A polynomial is a function described by an expression of the form \(f(x)=a_nx^n + \cdots + a_1x+a_0\text{,}\) where \(a_0, a_1,\ldots,a_n\) are real numbers and \(n\geq0\) is an integer.
The expressions \(a_n x^n, a_{n-1}x^{n-1}, \ldots, a_1x, a_0\) are called the terms of the polynomial \(f\text{.}\) For each integer \(k = 0, 1, \ldots, n\text{,}\) the number \(a_k\) is called the coefficient of \(x^k\) in the polynomial \(f\text{.}\) The number \(a_0\) is called the constant term of \(f\text{.}\) Notice that \(a_0 = a_0 x^0\text{,}\) so that \(a_0\) is the coefficient of \(x^0\text{.}\)
Let \(f(x) =a_nx^n + \cdots + a_1x+a_0\) be a polynomial in which at least one of the coefficients is not zero. Then, the degree of \(f\) is the largest integer \(k\) such that the coefficient of \(x^k\) is not zero. (Notice that such an integer \(k\) is guaranteed to exist.) In this case, \(a_kx^k\) is the leading term of \(f\text{,}\) and \(a_k\) is the leading coefficient of \(f\text{.}\) Since \(a_0 = a_0 x^0\text{,}\) it is possible for a polynomial to have degree zero.
Example2.5.1.Understanding a Polynomial.
Let \(f(x)=2x^3-10x+9\text{.}\) Then,
the terms of \(f\) are \(2x^3\text{,}\)\(0x^2\text{,}\)\(-10x\text{,}\) and \(9\text{;}\)
the coefficients of \(f\) are the numbers \(2, 0, -10\text{,}\) and \(9\text{;}\)
the degree of \(f\) is 3, the leading term of \(f\) is \(2x^3\text{,}\) and the leading coefficient of \(f\) is 2;
the constant term of \(f\) is 9.
Example2.5.2.Understanding another Polynomial.
Let \(f(x) = 14\text{.}\) Then,
the only term of \(f\) is \(14 = 14x^0\text{;}\)
the only coefficient of \(f\) is \(14\text{;}\)
the degree of \(f\) is 0, the leading term of \(f\) is \(14 = 14x^0\text{,}\) and the leading coefficient of \(f\) is 14;
the constant term of \(f\) is 14.
The polynomial in which all coefficients are zero, that is \(f(x) = 0 + 0x + 0x^2 \cdots\text{,}\) is called the zero polynomial. It is the only polynomial for which the degree is undefined.
An important fact is that if \(f(x)\) is a polynomial, then \(f(a)=0\) if and only if \((x-a)\) divides \(f(x)\text{.}\) In other words, \(a\) is a root of the polynomial \(f(x)\) if and only if \((x-a)\) is a factor of \(f(x)\text{.}\) Hence to check whether \(x-a\) is a factor of \(f(x)\text{,}\) check whether \(f(a) = 0\text{.}\) Conversely, if \(f(a) = 0\) then \((x-a)\) is a factor of \(f(x)\text{.}\)
Example2.5.3.Testing for Roots of a Polynomial.
Let \(f(x)=2x^3+x^2+1\text{.}\) Then,
\(x-1\) is not a factor of \(f(x)\) because we can calculate \(f(1) = 4 \neq 0\) and so \(1\) is not a root of \(f(x)\text{.}\)
\(x+1 = x- (-1)\) is a factor of \(f(x)\) because \(f(-1) = 0\text{.}\) In fact, \(f(x) = (x+1)(2x^2 - x + 1)\text{.}\) We can check this by multiplication \((x+1)(2x^2 - x + 1) = 2x^3 - x^2 + x + 2x^2 - x + 1 = 2x^3+x^2+1\text{.}\) (The factors in this expression can be found by long division, as discussed below.)
We can use the quadratric formula to verify that \(2x^2-x+1\) has no roots, which means there are no more roots of \(f(x)\text{.}\)
A useful fact is that if a polynomial \(f(x)\) has leading coefficient equal to 1, then an integer \(a\) can be a root of \(f(x)\) only if \(a\) or \(-a\) is a divisor of the constant term \(a_0\text{.}\) It may be that all, some, or none of these are actually roots of \(f\text{.}\) Also, \(f\) may have roots that are not integers.
