots as follows
These are all the possible combinations of positive real, negative real and complex roots:

+Real -Real Complex Total
3 2 0 5
3 0 2 5
1 2 2 5
1 0 4 5
Determine the possible combinations of possible real roots, negative real roots and complex roots.
2x5 + 10x4 − 3x3 − 15x2 − 35x − 175 = 0
  • Type of Root: Total Roots
    How to Find It: You will find the total number of roots by the Degree of the Polynomial
  • Type of Root: Total Positive Real Roots
    How to Find It: Find the number of sign changes or is less than this by an even number. If number of sign changes is 5, then there could be, 5, 3, 1 positive real roots.
  • Type of Root: Total Negative Real Roots
    How to Find It: Compute f(-x). Count the number of sign changes, or is less than this by an even number.
  • Type of root: Complex Roots
    How to Find It: ComplexRoots = Total Roots - Positve Real Roots - Negative Real Roots
  • What is the total possible number of roots?
  • Total Possible roots is 5 because Degree is 5.
  • How Many Positive Real Root? How many Sign Changes are there?
  • According to Descarte's Change of Signs Rule there could be 1 Positive Real Roots, or less than an even number
  • Positive Real Roots = 1
  • How many Negative Real Roots are there? Compute f( − x) then count the number of sign changes
  • f( − x) = 2( − x)5 + 10( − x)4 − 3( − x)3 − 15( − x)2 − 35( − x) − 175
  • f( − x) = − 2(x)5 + 10(x)4 + 3(x)3 − 15(x)2 + 35(x) − 175
  • There could be 4 Negative Real Roots because there were only 4 sign changes, or less than that by an even number.
  • Negative Real Roots, 4, 2 or 0
  • Complex Roots can be found by adding the different combinations of + Real and - Real Roots as follows
These are all the possible combinations of positive real, negative real and complex roots:

+Real -Real Complex Total
1 4 0 5
1 2 2 5
1 0 4 5
Determine the possible combinations of possible real roots, negative real roots and complex roots.
10x5 − 4x4 + 5x3 − 2x2 − 50x + 20 = 0
  • Type of Root: Total Roots
    How to Find It: You will find the total number of roots by the Degree of the Polynomial
  • Type of Root: Total Positive Real Roots
    How to Find It: Find the number of sign changes or is less than this by an even number. If number of sign changes is 5, then there could be, 5, 3, 1 positive real roots.
  • Type of Root: Total Negative Real Roots
    How to Find It: Compute f(-x). Count the number of sign changes, or is less than this by an even number.
  • Type of root: Complex Roots
    How to Find It: ComplexRoots = Total Roots - Positve Real Roots - Negative Real Roots
  • What is the total possible number of roots?
  • Total Possible roots is 5 because Degree is 5.
  • How Many Positive Real Root? How many Sign Changes are there?
  • + 10x5 − 4x4 + 5x3 − 2x2 − 50x + 20 = 0
  • According to Descarte's Change of Signs Rule there could be 4 Positive Real Roots, or less than an even number
  • Positive Real Roots = 4 or 2 or 0
  • How many Negative Real Roots are there? Compute f( − x) then count the number of sign changes
  • f( − x) = 10( − x)5 − 4( − x)4 + 5( − x)3 − 2( − x)2 − 50( − x) + 20
  • f( − x) = − 10(x)5 − 4(x)4 − 5(x)3 − 2(x)2 + 50(x) + 20
  • There could be 1 Negative Real Roots because there were only 1 sign changes, or less than that by an even number.
  • Negative Real Roots: 1
  • Complex Roots can be found by adding the different combinations of + Real and − Real Roots as follows
These are all the possible combinations of positive real, negative real and complex roots:

