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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (7)

1 answer

Last reply by: Dr Carleen Eaton
Sat Nov 7, 2015 5:50 PM

Post by Fadumo Kediye on October 13, 2015

P(x) = 2x^3 + ax^2 +bx + 6 is divided by x + 2, the remainder is -12. If x - 1 is a factor of the polynomial, find the values of a and b.

2 answers

Last reply by: Fadumo Kediye
Tue Oct 13, 2015 11:42 PM

Post by enya zh on September 29, 2012

At about 25:49, you didn't list EVERY POSSIBLE factor. Isn't all the factors 1,(x-2), (x+3), (x-4), (x^2-x-12), (x^2+x-6),(x^2-6x+8),&(x^3-3x^2-10x+24)? 1 and itself would always be factors and I got three additional factors with the degree of two by multiplying the binomial factors.

1 answer

Last reply by: Huseyin Kayahan
Fri Oct 14, 2011 5:45 AM

Post by Huseyin Kayahan on October 14, 2011

In the f(a) example, why the f(2) is equal to remainder?
what is the f(3) with the division method?

Remainder and Factor Theorems

  • Use synthetic substitution to evaluate a polynomial of degree 4 or more for a specific value.
  • Use synthetic division to find the factors of a polynomial of degree 3 or more. Guess and then check your guess by synthetic division. Once you find one factor, use synthetic division to find a factor of the quotient. Keep going until the quotient is quadratic. At this point, use the quadratic formula to factor the quadratic.

Remainder and Factor Theorems

Using synthetic division find f(1) if f(x) = x3 − x2 − x + 1
  • Recall that if you divide f(x) by x − 1 you will get:
  • [f(x)/(x − 1)] = Q + R; the remainder R is f(1).
  • Proceed to do synthetic division, notice that there are no missing terms.
  • 1 1 -1 -1 1
        1 0 -1
      1 0 -1
    0
In this case, f(1) = 0
Using synthetic division find f( − 8) if f(x) = x3 + 7x2 − 13x − 44
  • Recall that if you divide f(x) by x − 8 you will get:
  • [f(x)/(x + 8)] = Q + R; the remainder R is f(8).
  • Proceed to do synthetic division, notice that there are no missing terms.
  • -8 1 7 -13 -44
        -8 8 40
      1 -1 -5
    -4
In this case, f( − 8) = − 4
Using synthetic division to find f(7) if f(x) = x3 − 9x2 + 12x + 22
  • Recall that if you divide f(x) by x + 7 you will get:
  • [f(x)/(x − 7)] = Q + R; the remainder R is f(7).
  • Proceed to do synthetic division, notice that there are no missing terms.
  • 7 1 -9 12 22
        7 -14 -14
      1 -2 -2
    8
In this case, f(7) = 8
Using synthetic division to find f(4) if f(x) = x4 − 4x3 − 7x + 23
  • Recall that if you divide f(x) by x − 4 you will get:
  • [f(x)/(x − 4)] = Q + R; the remainder R is f(4).
  • Proceed to do synthetic division, notice that the square term is missing, add 0 as place holder
  • 4 1 -4 0 -7 23
        4 0 0 -28
      1 0 0 -7
    -5
In this case, f(4) = − 5
Using synthetic division to find f( − 4) if f(x) = 2x4 + 8x3 + 8x + 24
  • Recall that if you divide f(x) by x + 4 you will get:
  • [f(x)/(x + 4)] = Q + R; the remainder R is f( − 4).
  • Proceed to do synthetic division, notice that the square term is missing, add 0 as place holder
  • -4 2 8 0 8 24
        -8 0 0 -32
      2 0 0 8
    -8
In this case, f( − 4) = − 8
Check if (x + 6) is a factor of x3 + 15x2 + 49x − 30
  • Recall that in order for a binomial to be a factor of the given polynomial the Remainder must equal to zero.
  • Divide using synthetic division. Always check for missing terms.
  • -6 1 15 49 -30
        -6 -54 30
      1 9 -5
    0
Because the remainder is zero, (x + 6) is a factor.
Check if (x + 4) is a factor of 8x3 + 29x2 − 17x − 26
  • Recall that in order for a binomial to be a factor of the given polynomial the Remainder must equal to zero.
  • Divide using synthetic division. Always check for missing terms.
  • -4 8 29 -17 -26
        -32 12 20
      8 -3 -5
    -6
Because the remainder is not zero, (x + 4) is not a factor.
Check if (a − 10) is a factor of a5 − 10a4 + 8a − 80
  • Recall that in order for a binomial to be a factor of the given polynomial the Remainder must equal to zero.
  • Divide using synthetic division. Always check for missing terms, in this case, cube and
  • 10 1 -10 0 0 8 -80
        10 0 0 0 80
      1 0 0 0 8
    0
Because the remainder is zero, (a − 10) is a factor.
Show that (x − 3) is a factor of x3 − x2 − 14x + 24. Find the other factors.
  • To check that the binomial is a factor, Divide the Polynomial using synthetic division. If the Remainder is 0,
  • then the binomial is a factor. As always, check for any missing terms.
  • To find the other factors, try any methods covered in lessons before.
  • 3 1 -1 -14 24
        3 6 -24
      1 2 -8
    0
  • (x − 3) is a factor. Now find the other factors by factoring x2 + 2x − 8
  • x2 + 2x − 8 = (x + 4)(x − 2)
  • Factored completely would be
x3 − x2 − 14x + 24 = (x − 3)(x + 4)(x − 2)
Show that (x + 5) is a factor of x3 + 12x2 + 47x + 60. Find the other factors.
  • To check that the binomial is a factor, Divide the Polynomial using synthetic division. If the Remainder is 0,
  • then the binomial is a factor. As always, check for any missing terms.
  • To find the other factors, try any methods covered in lessons before.
  • -5 1 12 47 60
        -5 -35 -60
      1 7 12
    0
  • (x + 5) is a factor. Now find the other factors by factoring x2 + 7x + 12
  • x2 + 7x + 12 = (x + 4)(x + 3)
  • Factored completely would be:
x3 + 12x2 + 47x + 60 = (x + 5)(x + 4)(x + 3)