Example2.5.4.Finding the Roots of a Polynomial.
Let \(f(x)=x^3-x^2-2x+2\text{.}\) Since the leading coefficient of \(f\) is 1, the only possible integers which could possibly be roots of \(f\) are the divisors of 2 and their negatives, that is \(1, -1, 2\text{,}\) and \(-2\text{.}\) Of these possibilities, only \(1\) is a root (because \(f(1) = 0\) whereas \(f(-1), f(2)\) and \(f(-2)\) are all non-zero). In fact, \(f(x) = (x-1)(x^2 - 2)\)
Once we know a root of a polynomial, we can factor the polynomial using long division. Polynomials are divided just like numbers. We saw in Example 2.5.4 that 1 is a root of \(f(x)=x^3-x^2-2x+2\text{,}\) so we can divide \(f(x)\) by \(x-1\) (see Figure 2.5.5). Because the remainder is \(0\text{,}\) this means that \(x^3-x^2-2x+2=(x-1)(x^2-2)\text{.}\) We can factor further, because \((x^2-2) = (x-\sqrt{2})(x+\sqrt{2})\text{,}\) so:
Since a product of numbers equals zero if and only if one of them is zero, \(f(x)=0\) if and only if \(x-1=0\) or \(x-\sqrt{2}=0\) or \(x+\sqrt{2}=0\text{,}\) and so the roots of \(f(x)\) are \(-1\text{,}\)\(\sqrt{2}\) and \(-\sqrt{2}\text{.}\)
Polynomial long division of \(x^2-x^2-2x+2\) by \(x-1\text{.}\)
Figure2.5.5.
For any number \(x\) in its domain, and expression represents a number. If we remember that \(x\) just represents a number in the domain, then expressions involving a variable can be factored (using the commutative, associative and distributive laws).
Notice that, in Examples 2.5.6 and Example 2.5.7, factoring would have been much more difficult if the expression had been multiplied out. Sometimes factoring requires creatively grouping the terms.
State the terms, the coefficients, the degree, the leading term, the leading coefficient, and the constant term of the following functions.
\(\displaystyle f(x)=3x^2-10x\)
\(\displaystyle g(x)=10x^5+x^3+15\)
Answer1.
The terms of \(f\) are \(3x^2\text{,}\)\(-10x\text{,}\) and \(0\text{;}\) the coefficients of \(f\) are the numbers \(3, -10\text{,}\) and \(0\text{;}\) the degree of \(f\) is 2, the leading term of \(f\) is \(3x^2\text{,}\) and the leading coefficient of \(f\) is 3; the constant term of \(f\) is 0.
Answer2.
The terms of \(g\) are \(10x^5\text{,}\)\(0x^4\text{,}\)\(x^3\text{,}\)\(0x^2\text{,}\)\(0x\text{,}\) and \(15\text{;}\) the coefficients of \(f\) are the numbers \(10,0,1,0,0\text{,}\) and \(15\text{;}\) the degree of \(f\) is 5, the leading term of \(f\) is \(10x^5\text{,}\) and the leading coefficient of \(f\) is 10; the constant term of \(f\) is 15.
2.
Factor the following expressions:
\(\displaystyle 2x(x^2+1)^3-16(x^2+1)^5\)
\(\displaystyle 8x+8x^3+x^4+x^6\)
Answer1.
\(2(x^2+1)^3(x-8(x^2+1)^2)\)
Answer2.
\(x(1+x^2)(8+x^3) = x(1+x^2)(x+2)(x^2-2x+4)\)
3.
Find all possible integer roots of the function and of those possibilities, determine which are roots of the function.
\(f(x)=x^2+x-6\)
Answer.
Possible roots: \(1, -1,2,-2,3,-3,6,-6\text{.}\) Since \(f(2)=0\) and \(f(-3)=0\text{,}\)\(2\) and \(-3\) are roots of \(f\text{.}\)