+Real -Real Complex Total
4 1 0 5
2 1 2 5
0 1 4 5
Determine the possible combinations of possible real roots, negative real roots and complex roots.
64x6 − 1 = 0
  • Type of Root: Total Roots
    How to Find It: You will find the total number of roots by the Degree of the Polynomial
  • Type of Root: Total Positive Real Roots
    How to Find It: Find the number of sign changes or is less than this by an even number. If number of sign changes is 5, then there could be, 5, 3, 1 positive real roots.
  • Type of Root: Total Negative Real Roots
    How to Find It: Compute f(-x). Count the number of sign changes, or is less than this by an even number.
  • Type of root: Complex Roots
    How to Find It: ComplexRoots = Total Roots - Positve Real Roots - Negative Real Roots
  • What is the total possible number of roots?
  • Total Possible roots is 6 because Degree is 6.
  • How Many Positive Real Root? How many Sign Changes are there?
  • + 64x6 − 1 = 0
  • According to Descarte's Change of Signs Rule there could be 1 Positive Real Roots, or less than an even number
  • Positive Real Roots = 1
  • How many Negative Real Roots are there? Compute f( − x) then count the number of sign changes.
  • f( − x) = 64( − x)6 − 1
  • f( − x) = 64x6 − 1
  • There could be 1 Negative Real Roots because there were only 1 sign changes, or less than that by an even number.
  • Negative Real Roots: 1
  • Complex Roots can be found by adding the different combinations of + Real and − Real Roots as follows
These are all the possible combinations of positive real, negative real and complex roots:

+Real -Real Complex Total
1 1 4 6

*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

Roots and Zeros

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
  • Number of Roots 0:08
    • Not Nature of Roots
    • Example: Real and Complex Roots
  • Descartes' Rule of Signs 2:05
    • Positive Real Roots
    • Example: Positve
    • Negative Real Roots
    • Example: Negative
  • Finding the Roots 9:59
    • Example: Combination of Real and Complex
  • Conjugate Roots 13:18
    • Example: Conjugate Roots
  • Example 1: Solve Polynomial 16:03
  • Example 2: Solve Polynomial 18:36
  • Example 3: Possible Combinations 23:13
  • Example 4: Possible Combinations 27:11

Transcription: Roots and Zeros

Welcome to Educator.com.0000

We are going to continue our lesson on polynomials by discussing roots and zeroes.0002

The number of roots can be determined by looking at the degree of the polynomial equation,0008

because a polynomial equation of degree n has n roots.0014

This tells you only the number of roots; it doesn't tell you about the nature of the roots; the roots may be real or complex.0018

For example, if I am given a polynomial such as 5x6 + 2x3 - x2 + 9x - 2 = 0,0025

a polynomial equation, I see that the degree equals 6; this tells me that there are 6 roots for this equation.0038

Now, they can be real; and for example, these roots could be something like 1, -2...those are rational numbers.0053

They actually could be irrational numbers, like the square root of 3.0065

You also can have complex numbers as roots.0069

Recall that complex numbers have two parts--something like 2 + 3i; and the first part here is real, and the second part is imaginary.0073

And these are in the form a + bi.0086

We are going to talk, in a few minutes, about how these complex roots occur as conjugate pairs.0091

For example, 2 + 3i and 2 - 3i would be a conjugate pair--the same values here, but opposite signs; or 4 + i and 4 - i.0097

So, I know that here I have 6 roots; but I have no idea if they are real, rational, irrational, or complex, or what combination of those.0111

However, I can determine at least some of that through using Descartes' Rule of Signs.0119

And there are two sections to this rule: the first section helps you determine the possible numbers of positive real roots.0125

Then, we will talk about determining the possible number of negative real roots.0136

So, first we are just looking at the positive real roots: if we let p(x) be a polynomial with real coefficients (no imaginary coefficients--0141

just real coefficients), arranging in descending powers, recall that descending powers would be something like0154

f(x) = x6 + 5x5 + 7x4 - x3 + 6x2 - 2x + 4.0164

This is a degree 6 polynomial; and I have 6, 5, 4, and on down.0180

If you are trying to work with Descartes' Rule of Signs, the first thing to do is check the polynomial.0187