*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

Remainder and Factor Theorems

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
  • Remainder Theorem 0:07
    • Checking Work
    • Dividend and Divisor in Theorem
    • Example: f(a)
  • Synthetic Substitution 5:43
    • Example: Polynomial Function
  • Factor Theorem 9:54
    • Example: Numbers
    • Example: Confirm Factor
  • Factoring Polynomials 14:48
    • Example: 3rd Degree Polynomial
  • Example 1: Remainder Theorem 19:17
  • Example 2: Other Factors 21:57
  • Example 3: Remainder Theorem 25:52
  • Example 4: Other Factors 28:21

Transcription: Remainder and Factor Theorems

Welcome to Educator.com.0000

In today's lesson, we will be covering the remainder and factor theorems.0002

Now, today, with the remainder theorem, this is going to be familiar; we used it earlier on, without giving it a name.0007

And we are going to talk about it and put it to a different use today.0014

So, recall that, when we were dividing polynomials, either by long division or synthetic division,0017

a way to check our work was to recall that the dividend equals the quotient times the divisor, plus the remainder.0023

And recall that, if I have, say, 48 divided by 2, this is the dividend, and the term that you are dividing by is the divisor.0042

And of course, the solution is the quotient, and whatever is left over is the remainder.0055

So, sometimes when we did long division, I checked it by multiplying my quotient by the divisor,0059

adding the remainder, and making sure that I came up with the dividend.0066

Now, looking back at all this, this is really just saying the same thing.0070

If the polynomial, f(x), is divided by x - a (here I have the dividend, the polynomial f(x), and the divisor;0075

this is the dividend, f(x); here, the divisor is x - a), I have that the dividend equals the divisor,0084

times...and it tells me that g(x) is the quotient, so times the quotient, plus f(a) (f(a) is the remainder).0096

This is just saying the same thing: if I am working with polynomials, the polynomial equals the quotient times the divisor, plus the remainder.0109

Now, we are going to put this to a different use; and we are going to use it to find f(a).0119

Looking at an example: let's take the polynomial function f(x) = 2x3 + x2 - 3x + 4.0125

And let's say that I wanted to find f(2): well, I could use my substitution method (the usual),0136

where I would just say, "OK, 2 cubed, plus 2 squared" substituting 2 in for x, "minus 3x, plus 4; and this gives me...0145

