- Inicio
- Polynomials
- Division of polynomials
Division of polynomials
Now we will explain a method to divide polynomials of one variable. We will use an example to illustrate the procedure:
Let's consider,
$$p(x)=x^5-3x^3+2x-1$$
$$q(x)=x^2-1-2x$$
Calculate this quotient $\dfrac{p(x)}{q(x)}$.
- Complete and put in order both polynomials.
In our case,
$$p(x)=x^5+0x^4-3x^3+0x^2+2x-1$$
$$q(x)=x^2-2x-1$$
- 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:
| $x^5$ | $0$ | $-3x^3$ | $0$ | $2x$ | $-1$ | $x^2-2x-1$ |
- Divide the first monomial of the dividend by the first monomial of the divisor.
In our case: $\dfrac{x^5}{x^2}=x^3$
- 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:
| $x^5$ | $0$ | $-3x^3$ | $0$ | $2x$ | $-1$ | $x^2-2x-1$ |
| $-x^5$ | $+2x^4$ | $+x^3$ | $0$ | $0$ | $0$ | $x^3$ |
| $0$ | $+2x^4$ | $-2x^3$ | $0$ | $2x$ | $-1$ |
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.
- 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: $\dfrac{2x^4}{x^2}=2x^2$
$$2x^2(x^2-2x-1)=2x^4-4x^3-2x^2$$
| $x^5$ | $0$ | $-3x^3$ | $0$ | $2x$ | $-1$ | $x^2-2x-1$ |
| $-x^5$ | $+2x^4$ | $+x^3$ | $0$ | $0$ | $0$ | $x^3+2x^2$ |
| $0$ | $+2x^4$ | $-2x^3$ | $0$ | $2x$ | $-1$ | |
| $-2x^4$ | $4x^3$ | $2x^2$ | $0$ | $0$ | ||
| $0$ | $2x^3$ | $2x^2$ | $2x$ | $-1$ |
Now, we have a $0$ in the degree $4$ monomial. Let's continue:
$$\dfrac{2x^3}{x^2}=2x$$
$$2x(x^2-2x-1)=2x^3-4x^2-2x$$
| $x^5$ | $0$ | $-3x^3$ | $0$ | $2x$ | $-1$ | $x^2-2x-1$ |
| $-x^5$ | $+2x^4$ | $+x^3$ | $0$ | $0$ | $0$ | $x^3+2x^2+2x$ |
| $0$ | $+2x^4$ | $-2x^3$ | $0$ | $2x$ | $-1$ | |
| $-2x^4$ | $4x^3$ | $2x^2$ | $0$ | $0$ | ||
| $0$ | $2x^3$ | $2x^2$ | $2x$ | $-1$ | ||
| $-2x^3$ | $+4x^2$ | $+2x$ | $0$ | |||
| $0$ | $6x^2$ | $4x$ | $-1$ |
We can see, again, that we find a $0$ in the degree $3$ monomial. We repeat the operation:
$$\dfrac{6x^2}{x^2}=6$$
$$6(x^2-2x-1)=6x^2-12x-6$$
| $x^5$ | $0$ | $-3x^3$ | $0$ | $2x$ | $-1$ | $x^2-2x-1$ |
| $-x^5$ | $+2x^4$ | $+x^3$ | $0$ | $0$ | $0$ | $x^3+2x^2+2x+6$ |
| $0$ | $+2x^4$ | $-2x^3$ | $0$ | $2x$ | $-1$ | |
| $-2x^4$ | $+4x^3$ | $+2x^2$ | $0$ | $0$ | ||
| $0$ | $2x^3$ | $2x^2$ | $2x$ | $-1$ | ||
| $-2x^3$ | $+4x^2$ | $+2x$ | $0$ | |||
| $0$ | $6x^2$ | $4x$ | $-1$ | |||
| $-6x^2$ | $+12x$ | $+6$ | ||||
| $0$ | $16x$ | $+5$ |
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) < degree(divisor)
So, in the example
$3$=degree($x^3+2x^2+2x+6$)=degree($x^5-3x^3+2x-1$)-degree($x^2-2x-1$)=$5-2=3$
degree($16x+5$)=$1 < 2$=degree($x^2-2x-1$)
Calculate the quotient $\dfrac{p(x)}{q(x)}$ where $p(x)=1-x^3$ and $q(x)=x+2$.
- We complete and put in order
$$p(x)=-x^3+0x^2+0x+1$$
$$q(x)=x+2$$
- We define the initial table
| $-x^3$ | $0$ | $0$ | $1$ | $x+2$ |
Continuing with the operation: $$\dfrac{-x^3}{x}=-x^2$$ $$-x^2(x+2)=-x^3-2x^2$$
| $-x^3$ | $0$ | $0$ | $1$ | $x+2$ |
| $+x^3$ | $+2x^2$ | $0$ | $0$ | $-x^2$ |
| $0$ | $+2x^2$ | $0$ | $1$ |
And then, the next step: $$\dfrac{2x^2}{x}=2x$$ $$2x(x+2)=2x^2+4x$$
| $-x^3$ | $0$ | $0$ | $1$ | $x+2$ |
| $+x^3$ | $+2x^2$ | $0$ | $0$ | $-x^2+2x$ |
| $0$ | $+2x^2$ | $0$ | $1$ | |
| $-2x^2$ | $-4x$ | $0$ | ||
| $0$ | $-4x$ | $1$ |
Third step: $$\dfrac{-4x}{x}=-4$$ $$-4(x+2)=-4x-8$$
| $-x^3$ | $0$ | $0$ | $1$ | $x+2$ |
| $+x^3$ | $+2x^2$ | $0$ | $0$ | $-x^2+2x-4$ |
| $0$ | $+2x^2$ | $0$ | $1$ | |
| $-2x^2$ | $-4x$ | $0$ | ||
| $0$ | $-4x$ | $1$ | ||
| $+4x$ | $+8$ | |||
| $0$ | $9$ |
Now, we can see that
degree$(9)=0 < 1=$degree$(x+2)$
With this operation, the process is finished. We verify the process:
$$(-x^2+2x-4)(x+2)+(9)$$
We do the multiplication:
$$-x^2\cdot(x+2)=-x^3-2x^2$$
$$+2x\cdot(x+2)=2x^2+4x$$
$$-4\cdot(x+2)=-4x-8$$
$$(-x^3-2x^2)+(2x^2+4x)+(-4x-8)=-x^3-8$$
And, adding the remainder, we obtain the dividend:
$$(-x^3-8)+9=-x^3+1$$
Concerning the degrees, we can see that:
$$2=\mbox{degree}(-x^2+2x-4)=\mbox{degree}(x^3+1)-\mbox{degree}(x+2)=3-1=2$$
degree$(9)=0 < 1=$degree$(16x+5)$