And if it is not arranged in descending powers, you need to put it that way.0192

OK, so I have this arranged in descending powers; the number of positive real roots0196

is the number of changes in sign of the coefficients, or is less than this by an even number.0202

Let's look at what that means: I want to look for the number of sign changes of the coefficient.0209

So, here I have...this coefficient is a 1, and that is positive; 5 is positive; 7 is positive; -1--so there is a sign change.0217

This is positive; this is negative; so that is one sign change.0230

OK, now here, I am going from -1 to positive 6; again, I have a sign change.0237

Here, I am going from a positive to a negative--another sign change; negative to positive--another sign change.0245

That gives me 1, 2, 3, 4--4 sign changes means there are 4 positive real roots, or less than this by an even number.0254

So, less than 4 by an even number would mean 4 - 2 (would give me 2), or another even number, 4: 4 - 4 would give me 0.0275

There are 4, 2, or 0 positive real roots.0287

So again, take the polynomial; arrange it in descending powers; and then look at the number of sign changes.0295

I have 1, 2, 3, 4 sign changes; that tells me that I will have, at most, four positive real roots.0302

However, I may have less than this by an even number (4 - 2: 2; 4 - 4; 0).0312

So, I have three possibilities: I may have 4 positive real roots, 2 positive real roots, or 0 real roots.0319

And I have, since the degree equals 6, a total of 6 roots.0327

I have a total of 6; of these, 4, 2, or 0 may be positive real roots.0338

OK, the second part of this is looking at the number of negative real roots.0343

The number of negative real roots is the number of changes in sign of the coefficients of the terms p(-x), or is less than this by an even number.0350

Let's continue on with the example that we just looked at, where we were given0361

f(x) = x6 + 5x5 + 7x4 - x3 + 6x2 - 2x + 4.0366

Now, I found there were four sign changes, which means 4, 2, or 0 positive real roots.0383

For the negative real roots, I have to look at f(-x).0388

So, I need to change these x's to -x and be very careful with the signs; OK.0394

So, this is going to give me a coefficient here of -1.0416

But if I take a negative to an even power, it is going to become positive.0421

-1 to the sixth power is going to become positive, so this is going to give me f(-x); here it is just going to be x6.0425

Here, I have -1 times x; you can look at it that way--it is -1 times x5.0435

Well, if I take a -1 to an odd power, it is going to remain negative; so, -x5 times 5 is going to give me -5x5.0443

Here, I have a negative coefficient to an even power; it is going to become positive, so this is really 7x4.0457

Here, I have a negative coefficient to an odd power, so it will remain negative; so this is -x3,0467

but it is times a negative: so -x3 times -1 becomes + x3.0475

OK, I have a -x; this actually should be outside...(-x)2 is going to give me -x times -x; that is going to give me a positive.0485

So, this is going to be plus 6x2.0501

Here, I have -x times -2; that is +2x; and my constant remains positive.0507

OK, number of sign changes: I am going to look for the changes in sign.0516

This is positive out here; so a positive to a negative--that is 1; a negative to a positive--that is 2.0523

This stays positive, positive, positive, positive; the number of sign changes equals 2.0533

OK, so the number of negative real roots is the number of changes in sign of the coefficients of the term p(-x), or less than this by an even number.0539

Therefore, I am going to have two negative real roots, or less than this by an even number.0552

Well, 2 minus 2 is 0; I can't go any lower than that for the number of roots; so there are 2 or 0 negative real roots.0559

Again, this power is 6; I have 6 total roots; I have 2 or 0 negative real roots; and last slide, we talked about having 4, 2, or 0 positive real roots.0575

So, we covered the real roots; there also may be complex roots, so let's talk now about the total roots and the different combinations that you could have.0589

We can use Descartes' Rule of Signs to determine the possible combinations of the real and complex roots.0600

