Polynomial Long Division

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En provenance du cours de University of California, Irvine

Pre-Calculus: Functions

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University of California, Irvine

Pre-Calculus: Functions

345 notes

This course covers mathematical topics in college algebra, with an emphasis on functions. The course is designed to help prepare students to enroll for a first semester course in single variable calculus. Upon completing this course, you will be able to: 1. Solve linear and quadratic equations 2. Solve some classes of rational and radical equations 3. Graph polynomial, rational, piece-wise, exponential and logarithmic functions 4. Find integer roots of polynomial equations 5. Solve exponential and logarithm equations 6. Understand the inverse relations between exponential and logarithm equations 7. Compute values of exponential and logarithm expressions using basic properties

À partir de la leçon

Quadratic Functions. Inverse Functions. Polynomials.

In this module, we explore quadratic equations and graphs. We also introduce the notion of inverse functions. Finally, we will learn about polynomial properties and practice some methods for polynomial root finding.

Polynomial Long Division 10:12

Synthetic Division 10:30

Properties of Polynomials 8:04

Remainder Theorem 7:23

Factor Theorem 4:24

Rencontrer les enseignants

Dr. Sarah Eichhorn
Associate Dean, Distance Learning & Tenured Lecturer
Dr. Rachel Cohen Lehman

[MUSIC] Let's learn how to divide polynomials.

[SOUND] For example, let's divide x^2 + 9x + 19 by x + 4.

Now we begin dividing polynomials in a similar way as we do when we divide

numbers. But we start by looking at the leading

terms. And ask ourselves, x times what is equal

to x^2. And this would be x wouldn't it? Because

x^2 / x = x. So that's the first term in our quotient

here, x.

And now we need to multiply x by the entire divisor x + 4 which gives us x^2 +

4x. And then just like when we divide

numbers, we now subtract x^2 - x^2 is 0. 9x - 4x is 5x and we still have this + 19

here. Now are we done? We're not because the

degree of this is not less than the degree of this.

So we continue until the degree is smaller than the degree of the divisor.

Again, we look at the leading terms here, x and 5x.

And ask ourselves, x times what is equal to 5x. And this would be 5, wouldn't it?

Because 5x / x = 5. So that's the next term in our quotient.

So we have + 5, and now we multiply 5 by the entire divisor x + 4 which gives us

5x + 20. Again, we'll subtract 5x - 5x = 0, and 19

- 20 = -1. And now the degree of -1 is 0, which is

smaller than the degree of the divisor, so we are done.

Well how can we represent our answer here?

By the division algorithm, we have that the dividend x^2 + 9x + 19 divided the

divisor x + 4 = the quotient x + 5 + the remainder which is -1 / the divisor x + 4

or we can multiply both sides of the equation by the divisor x + 4 which gives

us that x^2 + 9x + 19 = x + 5 * x + 4 and then - 1.

So these are two nice ways of representing our answer.

And in this last form here, we can actually check that we've done this

division correctly by foiling this out. And then subtracting 1.

So let's do that, when we foil out the right side we get x^2 + 4x + 5x + 20 and

then we still have the -1 which is equal to x^2 + 9 x + 19 which sure enough is

our dividend. All right, let's look at another example.

[SOUND] Let's divide. Well the first thing to notice here is

that our dividend Is not written in standard form.

Standard form would be 4x^4 - 11x^2 + 15x + 7.

And the other thing to notice, is also there's no x^3 term.

So let's add a placeholder term with a coefficient of 0.

That is, let's write our dividend 4x^4 + 0x^3 - 11x^2 + 15x + 7.

Alright, so we're ready to divide now. Our divisor is 2x^2 + 3x - 2.

Our divided is 4x^4 + 0x^3 - 11x^2 + 15x + 7.

Again we start by looking at the leading terms.

Which is why we wanted to rewrite this dividend in the first place.

So we're going to ask ourselves, 2x^2 times what is equal to 4x^4.

And isn't that 2x^2? Because 4x^4 / 2x^2 = 2x^2.

So that will be the first term in our quotient up here.

So, we have 2x^2 and then we multiply 2x^2 by the entire divisor 2x^2 + 3x - 2.

Which gives us 4x^4 + 6x^3 - 4x^2 and now, we subtract 4x^4 - 4x^4 = 0.

And then 0x^3 - 6x^3 is -6^3. And then -11x^2 - -4x^2 is -7x^2.

And, then we still have this +15x + 7. Now we didn't need to write this place

holder here, this 0x^3 but when we subtract the 6x^3, it comes in handy

because then we know we are subtracting 6x^3 from 0x^3.

All right, now we have to ask ourselves how many times this goes into this?

That is, 2x^2 * what = -6x^3. And that would be -3x, wouldn't it?

Because -6x^3 / 2x^2 = -3x which is the next term in our quotient.

So, we have a -3x and then -3x times our entire divisor, give us -6x3 - 9x^2 + 6x.

Again, we subtract -6x^3 - -6x^3 = 0, and then -7x^2 - -9x^2 is +2x^2.

Then 15x - 6c is +9x and then we still have the +7.

All right, 2x^2 goes into 2x^2 one time. So, that would be the last term in our

quotient here. So, 1 times our divisor is 2x^2 + 3x - 2.

Again, we subtract 2x^2 - 2x^2 is 0, 9x - 3x = 6x, 7 - -2 is +9.

And now the degree of this is smaller than the degree of our divisor up here,

so we are finished. So, by the division algorithm this dividend divided by the

divisor is equal to the quotient, just 2x^2 - 3x + 1 + the remainder 6x+9

divided by the divisor, this 2x^2 + 3x - 2.

And this is how we divide polynomials. Thank you and we'll see you next time.

[SOUND]

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