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Fundamental Theorem of Algebra
If p(x) is a polynomial function of degree n with n > 0, then p(x) has at least one zero in the
complex number system.
Linear Factor Theorem
If p(x) is a polynomial function of degree n, where n > 0, then p(x) has exactly n linear factors
and p(x) = an(x – c1)(x – c2)…(x – cn) where an is a real number and ci represents a complex
number.
– ci may be a real number.
It is a real number if b = 0 in a + bi.
– Every polynomial can be factored into linear or irreducible quadratic factors.
A linear factor has a real zero.
An irreducible quadratic factor has no real zeros.
– Let k be the complex conjugate of c. If x – c is a factor, then x – k is also a factor.
Complex numbers occur in pairs.
Descartes’ Rule of Signs
Let p(x) be a polynomial function with a0 not 0.
– If a0 = 0, factor out xm and apply to the new polynomial.
1. The number of positive real zeros is either equal to the number of sign changes in p(x) or less
than by an even number.
2. The number of negative real zeros is either equal to the number of sign changes in p(-x) or
less than by an even number.
Examples
1. p(x) = 2x4 – 5x3 + 7x2 – 7x + 3
3. p(x) = x5 – 7x3 – 6x2
2. p(x) = 6x4 + 13x3 – 32x2 – 45x + 18
4. p(x) = 4x6 – 9x5 – 15x4 + 29x3 + 27x2 – 24x – 20
Rational Zero Test
If p(x) is a polynomial function of degree n, where n > 0, and ai are integers, then the rational
zeros of p(x) are of the form p/q where p is a factor of a0 and q is a factor of an.
– If an = 1, then the rational zeros are just the factors of a0.
Examples
1. p(x) = 2x4 – 5x3 + 7x2 – 7x + 3
3. p(x) = x5 – 7x3 – 6x2
2. p(x) = 6x4 + 13x3 – 32x2 – 45x + 18
4. p(x) = 4x6 – 9x5 – 15x4 + 29x3 + 27x2 – 24x – 20
Upper and Lower Bound Rules
Let p(x) be a polynomial with real coefficients and positive leading coefficient.
Let p(x) be divided by x – c using synthetic division.
1. If c > 0 and each number in the last row is positive or 0, then c is an upper bound for the
zeros of p(x).
2. If c < 0 and the numbers in the last row alternate between positive and negative (0 can be
considered either), then c is a lower bound for the zeros of f(x).
Examples
1. p(x) = 2x4 – 5x3 + 7x2 – 7x + 3 test x = -1 and 2
2. p(x) = x5 – 7x3 – 6x2
test x = -3 and 2
Finding Rational Zeros
1. Factor, if possible.
2. Use Descartes’ Rule of Signs to determine the number of positive and negative real zeros.
3. Use the Rational Zero Test to find the possible zeros.
– If there are no positive real zeros, only use the negative values.
– If there are no negative real zeros, only use the positive values.
4. Use synthetic division to test one of the possible zeros, call it c.
a. Use the Upper and Lower Bound Rules to determine if it is an upper or lower bound.
i. If it is an upper bound, exclude larger possible zeros.
ii. If it is a lower bound, exclude smaller possible zeros.
b. If x = c is a zero, then rewrite p(x) = (x – c)q(x).
i. Update the number of positive or negative real zeros.
ii. Update the possible rational zeros using Descartes Rule of Signs on q(x).
– If there are no positive real zeros left, only use the negative values.
– If there are no negative real zeros left, only use the positive values.
iii. Repeat testing possible zeros until p(x) has been factored into n – 2 linear factors and
a quadratic factor.
– Test c again, because the multiplicity may be more than 1.
c. Factor or use the quadratic formula to find the zeros of the quadratic factor.
Examples
1. p(x) = 2x4 – 5x3 + 7x2 – 7x + 3
3. p(x) = x5 – 7x3 – 6x2
2. p(x) = 6x4 + 13x3 – 32x2 – 45x + 18
4. p(x) = 4x6 – 9x5 – 15x4 + 29x3 + 27x2 – 24x – 20
Intermediate Value Theorem
Let a < b and p(x) be a polynomial function with p(a) ≠ p(b). On the interval [a, b], p(x) will
take on all values between p(a) and p(b).
– Since p(x) is continuous, to get from p(a) to p(b) you must go through all the values between
them.
Approximating Real Zeros
1. Find the rational zeros, if there are any.
2. Find two consecutive integers, m and m + 1, such that p(m) and p(m + 1) have opposite
signs.
– Since one is positive and the other is negative, the Intermediate Value Theorem implies
p(x) must be zero somewhere in between m and m + 1.
3. Find the two tenths between m and m +1 that have opposite signs when p(x) is evaluated.
4. Continue to the desired accuracy.
Example
p(x) = 2x6 + 5x5 – 7x4 – 3x3 – x2 + 6x – 1 on [0, 1] and [-3, -4]