Polynomial Long Division
By Catalin David
Introduction
The general form of a monomial
f(x)=axn, where:
-a is the coefficient and can be part of the sets N, Z, Q, R, C
-x is the variable
-n is the degree and is part of N
Two monomials are equal if they have the same variable and the same degree.
Example: 3x2 and -5x2; ½x4 and 2√3x4
The sum of unequal monomials is called a polynomial. In this case, the monomials will be the terms of the polynomial. A polynomial formed of two terms
is called a binomial.
Example: p(x)=3x2-5; h(x)=5x-1
A polynomial formed of three terms is called a trinomial.
The general form of the polynomial with only one variable
p(x)=anxn+an-1xn-1+...+a1x1
+a0
where:
- an,an-1,an-2,...,a1,a0 are the coefficients of the polynomial. They can be natural numbers, integers, rational numbers, real numbers or complex numbers.
- an is the coefficient of the term of the largest degree(the leading coefficient)
- a0 is the coefficient of the term of the smallest degree(the constant)
- n is the degree of the polynomial
Example 1
p(x)=5x3-2x2+7x-1
- third-degree polynomial with the coefficients 5, -2, 7 and -1
- 5 is the leading coefficient
- -1 is the constant
- x is the variable
Example 2
h(x)=-2√3x4+½x-4
- fourth-degree polynomial with coefficients -2√3,½ and -4
- -2√3 is the leading coefficient
- -4 is the constant
- x is the variable
Dividing Polynomials
p(x) and q(x) are two polynomials:
p(x)=anxn+an-1xn-1+...+a1x1
+a0
q(x)=apxp+ap-1xp-1+...+a1x1
+a0
To find out the quotient and the remainder of dividing p(x) by q(x) we need to use the following algorithm:
- The degree of p(x) has to be either equal to or greater than the degree of q(x).
- We write the terms of each polynomial in descending order of the degrees. If p(x) has a missing term, it will be written with the coefficient 0.
- The leading term of p(x) is divided by the leading term of q(x) and the result is written under the line of the divisor (denominator).
- We multiply the result with all the terms of q(x) and the results are written with changed sign under the terms of p(x) with corresponding degrees.
- We add up the terms with the same degrees.
- Next to the results, we write the other terms of p(x).
- We divide the leading term of the new polynomial by the first term of q(x) and repeat steps 3-6.
- We repeat all these steps until the new polynomial will be of a smaller degree than the one of q(x). This will be the remainder of the division.
- The polynomial written under the line of the divisor will be quotient.
Example 1
Step 1 and 2) $p(x)=x^5-3x^4+2x^3+7x^2-3x+5 \\ q(x)=x^2-x+1$
Answer: p(x)=x5 - 3x4 + 2x3 + 7x2- 3x + 5 = (x2 - x + 1)(x3 - 2x2 - x + 8) + 6x - 3
Example 2
p(x)=x4+3x2+2x-8
q(x)=x2-3x
Answer: x4 + 3x2 + 2x - 8 = (x2 - 3x)(x2 + 3x + 12) + 38x - 8
Dividing by a first-grade polynomial
It can be done by using the algorithm mentioned before, or in a faster way by using Horner's rule.
If f(x)=anxn+an-1xn-1
+...+a1x+a0, the polynomial can be written f(x)=a0+x(a1+x(a2+...+x(an-1+anx)...))
q(x) is of the first degree ⇒ q(x)=mx+n
The quotient polynomial will be of the degree n-1.
Following Horner's rule, $x_0=-\frac{n}{m}$.
bn-1=an
bn-2=x0.bn-1+an-1
bn-3=x0.bn-2+an-2
...
b1=x0.b2+a2
b0=x0.b1+a1
r=x0.b0+a0
where bn-1xn-1+bn-2xn-2+...+b1x+b0 is the quotient.
The remainder will be a zero degree polynomial because the degree of the remainder must be smaller than the degree of the divisor.
Euclidean division ⇒
p(x)=q(x).c(x)+r ⇒ p(x)=(mx+n).c(x)+r if $x_0=-\frac{n}{m}$
We can observe that p(x0)=0.c(x0)+r ⇒ p(x0)=r
Example 3
p(x)=5x4-2x3+4x2-6x-7
q(x)=x-3
p(x)=-7+x(-6+x(4+x(-2+5x)))
x0=3
b3=5
b2=3.5-2=13
b1=3.13+4=43 ⇒ c(x)=5x3+13x2+43x+123; r=362
b0=3.43-6=123
r=3.123-7=362
5x4-2x3+4x2-6x-7=(x-3)(5x3+13x2+43x+123)+362
Example 4
p(x)=-2x5+3x4+x2-4x+1
q(x)=x+2
p(x)=-2x5+3x4+0x3+x2-4x+1
q(x)=x+2
x0=-2
p(x)=1+x(-4+x(1+x(0+x(3-2x))))
b4=-2 b1=(-2).(-14)+1=29
b3=(-2).(-2)+3=7 b0=(-2).29-4=-62
b2=(-2).7+0=-14 r=(-2).(-62)+1=125
⇒ c(x)=-2x4+7x3-14x2+29x-62; r=125
-2x5+3x4+x2-4x+1=(x+2)(-2x4+7x3-14x2+29x-62)+125
Example 5
p(x)=3x3-5x2+2x+3
q(x)=2x-1
$x_0=\frac{1}{2}$
p(x)=3+x(2+x(-5+3x))
b2=3
$b_1=\frac{1}{2}\cdot 3-5=-\frac{7}{2}$
$b_0=\frac{1}{2}\cdot \left(-\frac{7}{2}\right)+2=-\frac{7}{4}+2=\frac{1}{4}$
$r=\frac{1}{2}\cdot \frac{1}{4}+3=\frac{1}{8}+3=\frac{25}{8} \Rightarrow c(x)=3x^2-\frac{7}{2}x+\frac{1}{4}$
$\Rightarrow 3x^3-5x^2+2x+3=(2x-1)(3x^2--\frac{7}{2}x+\frac{1}{4})+\frac{25}{8}$
Conclusion
If we divide by a polynomial of a degree higher than one, to find out the quotient and the remainder we use steps 1-9.
If
we divide by a first-degree polynomial mx+n, to find out the quotient and the remainder we use Horner's rule where $x_0=-\frac{n}{m}$.
If we only have to find out the remainder of a division by a first-degree polynomial, we find out p(x0).
Example 6
p(x)=-4x4+3x3+5x2-x+2
q(x)=x-1
x0=1
r=p(1)=-4.1+3.1+5.1-1+2=5
r=5