Professor P. Bishop
MAC1147 (MDN)
R.6 Synthetic Division (Condensed)
Synthetic division is used to divide a polynomial of degree 1 or higher by a binomial g(x) of the form
x – c.
Use synthetic division to determine whether x – c is a factor of f(x):
1. f  x   x 4  x 3  8x 2  15x  4; g  x   x  4
2. f  x   x 2  3x  5; g  x   x  4
5.5 The Real Zeros of a Polynomial Function
Remainder Theorem: If f(x) is divided by x-c, then the remainder is f(x). Factor theorem: If f is a polynomial
function, then x – c is a factor of f(x) if any only if f(c) = 0. Number of Real Zeros Theorem: A polynomial function
cannot have more real zeros than its degree. Descartes’ Rule of Signs: If f is a polynomial function in standard form,
then (a) the number of positive real zeros of f either equals the number of variations in the sign of the nonzero coefficients
of f(x) or else equals that number less an even integer; (b) ) the number of negative real zeros of f either equals the
number of variations in the sign of the nonzero coefficients of f(-x) or else equals that number less an even integer.
Rational Zero Theorem: Given f is a polynomial function of degree 1 or higher in descending order and each coefficient
is an integer, if p/q, in lowest terms, is a rational zero of f, then q is a factor of the lead coefficient ( ≠0) and p is a factor is
a factor of last coefficient (≠0). Bounds on Zeros: Let f be a polynomial function whose leading coefficient is 1: f(x) =
xn + an-1xn-1 + … + a1x + a0. The bound M on the zeros of f is the smaller of the sum of the absolute values of a0 through
an-1 or 1 + the coefficient with the largest absolute value (see pg. 367 for mathematical symbols). All the zeros of the
polynomial function will fall between ±M. Intermediate Value Theorem: If f is a polynomial function and a < b and
f(a) and f(b) have opposite signs, then there is a zero between a and b.
Determine if g(x) is a factor of f(x) = x4 – 5x3 - 3x2 + 20x - 25 by finding f(c):
1. g(x) = x + 3
2. g(x) = x – 5
Find the real zeros of each polynomial and factor over the real numbers:
3. f(x) = 2x6 + 3x5 – 3x4 + 10x3 - 24x2 + 3x + 9
Max. Zeros:
Positive Zeros:
Negative Zeros:
Potential Zeros:
Real Zeros:
4. f(x) = 4x5 + 12x4 - x - 3
Max. Zeros:
Positive Zeros:
Potential Zeros:
Real Zeros:
Negative Zeros:
Solve the equation in the real number system:
3
5. x 3  x 2  3 x  2  0
2
Max. Zeros:
Positive Zeros:
Negative Zeros:
Potential Zeros:
Real Zeros:
Find a possible graph for the following:
6. f(x) = 4x5 + 12x4 - x - 3
7.
f(x) = 2x6 + 3x5 – 3x4 + 10x3 - 24x2 + 3x + 9
Find the bound on the real zeros of the polynomial function:
8. f(x) = x4 – 5x2 – 36
9.
f(x) = 3x4 – 3x3 - 5x2 + 27x – 36
Use the Intermediate Value Theorem to show that the polynomial has a zero in the given interval:
10. f(x) = 2x4 + x3 – 24x2 + 20x + 16; [0, -1)
Given that the equation has a solution r in the given interval, approximate the solution correct to two decimal places:
11. x4 - 4x3 - x2 - 2x – 3 = 0
[4, 5]
The following equation has exactly one positive zero. Approximate its value to two decimal places:
12. x5 - 2x3 - x2 - 2x – 7 = 0
Professor P. Bishop
MAC1147 (MDN)
Try these (R.6, 5.5)
R.6
1. f(x) = x5 + 1; g(x) = x + 1
2.
Determine if x + 1/3 is a factor 3x4 + x3 – 3x + 1
3.
Find the sum of a, b, c, and d if:
x3  2 x2  3 x  5
d
 ax 2  bx  c 
x2
x2
4.6
4. Solve: 2x4 + x3 – 24x2 + 20x + 16 = 0
Max. Zeros:
Potential Zeros:
Real Zeros:
5.
Positive Zeros:
Graph: f(x) = x4 – x3 – 6x2 + 4x + 8
Negative Zeros: