The sum of the coefficients is the total obtained by adding every numerical coefficient in a polynomial or algebraic expression. The quickest method is usually to replace each variable with 1 and simplify the resulting expression. To give you an idea, the coefficients of (4x^3-2x^2+5x-7) add to (4+(-2)+5+(-7)=0), while evaluating the polynomial at (x=1) produces the same result.
Introduction
When someone asks, “what is the sum of the coefficients?”, they are usually referring to a polynomial such as
[ p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+ax+b. ]
The answer depends on which polynomial is being discussed, but the general rule is remarkably simple: substitute (x=1). Since every power of 1 equals 1, each term becomes equal to its coefficient Less friction, more output..
This technique applies not only to ordinary polynomials but also to expanded binomials, expressions with several variables, and certain algebraic equations. Understanding why it works helps prevent common mistakes involving negative coefficients, missing powers, constants, and expressions that have not yet been expanded.
What Are Coefficients?
A coefficient is the numerical factor attached to a variable term. In the polynomial
[ 6x^4-3x^2+8x-10, ]
the coefficients are:
- (6) for (6x^4)
- (-3) for (-3x^2)
- (8) for (8x)
- (-10) for the constant term (-10)
The constant term is also included in the sum because it can be viewed as a coefficient multiplied by (x^0). Since (x^0=1) whenever (x\neq0), we can write
[ -10=-10x^0. ]
Thus, its coefficient is (-10) That's the part that actually makes a difference..
A missing power does not mean that a coefficient is absent or that zero should be ignored. Take this case:
[ x^5-4x+2 ]
has coefficients (1), (0) for (x^4), (0) for (x^3), (0) for (x^2), (-4), and (2). Their sum is still
[ 1+0+0+0+(-4)+2=-1. ]
Adding zero does not change the result, so it is usually sufficient to add the coefficients that are explicitly written.
The Core Rule: Substitute 1 for the Variable
Consider a polynomial in one variable:
[ p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0. ]
Now replace (x) with (1):
[ p(1)=a_n(1)^n+a_{n-1}(1)^{n-1}+\cdots+a_2(1)^2+a_1(1)+a_0. ]
Because (1^k=1) for every nonnegative integer (k), this becomes
[ p(1)=a_n+a_{n-1}+\cdots+a_2+a_1+a_0. ]
That expression is exactly the sum of all coefficients. So,
[ \boxed{\text{Sum of coefficients}=p(1)}. ]
This rule is efficient because it avoids writing out every coefficient separately, especially when the polynomial has many terms or a high degree And that's really what it comes down to..
Step-by-Step Method
To find the sum of the coefficients of a polynomial, follow these steps:
- Identify the polynomial expression. Make sure the expression has been simplified.
- Replace every variable with 1.
- Simplify powers of 1. Every power equals 1.
- Perform the arithmetic. Include positive and negative signs correctly.
- Check the result against the coefficients, if the polynomial is short enough to list them.
To give you an idea,
take
[ p(x)=4x^5-7x^3+2x-9. ]
Instead of listing every coefficient, evaluate the polynomial at (x=1):
[ p(1)=4(1)^5-7(1)^3+2(1)-9. ]
Since each power of (1) is (1),
[ p(1)=4-7+2-9=-10. ]
So the sum of the coefficients is
[ \boxed{-10}. ]
Using the Rule with Factored Expressions
The polynomial does not need to be expanded first. Here's one way to look at it: suppose you want the sum of the coefficients of
[ (2x-3)^4. ]
Substitute (x=1):
[ (2(1)-3)^4=(2-3)^4=(-1)^4=1. ]
So, the sum of the coefficients is
[ \boxed{1}. ]
This is often much faster than expanding the expression.
Products of Polynomials
The same idea works for products. If
[ q(x)=p(x)r(x), ]
then
[ q(1)=p(1)r(1). ]
So the sum of the coefficients of a product is the product of the sums of the coefficients of the factors.
As an example, find the sum of the coefficients of
[ (x^2+3x-2)(4x-5). ]
Evaluate each factor at (x=1):
[ 1^2+3(1)-2=2 ]
and
[ 4(1)-5=-1. ]
So,
[ 2(-1)=-2. ]
So the sum of the coefficients is
[ \boxed{-2}. ]
Expressions with More Than One Variable
For polynomials with several variables, substitute (1) for every variable.
Here's one way to look at it: consider
[ 3x^2y-4xy^2+7x-2y+9. ]
Substitute (x=1) and (y=1):
[ 3(1)^2(1)-4(1)(1)^2+7(1)-2(1)+9. ]
This simplifies to
[ 3-4+7-2+9=13. ]
So the sum of all coefficients is
[ \boxed{13}. ]
In general, for a polynomial (P(x,y)), the sum of its coefficients is
[ \boxed{P(1,1)}. ]
For a polynomial in three variables, (P(x,y,z)), use
[ \boxed{P(1,1,1)}. ]