2 times 23 is 8, plus 4...actually putting in a 2 right here...minus 3 times 2 is 6, plus 4, equals 16 + 4 - 6 + 4."0156

So, this gives me 20 = 2, or 18.0173

OK, that wasn't too bad; but if this had been a polynomial of, say, degree 5 or 6,0179

and I was asked to find something like f(5), that is a lot of arithmetic, working with large numbers.0185

So, there is another way to approach this.0191

What this is saying is that, if I divide a polynomial by x - a, the remainder is f(a).0195

So, instead of finding f(2) by substitution, I could find f(2) a different way: I could say that f(2) equals the remainder when f(x) is divided by x - a.0202

And in this case, I have f(2), so a equals 2.0225

This remainder theorem gives me another method for finding the value of a function at a certain value.0235

So, if I am asked to find f(2), I could just say, "OK, this equals 2x3 + x2 - 3x + 4, divided by x - 2."0246

And if I were to divide all of this out, I would actually find that the remainder is 18.0261

So, I would end up getting a polynomial, f(2), and I would end up getting some quotient;0270

when I did my division, I would get some quotient; plus I would get a remainder after this division.0283

And this remainder is going to be equal to f(2).0291

So, there are two ways for finding the value of a function: if I am asked to find, say, f(2),0298

I can either substitute in, or I can divide the polynomial by x, minus that value (which is 2).0304

And this actually turns out to be 2x2 + 5x + 7, with a remainder of 18.0313

So, if you were to divide this out, you would get this (which is g(x)), and the remainder would be 18.0324

And this remainder of 18 equals f(2).0333

OK, so we are using the remainder theorem to find values for functions.0337

Now, I mentioned that, in order to do this, you are going to have to divide the polynomial by x - a.0344

Long division is a lot of work; so the quicker way to go, since the divisor is in the form x - a constant, is to use synthetic division.0352

And here, we are calling it synthetic substitution only because we are using it to find the value of a polynomial function.0361

It is the same thing: we are using synthetic division, but when we use it for this purpose, we call it synthetic substitution.0368

So, for example, if I was given a polynomial function, f(x) = 4x4 - 2x3 + x - 1,0375

and I was asked to find f(2), I could substitute 2 in here; or I can use this new method.0386

The way I am going to find this is: I am going to take 4x4 - 2x3 + x - 1, and I am going to divide it by x - 2.0397

And then, I am going to find a quotient, plus a remainder; and that remainder is going to equal f(2).0409

Now, reviewing synthetic division (which is much quicker than long division), recall that, in synthetic division,0416

you are going to take this constant here, and put it here, but with the opposite sign.0422

I have a negative 2; I am going to make this a 2.0427

Then, I am going to take these coefficients and put them here.0430

It is very important, before you do that, though, to check for missing terms.0433

So, here, I have x4, x3...I do not have an x2 term; that is missing.0437

So, I am actually going to rewrite this as f(x) = 4x4 - 2x3...and for the missing term,0445

I am going to use a coefficient of 0; that is going to be my placeholder when I do my division...+ x - 1.0452

My coefficients are 4, -2, 0, 1, and -1; with synthetic division, your first step is to bring down that first term.0462

So, this is going to give me 4; then multiply 4 times this divisor, 2, to get 8.0477

Then add: -2 plus 8 gives me 6; multiply again: 6 times 12 gets 12; add: 12...0485

And you see, if I hadn't had this 0 here as a placeholder, I would have gotten a completely different answer.0497

So, it is very important, before you even start writing down the coefficients, to check and see if there are any missing terms.0501

OK, the next step: 12 times 2 gives me 24; add that to 1: 25.0507

25 times 2 gives me 50, plus -1 is 49.0516

Now, what this is giving me is 4x3 (because the variable here is going to have a degree0522

one less than the degree of the dividend, so this is 4x3) + 6x2 + 12x + 25.0531

So, this is the quotient; this last term is the remainder.0541

And this remainder of 49 equals f(2).0553

Again, when you are working with a high-degree polynomial and a large value that you are looking for in the function,0559

then this is a much quicker method than substitution.0568

So again, I handled this by saying I wanted to find f(2), so I am going to divide this polynomial by x - 2,0572

using synthetic division, being careful to include a 0 coefficient as a placeholder for missing terms.0579

