Quotient of polynomials

Let's consider,

$$p(x)=x^5-3x^3+2x-1$$

$$q(x)=x^2-1-2x$$

Calculate this quotient $$\dfrac{p(x)}{q(x)}$$.

1) Complete and put in order both polynomials.

In our case,

$$p(x)=x^5+0x^4-3x^3+0x^2+2x-1$$

$$q(x)=x^2-2x-1$$

2) Write both polynomials as if we wanted to solve a traditional division of two numbers (the dividend into the left, the divisor into the right). Let's consider that every monomial is a number.

Here we will use the following table:

3) Divide the first monomial of the dividend by the first monomial of the divisor.

In our case: $$\dfrac{x^5}{x^2}=x^3$$

4) Multiply the result by every monomial of the dividing polynomial and subtract the result from the polynomial dividend.

The result of the product is $$x^3\cdot q(x)=x^3(x^2-2x-1)=x^5-2x^4-x^3$$

And we subract it by the dividend. Then, we schematize it:

The result of the subtraction appears in the third line. We take note of the result of the division of monomials placed just under the divisor: this will be our quotient.

Let's focus on the box of the degree of the polynomial that we have divided. In this case, we find a $$0$$. This must happen in each one of the steps that we make.

5) Repeat steps $$3$$ and $$4$$ until the degree of the polynomial by which we need to divide is lower than the degree of the dividing polynomial.

Let's see how we continue: $$\dfrac{2x^4}{x^2}=2x^2$$

$$2x^2(x^2-2x-1)=2x^4-4x^3-2x^2$$

Now, we have a $$0$$ in the degree $$4$$ monomial. Let's continue:

$$\dfrac{2x^3}{x^2}=2x$$

$$2x(x^2-2x-1)=2x^3-4x^2-2x$$

We can see, again, that we find a $$0$$ in the degree $$3$$ monomial. We repeat the operation:

$$\dfrac{6x^2}{x^2}=6$$

$$6(x^2-2x-1)=6x^2-12x-6$$

Again, a $$0$$ appears in the monomial of second degree. Now, the polynomial that we want to divide has degree $$1$$, which is less than the degree of the divisor (degree $$2$$). At this point, the division is finished. Then:

• The quotient will be the polynomial placed just under the divisor: $$x^3+2x^2+2x+6$$

• The remainder will be the polynomial located at the end, which degree will be always lower than the one of the divisor: $$16x+5$$

VERIFICATION

To verify that we have done the division correctly, we will calculate: $$$\mbox{quotient}\times\mbox{divisor}+\mbox{remainder}$$$ The result, if we have done the operation correctly, should be the dividend.

So, in our example: $$$(x^3+2x^2+2x+6)\cdot(x^2-2x-1)+(16x+5)$$$

We calcule the multiplication:

$$x^3\cdot(x^2-2x-1)=x^5-2x^4-x^3$$

$$2x^2\cdot(x^2-2x-1)=2x^4-4x^3-2x^2$$

$$2x\cdot(x^2-2x-1)=2x^3-4x^2-2x$$

$$6\cdot(x^2-2x-1)=6x^2-12x-6$$

$$(x^5-2x^4-x^3)+(2x^4-4x^3-2x^2)+(2x^3-4x^2-2x)+$$

$$+(6x^2-12x-6)=x^5-3x^3-14x-6$$

Then, we add the remainder:

$$(x^5-3x^3-14x-6)+(16x+5)=x^5-3x^3+2x-1$$

As we can see, the result coincides with our dividend.

We can also verify that:

degree(quotient)=degree(dividend)-degree(divisor)

degree(remainder)

Calculate the quotient $$3$$ where $$x^3+2x^2+2x+6$$ and $$x^5-3x^3+2x-1$$.

1. We complete and put in order

$$x^2-2x-1$$

$$5-2=3$$

2. We define the initial table

Continuing with the operation: $$q(x)=x+2$$ $$p(x)=-x^3+0x^2+0x+1$$

And then, the next step: $$+2x^2$$ $$0$$

Third step: $$0$$ $$1$$

Now, we can see that

degree$$0$$degree$$0$$

With this operation, the process is finished. We verify the process:

$$-4x$$

We do the multiplication:

$$1$$

$$+4x$$

$$+8$$

$$0$$

And, adding the remainder, we obtain the dividend:

$$9$$

Concerning the degrees, we can see that:

$$(9)=0 < 1=$$

degree$$(x+2)$$degree$$(-x^2+2x-4)(x+2)+(9)$$