So, in that example above, f(x) = x6 + 5x5 + 7x4 - x3 + 6x2 - 2x + 4,0607

the total roots (this is real and complex) equal 6.0624

Positive real roots: using Descartes' Rule of Signs, I found that I could have 4, 2, or 0.0632

Negative real roots: I found f(-x) and looked for the sign changes, and found two of those.0645

So, the number of negative real roots could be 2 or 0.0650

Now, I can figure out the combinations: I know that they need to total 6, and I have0661

my positive real roots' possibilities and my negative real roots; and the last part of this is complex roots.0669

So, I have positive real roots, the number of negative, and the number of complex; and these need to total 6; the total must equal 6.0681

So, I said that, for positive real roots, I could have 4; then, I may have 2 negative real roots.0693

4 and 2 is 6, so that leaves me with no complex.0700

I could have 4 positive real roots; I could have 0 negative real roots; 4 and 0 is 4; to total 6, I will have to have 2 complex roots.0703

OK, another possibility: I have 2 positive real roots and 2 negative real roots: 2 and 2 is 4; 2 more complex roots will give me 6.0717

Or I could have 2 positive real roots and 0 negative real roots.0731

2 and 0 is 2; to total 6, I will have 4 complex roots.0737

OK, finally, I may have 0 positive real roots and 2 negative real roots.0744

0 and 2 is 2, so I have to have 4 complex roots.0753

Then, I may have 0 positive real roots and 0 negative real roots, and that gives me 0 and 0.0760

So, to total 6, I would have to have 6 right here.0774

And since 3 times 2 is 6, I expect 6 combinations, and that is what I have here.0780

So, Descartes' Rule of Signs, and knowing that the degree is 6 (so I have 6 total roots)0785

allows me to figure out the possible combinations of the roots of a polynomial.0790

Conjugate roots: we talked a bit about complex conjugates and the fact that there are complex roots.0798

But getting into a bit more detail: if the polynomial p(x) has real coefficients...0806

the coefficients are not imaginary, which is what we will be working with--real coefficients--for now...0812

the complex roots of p(x) occur as complex conjugates.0819

And you saw this earlier on, when we worked with quadratic equations and quadratic functions.0823

Thinking about something like this: x2 + 4 = 0, if I wanted to find the solutions for this, I could say, "OK, x2 = -4."0830

Now, using the square root property, I take the square root of both sides; and this gives me x = ±√-4.0851

Recall that the square root of -1 equals i; then I can rewrite this as x = ±√-1, times √4,0861

or x = ±i√4, or x equals ±2i.0875

And we can look at this, also, in a different way: we can say x = 0 + 2i, or x = 0 - 2i.0886

And you can then see that this is a pair of complex conjugates.0896

And it is because of this, where we are finding the square root, that we end up with plus or minus some number.0902

And therefore, these complex roots of polynomials occur as complex conjugates.0909

Another example of complex conjugates would be, say, 8 + 5i and 8 - 5i--the same values, but switch the sign--or 2 + 7i and 2 - 7i.0918

And this explains why, when we talked about the number of negative real roots and positive real roots,0935

we said it was that number (say, 4), or less than that by 2.0943

And the reason it goes down by pairs is because the complex conjugates occur in pairs.0946

So, they could take up two of the spots for the roots; and so they are going to decrease the number of the real roots by 2 or multiples of 2.0951

Here, we are asked to solve this polynomial equation, x3 - x2 - 6x = 0.0965

I am going to solve this by factoring; and the first step is to factor out the greatest common factor.0975

And here, I see I have a common factor of x; there is an x in each of these.0980

So, factor that out; this is going to give me x times x2, minus x, minus 6, equals 0.0984

Don't forget to bring this x along as you factor this, because this is going to give us one of the solutions.0995

I have x2 - x - 6 that I need to factor; this is in the form (x + a constant) (x - a constant), because the sign is negative.1002

Factors of 6: 1 and 6, 2 and 3; and I need these to add up to -1, and their signs are going to be opposite.1015