I found the quotient and the remainder, and the remainder equals f(2).0585

OK, the factor theorem is a result of the remainder theorem.0594

Now, what this says is that x - a is a factor of the polynomial f(x) if and only if f(a) equals 0.0599

And this makes sense, because if you divide something by its factor, the remainder is going to be 0.0609

Just a very simple case--16: factors include 2 and, say, 4.0616

OK, so if I do 16 divided by 2, it equals 8; the remainder is 0, or 16 divided by 4 equals 4; the remainder is 0.0627

So, I know that x - a is a factor of the polynomial, if and only if the remainder is 0.0639

This tells me something else, too: it tells me that what I have here with a is a 0,0646

because if I have a value of the function, a (let's say I am given a value x = 4 for a polynomial),0653

and if I find that f(4) equals 0, that means that when x equals 4, the value of the function is 0.0662

So, this is a zero of the polynomial, which is very helpful information to have.0669

OK, now, how does this get put into play?0675

What this can help you do is: if you are trying to find the factors, or confirm that something is a factor, you can use synthetic division.0678

For example, let's say I have a polynomial function f(x) = 2x4 - 2x3 - 17x2 + 12x + 9.0686

And let's say that I am asked to determine if x - 3 is a factor of f(x).0700

I may be given a problem to work out, like this; or I may be just asked to factor this.0714

And if I am asked to factor this, I need to start out by just taking some guesses--0719

Is x - 2 a factor? Is x - 3 a factor? Is x - 4 a factor?--and then checking those guesses using synthetic division.0726

So, given f(x), I need to determine if x - 3 is a factor.0733

So, if I divide f(x) by x - 3, if the remainder equals 0, then x - 3 is a factor of f(x).0738

All right, again, using synthetic division is the easiest way to go about it.0760

So, I have x - 3, and I am going to take the opposite sign of -3; I am going to make that a 3 and put that here.0764

Now, I check, and I don't have any missing terms: x4, 3rd, 2nd, 1st, and a constant; I have no missing terms.0770

OK, 2, -2, -17, 12, and 9...0777

Synthetic division: bring that first term down; multiply 3 times 3 to get 6; add 6 and -2 to get 4.0788

Multiply again; 4 times 3 gives me 12; 12 minus 17 is -5; -5 times 3 is -15.0799

-15 plus 12 is -3; -3 times 3 is -9; 9 plus -9 is 0.0809

OK, what this is giving me is my quotient here and my remainder here.0819

So, what this is telling me is: I look up here, and the degree is 4; so the degree of the variable here is actually going to be 3.0826

So, this is telling me that what I have is x - 3, times 2x3 + 4x2 - 5x - 3.0833

And the remainder is 0 here, so x - 3 is a factor of f(x).0852

What I have done now is: I have pulled out this factor, x - 3; I have factored it out of the original; and what is left behind is this.0866

OK, so x - 3 is a factor of f(x); if this remainder had turned out to be anything other than 0, then x - 3 would not have been a factor.0878

So, factoring polynomials: synthetic division can be used to factor polynomials of a degree greater than 2.0888

And we just saw how the factor theorem can help us achieve that.0896

So, if we are working with polynomials that have a greater degree than a quadratic equation, we can use synthetic division to help.0900

For example, if I have something...f(x) = x3 - 5x2 - 2x2 + 24, what I want to do is find one factor.0907

And then, I want to pull that out through synthetic division, just as I did last time.0924

Sometimes you will be given that first factor and just asked to confirm, as I mentioned.0930

Other times, you won't; so then you just have to take guesses.0934

Let me give you an example: is x - 3 a factor? I am just taking a guess.0938

And so, what I am going to do is: I am going to use synthetic division; I am going to put a 3 here, and this coefficient is 1, -5...0947

this x should actually be...let's make that -2x...-2, and 24.0959

OK, so x3, second, first, constant...there are no missing terms.0969

And I am checking to see if x - 3 is a factor; I am going to bring down the 1.0974

1 times 3 is 3; 3 + -5 is -2; -2 times 3 is -6; -6 plus -2 is -8; -8 times -3 is -24.0978

