Adding and Subtracting Polynomials
Definition and basic properties
Let $R$ be a commutative ring and $R [ x ]$ the ring of polynomials in one indeterminate over $R$. Consider two polynomials $P ( x )$ and $Q ( x )$, where any missing coefficients are understood to be zero:
\[P ( x ) = \sum_{k = 0}^{n} a_{k} x^{k} Q ( x ) = \sum_{k = 0}^{m} b_{k} x^{k}\]Addition is performed by summing the coefficients of terms with the same degree:
\[P ( x ) + Q ( x ) = \sum_{k = 0}^{max ( n , m )} ( a_{k} + b_{k} ) x^{k}\]The sum is the Cauchy addition of the coefficient sequences of $P$ and $Q$, which ensures that polynomial addition is well-defined and independent of the particular representation chosen for each polynomial. The result is again a polynomial, and $R [ x ]$ forms an abelian group under addition.
An abelian group is a set equipped with a binary operation that is commutative, associative, admits a neutral element, and for which every element has an inverse; $( R [ x ] , + )$ satisfies all four conditions, with the zero polynomial as neutral element and $- P ( x )$ as the inverse of $P ( x )$.
Subtraction is defined analogously, with each $b_{k}$ replaced by $- b_{k}$.
\[P ( x ) - Q ( x ) = \sum_{k = 0}^{max ( n , m )} ( a_{k} - b_{k} ) x^{k}\]How these operations affect the degree depends on whether the leading terms cancel. Denoting $deg P = n$ and $deg Q = m$, one has:
\[deg ( P ( x ) \pm Q ( x ) ) \leq max \{ n , m \}\]If $n \neq m$, the leading term of the polynomial with higher degree remains unchanged, so the degree of the result is exactly $max n , m$. When $n = m$, the degree may decrease: if $a_{n} + b_{n} = 0$ under addition, or $a_{n} = b_{n}$ under subtraction, the leading term cancels and the degree falls. If every term cancels, the result is the zero polynomial, assigned degree $- \infty$ by convention, so that the rule $deg ( P Q ) = deg P + deg Q$ holds without treating zero as a special case.
That multiplicative rule extends to $deg ( P^{k} ) = k deg P$ for any integer $k \geq 1$, provided $R$ is an integral domain. Over a general commutative ring this need not hold, as zero divisors can cause the leading coefficient of a product to vanish even when neither factor is zero.
Example 1
Take the following polynomials of degree 2: \(P ( x ) = x^{2} + 3 x - 1\) \(Q ( x ) = 2 x^{2} - x + 5\)
Their sum is:
\[P ( x ) + Q ( x ) & = ( x^{2} + 3 x - 1 ) + ( 2 x^{2} - x + 5 ) \\ & = ( 1 + 2 ) x^{2} + ( 3 - 1 ) x + ( - 1 + 5 ) \\ & = 3 x^{2} + 2 x + 4\]The leading coefficients sum to $3 \neq 0$, so the degree is unchanged and the result is a polynomial of degree 2.
Example 2
Consider the following polynomials: \(P ( x ) = 2 x^{2} + 3 x - 1\) \(Q ( x ) = 2 x^{2} - x + 5\)
Their difference is computed as follows:
\[P ( x ) - Q ( x ) & = ( 2 x^{2} + 3 x - 1 ) - ( 2 x^{2} - x + 5 ) \\ & = ( 2 - 2 ) , x^{2} + ( 3 + 1 ) , x + ( - 1 - 5 ) \\ & = 4 x - 6\]Since both polynomials share the same leading coefficient, the degree-2 term cancels, reducing the result to a polynomial of degree 1. This is a case where the following bound is strict:
\[deg ( P - Q ) \leq max \{ deg P , deg Q \}\]Example 3
Consider the following polynomials:
\(P ( x ) = x^{2} + 3 x - 1\) \(Q ( x ) = 2 x^{4} - x + 5\)
Their sum is computed as follows:
\[P ( x ) + Q ( x ) & = ( x^{2} + 3 x - 1 ) + ( 2 x^{4} - x + 5 ) \\ & = 2 x^{4} + x^{2} + ( 3 - 1 ) x + ( - 1 + 5 ) \\ & = 2 x^{4} + x^{2} + 2 x + 4\]Since the two polynomials have different degrees, the leading term of $Q ( x )$ has no counterpart in $P ( x )$ to cancel against. The result is a polynomial of degree 4, consistent with $deg ( P + Q ) = max { 2 , 4 } = 4$.