These (2 and 3) are close together, so if I make the larger one negative and the smaller one positive, I am going to get a -1.1026

Therefore, 3 is negative, and 2 is positive.1037

Now, according to the zero product property, if any of these terms (x, x + 2, or x - 3) is 0, then this product will be 0.1042

And it will equal the right side of the equation.1053

Therefore, x equals 0 (and make sure you don't leave this one out, or you will be missing one of the solutions), x + 2 = 0, or x - 3 = 0.1056

So here, I don't have to do anything further with this.1068

I just have that one of the solutions here is x = 0.1070

Here, I have to subtract 2 from both sides to get x = -2; and here, I need to add 3 to both sides.1075

So, solutions: x = 0, x = -2, and x = 3; these are all solutions to this equation.1083

And I solved this by factoring out the greatest common factor, x, factoring this trinomial,1098

and then using the zero product property to solve for x in each of these expressions' terms.1107

Here, I am asked to solve x4 - 256 = 0.1119

This is actually the difference of two squares; this is in the form a2 - b2.1126

So, it is going to factor out to (a + b) (a - b).1133

And if you look at it and think about it this way, x2, squared, is x4.1137

And if you look at this one, 256, and take the square root of that, it is actually 16.1143

Therefore, this is telling me that a equals x2, and b equals 16.1155

So, I can factor it as follows: a + b (that is x2 + 16), times a - b (or x2 - 16).1161

OK, and looking at what I have, I can't do anything else with x2 + 16.1177

But I recognized again, here: I have the difference of two squares; this time, a equals x, and b equals 4.1182

So, it is going to factor out to (x + 4) (x - 4); and these are set equal to 0.1189

Now, I use the zero product property, which is going to tell me that I could have x2 + 16 = 0, x + 4 = 0, or x - 4 = 0.1200

And let's work with these simpler ones first.1214

Simply subtract 4 from both sides to give me x = -4.1217

Add 4 to both sides: x = 4; I have two of my solutions.1224

Now, looking over here, it is a little bit more complex: subtract 16 from both sides--that gives me x2 = -16.1229

Now, I am going to take the square root of both sides, and you can see that this ends up being ±√-16.1239

And this is a negative number; and since I know that √-1 equals i, I can rewrite this as x = ±i√16.1247

Well, the square root of 16 is 4, so this gives me a complex conjugate pair, plus or minus 4i.1264

So, I have four solutions: x = 4i, x = -4i, x = 4, and x = -4.1276

And let's just think about Descartes' Rule of Signs and show that it predicted the possibilities for the type of roots that I could get.1293

Since I have x4 - 256, if I look at the number of sign changes for this, this is positive to negative (one sign change).1303

This tells me that I am going to have one positive real root, or less than that by an even number;1320

but I can't go there, because then I would be going into negative numbers,1327

and I can't say there are -1 real roots; that wouldn't make sense.1330

So, it is just one positive real root.1333

Now, looking at f(-x): this gives me -x4 - 256.1340

Well, this -1, when you take it to the fourth power, is just going to become positive; so this gives me this.1350

And again, I have one sign change; so this tells me that I have one negative real root.1359

Since the degree here is 4, I have 4 total roots.1370

So, this is going to leave me with one positive real, one negative real, and two complex roots,1376

which is exactly what I see: a positive real, a negative real, and the set of complex conjugates.1387

Determine the possible combination of positive real roots, negative real roots, and complex roots.1395

We will use Descartes' Rule of Signs to determine this, and we will start out by thinking about the total.1402

Since the degree is 3, there are 3 total roots.1410

Now, using the rule of signs, I am going to look for f(x) and the sign changes: this is 2x3 - 3x2 + 4x - 5.1421

Number of sign changes: well, this is positive, and this is negative--that is one.1432

I am going from negative here to positive here; that is two; from positive to negative--that is 3.1440