OK, -24 and 24 is 0; my remainder is 0: "Yes," since the remainder is 0 when I divided this polynomial by x - 3.0996

Now, this is degree 3; that tells me that what I have here is going to be degree 2.1012

So, I pulled out this factor, x - 3, and what I have left behind is the quadratic expression x2 - 2x - 8.1019

Not only, now, do I know that this is a factor, but I have something left that I can work with--that I can factor further, using quadratic methods.1037

And in fact, if this were an equation--if I was given that this equals 0 or equals some other number--1046

then I could solve it using either factoring or the quadratic formula.1054

So, just continuing on with the factoring: I have gotten this far; I have pulled out this factor; here is what I have left behind, x - 3.1062

Now, factoring this out, this is just a general trinomial, and I know that it is going to be in this form: (x + something) (x - something), since this is negative.1070

Factors of 8 are 1 and 8, 2 and 4; and I want those to sum up to -2 when one is positive and the other is negative.1081

So, I know that these are too far apart; but if I take 2 and -4, I am going to get -2.1093

So, I am going to factor this out as such; OK, so now this is factored out as far as I can go.1105

Again, synthetic division was very useful, because I started out with this polynomial.1113

And then, I had to just take a guess, and I said, "OK, is x - 3 a factor?"1118

I confirmed that through synthetic division, finding that this remainder is 0.1124

So, this is a factor; so that tells me I have this factor, times what is left behind, which was a quadratic, which I could factor further.1130

If I was wrong, then I would have tried another guess; I would have tried x + 3 or x - 2 or x - 1, until I found one.1139

So, that first step is the longest; if you make the right guess and you get the right factor,1147

and then use synthetic division, then often what you are left with is easier to work with.1151

Applying these concepts to some examples: f(x) = 2x3 - x + 7; I am asked to find f(4).1158

I could find it by substitution; it actually wouldn't be that difficult, just substituting 4 for x wherever the x's appear.1165

But just to practice using the remainder theorem, let's use the remainder theorem and synthetic division, and find f(4) that way.1173

Recall that, if I want to find f(4), what I need to do is divide f(x) by x - 4, and then it is going to give me a quotient plus a remainder.1182

And the remainder is going to equal f(4), according to the remainder theorem.1198

OK, so one thing before I proceed with my division is: I am going to realize that I have a missing term.1204

I have 2x3; I don't have an x2 term, so I am going to represent that with a coefficient of 0.1211

I do have an x term, and then I have my constant.1217

And I am going to divide all that by x - 4.1221

Using synthetic division, I am going to put the 4 out here, and then my coefficients 2, 0, this is -1, and 7,1227

bringing the 2 down; 2 times 4 is 8; 8 plus 0 is 8; 8 times 4 is 32; 32 plus -1 is 31.1237

31 times 4 gives me 124; 124 plus 7 is 131; here is my quotient; here is my remainder.1256

The remainder equals 131; therefore, f(4) equals 131.1270

And you could check this by using the substitution method; and you would find, again, that you got f(4) is 131.1278

Again, I handled this by saying that, if I want to find f(4), I can do that by dividing the function by x - 4, finding the quotient, and finding the remainder.1285

The remainder, when I divide this, is equal to f(4).1295

I use synthetic division, being careful to include 0 for a coefficient for the missing x2 term.1301

I came up with this quotient and a remainder of 131; so, f(4) = 131.1307

Show that x - 2 is a factor of this polynomial, and find the other factors.1319

Recall that, if x - 2 is a factor of this polynomial, then if I divide (I am going to call this f(x)) f(x) by x - 2, the remainder will equal 0.1325

OK, I am supposed to show that this is a factor and go on to find the other factors.1350

So, I am going to do that by using synthetic division.1356

All right, I am dividing by 2; and I do not have any missing terms: cubed, squared, x to the first, constant--no missing terms.1362

So, this is a coefficient of 1, a coefficient of -3, a coefficient of -10, and a coefficient of 24...or actually, the constant is 24.1373

