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Polynomial Functions
Digital Lesson
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A polynomial function is a function of the form1
1 1 0( ) , 0n nn n nf x a x a x a x a a
where n is a nonnegative integer and each ai (i = 0, , n)
is a real number. The polynomial function has a leading coefficient an and degree n.
Examples: Find the leading coefficient and degree of each polynomial function.
Polynomial Function Leading Coefficient Degree5 3( ) 2 3 5 1f x x x x
3 2( ) 6 7f x x x x
( ) 14f x
– 2 5
1 3
14 0
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A real number a is a zero of a function y = f (x)if and only if f (a) = 0.
A polynomial function of degree n has at most n zeros.
Real Zeros of Polynomial Functions
If y = f (x) is a polynomial function and a is a real number then the following statements are equivalent.
1. x = a is a zero of f.
2. x = a is a solution of the polynomial equation f (x) = 0.
3. (x – a) is a factor of the polynomial f (x).
4. (a, 0) is an x-intercept of the graph of y = f (x).
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y
x–2
2
Example: Find all the real zeros of f (x) = x 4 – x3 – 2x2.
Factor completely: f (x) = x 4 – x3 – 2x2 = x2(x + 1)(x – 2).
The real zeros are x = –1, x = 0, and x = 2.
These correspond to the x-intercepts.
f (x) = x4 – x3 – 2x2
(–1, 0) (0, 0)
(2, 0)
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+ 2 2 3 1 2 xxx
Example: Divide x2 + 3x – 2 by x – 1 and check the answer.
x
x2 + x2x – 22x + 2
– 4
remainder
Check:
xx
xxx
22 1.
xxxx 2)1(2.
xxxxx 2)()3( 22 3.
22
2 x
xxx4.
22)1(2 xx5.
4)22()22( xx6.
(x + 2)
quotient
(x + 1)
divisor
+ (– 4)
remainder
= x2 + 3x – 2
dividend
Answer: x + 2 +1x
– 4
Dividing Polynomials
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16
Synthetic division is a shorter method of dividing polynomials.
This method can be used only when the divisor is of the form
x – a. It uses the coefficients of each term in the dividend.
Example: Divide 3x2 + 2x – 1 by x – 2 using synthetic division.
3 2 – 12
Since the divisor is x – 2, a = 2.
3
1. Bring down 3
2. (2 • 3) = 6
6
8 15
3. (2 + 6) = 8
4. (2 • 8) = 16
5. (–1 + 16) = 15coefficients of quotient remainder
value of a coefficients of the dividend
3x + 8Answer: 2
x15
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Remainder Theorem: The remainder of the division of a
polynomial f (x) by x – k is f (k).
Example: Using the remainder theorem, evaluate
f(x) = x 4 – 4x – 1 when x = 3.
9
1 0 0 – 4 – 13
1
3
3 9
6927
23 68
The remainder is 68 at x = 3, so f (3) = 68.
You can check this using substitution: f(3) = (3)4 – 4(3) – 1 = 68.
value of x
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Factor Theorem: A polynomial f(x) has a factor (x – k) if and
only if f(k) = 0.
Example: Show that (x + 2) and (x – 1) are factors of
f(x) = 2x 3 + x2 – 5x + 2.
6
2 1 – 5 2– 2
2
– 4
– 3 1
– 2
0
The remainders of 0 indicate that (x + 2) and (x – 1) are factors.
– 1
2 – 3 11
2
2
– 1 0
The complete factorization of f is (x + 2)(x – 1)(2x – 1).
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Rational Zero Test: If a polynomial f(x) has integer coefficients, every rational zero of f has the form
where p and q have no common factors other than 1.
rational zero = pq
Example: Find the rational zeros of f(x) = x3 + 3x2 – x – 3.
The possible rational zeros are ±1 and ±3.Synthetic division shows that the factors of f are (x + 3), (x + 1), and (x – 1).
• p is a factor of the constant term. • q is a factor of the leading coefficient.
q = 1 p = – 3
The zeros of f are – 3, – 1, and 1.
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A variation in sign means that two consecutive, nonzero
coefficients have opposite signs.
Descartes’s Rule of Signs: If f(x) is a polynomial with real
coefficients and a nonzero constant term,
1. The number of positive real zeros of f is equal to the number
of variations in sign of f(x) or less than that number by an
even integer.
2. The number of negative real zeros of f is equal to the
number of variations in sign of f(–x) or less than that number
by an even integer.
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The polynomial has three variations in sign.
Example: Use Descartes’s Rule of Signs to determine the
possible number of positive and negative real zeros of
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45.
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45
+ to –
– to +
+ to –
f(– x) = 2(– x)4 – 17(– x)3 + 35(– x)2 + 9(– x) – 45 =2x4 + 17x3 + 35x2 – 9x – 45
f(x) has either three positive real zeros or one positive real zero.
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45 = (x + 1)(2x – 3)(x – 3)(x – 5).
f(x) has one negative real zero.One change in sign
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Graphing Utility: Find the zeros of f(x) = 2x3 + x2 – 5x + 2.
Calc Menu:
The zeros of f(x) are x = – 2, x = 0.5, and x = 1.
– 10 10
10
– 10