So, the number of sign changes equals 3.1448

Therefore, the number of positive real roots is 3, or less than this by an even number (3 - 2 is 1).1450

You can't go any lower than that; if I subtracted by 2 again, I would get a negative number.1465

So, I have either 3 or 1 positive real roots.1470

All right, now let's look at the negative scenario for f(-x) to figure out the negative real roots.1477

2 times -x cubed, minus 3 times -x squared, plus 4 times -x, minus 5:1484

OK, this is going to give me f(-x) =...this is going to remain negative, so this is going to give me -2x3.1500

A negative and a negative (squared) is going to give me a positive, so this will become x2.1512

This is negative here, though, so it is -3x2.1519

4 times -x is -4x, minus 5.1524

OK, the number of sign changes: none, none, none: the signs are all negative--the coefficients of f(-x).1529

So, the number of sign changes equals 0; so there are 0 negative real roots.1542

OK, so let's figure out what we have going on here.1551

We have positive real roots, negative real roots, and complex.1554

And remember: these need to total 3.1567

Positive real: I could have 3; negative real: 0; I need them to total 3--this already does, so the complex is going to be 0, since this totals 3.1571

OK, another possibility is that I have one positive real root, 0 negative real roots.1586

1 and 0 is 1; I need it to total 3; so there must be 2 complex real roots.1594

So, the possible combination of positive real roots, negative real roots, and complex roots is 3, 0, 0, or 1, 0, 2.1601

And I know I have three total roots, because the degree is 3.1612

I use Descartes' Rule of Signs to tell me that I had either 3 or 1 positive real roots; I have 0 negative real roots.1615

And then, figuring out what is lacking to get my total of 3, I could fill it in with complex roots.1623

OK, determine the possible combinations of positive real roots, negative real roots, and complex roots.1632

Degree equals 4, so I have four total roots.1639

f(x) = -3x4 - 4x3 + 2x2 + 6x + 7.1649

So, positive real roots: let's look for sign changes between these coefficients.1659

A negative to a negative; a negative to a positive (that is 1); a positive to a positive--no sign change; a positive to a positive again.1667

So, the positive real roots equals 1; and I can't go by less than that,1678

because if I subtracted 2, I would go into negative numbers.1685

So, the number of positive real roots is 1.1688

Now, looking at negative real roots: I am going to take f(-x) to get1691

-3 times -x to the fourth, minus 4 times -x to the third, times 2 times -x squared, plus 6 times -x, plus 7.1700

OK, this gives me f(-x) =...well, -1 to the fourth power--this will become positive; so I have -3x41718

Here, -x cubed...this is going to remain negative as a coefficient...times -4; that is going to become positive, so this is going to give me + 4x3.1733

-x squared is going to give me positive x2 times 2; so that is + 2x2; 6 minus -x is -6x; plus 7.1748

OK, negative real roots is going to be determined by the sign changes of the coefficients of f(-x), according to Descartes' Rule of Signs.1760

A negative to a positive; that is one sign change; that stays positive; a positive to a negative: 2; a negative to a positive: 3.1772

So, here I have 3 or less than that by an even number: 3 - 2 is 1.1784

You can't go any less than that.1796

I have a total of 4 roots; I have one positive real root, and either 3 or 1 negative real roots.1799

So, let's look at the possibilities: positive real roots, negative real roots, and complex.1810

Positive real: 1; then I could have a negative real root totaling 3: 1 + 3 is 4; I only have 4; so complex must be 0.1825

Or I could have 1 positive real root and 1 negative real root; and I am going to total those to get 2.1838

But I should have 4; so in this case, there would be two complex roots--a pair of complex conjugates.1847

OK, so I determined there were four total roots, because the degree here is 4.1855

Using Descartes' Rule of Signs, the positive case, f(x), I found that there is one positive real root.1861

f(-x) gave me three sign changes, so there are either three or one negative real roots.1869

And then, to get a total of four, I had zero complex roots in this case, and two in this case.1875

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