OK, bring down the 1; 1 times 2 is 2; 2 and -3 gives me a -1.1387

-1 times 2 is -2; -2 and -10 is -12; -12 times 2 gives me -24.1396

OK, the remainder equals 0, so x - 2 is a factor of f(x).1414

All right, now this is a degree 3 polynomial; so what I am going to have down here is actually going to be the quadratic equation.1427

That is at 1x2 - 1x - 12.1438

Now, I am supposed to find the other factors; and I always want to remember that x - 2 is a factor.1442

So, x - 2 times this quotient will give me the polynomial back; I don't want to leave this off.1450

But to find the other factors, I am going to have to factor this out farther.1456

So, x - 2 is factored as far as it can be; now, this is going to be a general trinomial in the form (x + something) (x - something), because this is negative.1461

Looking at the factors of 12: the factors of 12 include 1 and 12, 2 and 6, 3 and 4.1479

The middle term here is -1, so I am going to look for factors that, when one of them is negative and the other is positive, add up to -1.1488

-1 is a small number, so I am going to look for factors that are close together; so I am going to focus on this.1497

And I am going to make the larger number negative, because this ends up being negative.1502

-4 + 3 does equal -1; therefore, I am going to make this 3, and this -4.1507

OK, so I did show that x - 2 is a factor, because the remainder is 0.1519

And then, the other factors, in addition to x - 2, are x + 3 and x - 4.1526

This is the complete factorization right here, the one they gave me, and the two other factors.1545

All right, here we are given a function, and we are asked to find f(3),1554

using the remainder theorem, which states that f(3) will be the remainder when I divide this polynomial here by x - 3;1558

I am going to get a quotient plus a remainder, and the remainder is going to equal f(3).1580

OK, using synthetic division: looking here, what I actually have are some missing terms.1586

So, I am going to rewrite this as f(x) = x4...my x3 term is missing,1595

so I will give that a 0 coefficient; I have an x2 term; my x term is missing,1601

so I will give that a 0 coefficient; and I have a constant.1608

So, this is going to give me a coefficient of 1; a coefficient of 0; -2; 0; and 6.1613

Bring down the 1; 1 times 3 is 3; 3 plus 0 is 3; 3 times 3 is 9; 9 plus -2 is 7.1624

7 times 3 is 21; 21 plus 0 is 21; 21 times 3 gives me 63; 63 plus 6 is 69.1639

OK, this is the quotient; this is the remainder.1654

Since the remainder is 69, f(3) equals 69.1660

Now, I could have solved this by substituting 3 in here; but this is actually an easier way to go about it.1669

Again, to find f(3), I am going to divide this polynomial by x - 3; I am going to find my quotient and my remainder; and f(3) equals that remainder of 69.1679

Always be careful to include 0's for coefficients for missing terms.1695

Show that x - 3 is a factor of this polynomial, and find the other factors.1703

If x - 3 is a factor, then when you divide f(x) by x - 3, the remainder will be 0.1711

So, what this is saying is that f(3) equals 0.1726

OK, I am looking, and I do not have any missing terms: cubed, squared, just x to the first, and then a constant.1734

So, I can go ahead with my division.1741

And this is a coefficient of 1; a coefficient of -3; -4; and 12.1745

OK, I am bringing down the 1 to give me 1 times 3 is 3; 3 minus 3 is 0; 0 times 3 is 0; -4 and 0 is -4; -4 times 3 is -12; -12 and 12 is 0.1754

OK, so this is giving me a remainder of 0, so x - 3 is a factor of f(x).1778

I am going to call this f(x).1796

So, I was first asked to show that x - 3 is a factor of this polynomial; and it is, because the remainder is 0.1801

Now, I am supposed to find the other factors.1807

Well, I pulled out this factor, x - 3; so that factor, times what is left over, will give me the polynomial back.1809

What I have here is x3: that means that, after dividing, I am going to be left with1818

1x2 (so x2) + 0x (so I can just leave that out) - 4.1824

All right, so what I have here, x2 - 4, is the difference of two squares: it is in the form a2 - b2.1833

So, I can factor that out to (a + b) (a - b).1841

This gives me (x - 2) (x + 2); this is the complete factorization of this polynomial function.1850

And the factor they gave me is x - 3; the other factors are x - 2 and x + 2.1860

So, I have three factors for this polynomial.1874

That concludes this lesson on Educator.com; and I will see you next time!1877