using our tools to find the zeros of polynomials 17 november 2014

Using Our Tools to Find the Zeros of Polynomials

17 November 2014


Do you feel like this?Fundamental Theorem

of Algebra

Quadratic Formula

Remainder Theorem

Synthetic Division

Factor Theorem

The Rational Root Theorem

Integral Root Theorem

Descartes’ Rule of Signs

Location Principle

Upper Bound Theorem

Lower Bound Theorem


Let organize our thinking.Fundamental Theorem

of Algebra

Quadratic Formula

Remainder Theorem

Synthetic Division

Factor Theorem

The Rational Root Theorem

Integral Root Theorem

Descartes’ Rule of Signs

Location Principle

Upper Bound Theorem

Lower Bound Theorem

These are all tools to help us solve polynomials


• Review– Fundamental Theorem of Algebra• Every polynomial equation with a degree greater than

zero has a least one root in the set of complex numbers• Based on the corollary, a polynomial equation of

degree n has exactly n complex roots.

– Quadratic Formula


• Review (Cont’d)– Remainder Theorem

• If a polynomial P(x) is divided by x-r, the remainder is a constant P(r) and P(x) = (x-r)(Q(x)) + P(r) where Q(x) is one degree less than the degree P(x)

– Factor Theorem• This means that we can try to find where P(r) = 0. In this case x-r is a

factor. Also Q(x) may be factorable.

– Synthetic Division• Shorthand process to divide P(x) by x-r . • Process is more efficient to try potential zeros (roots)• synthetic division is an easy way to get a depressed polynomial. (In the

case of a degree 3, finding one root opens up the use of factoring or the quadratic formula on the resulting degree 2 so the process is shortened considerably)


• Review (Cont’d)– Rational Root Theorem - If a polynomial function,

written in descending order of the exponents, has integer coefficients, then any rational zero must be of the form ± p/ q, where p is a factor of the constant term and q is a factor of the leading coefficient.• This means that we can very quickly create a “short” list

of possible solutions


• Review (Cont’d)• Descartes’ Rule of Signs– A method of determining the maximum number of

positive and negative real roots of a polynomial. – For positive roots, start with the sign of the coefficient

of the lowest (or highest) power. Count the number of sign changes n as you proceed from the lowest to the highest power (ignoring powers which do not appear). Then n is the maximum number of positive roots. Furthermore, the number of allowable roots is n, n-2, n-4, ....


• Review (Cont’d)• Descartes’ Rule of Signs– A method of determining the maximum number

of positive and negative real roots of a polynomial. – For negative roots, starting with a polynomial f(x),

write a new polynomial f(-x) with the signs of all odd powers reversed, while leaving the signs of the even powers unchanged. Then proceed as before to count the number of sign changes n. Then n is the maximum number of negative roots.


• Review (Cont’d)– Location Principle• if a continuous function has opposite signs for two

values of the independent variable, then it is zero for some value of the variable between these two values.

– Upper Bound Theorem• If c is a positive real number and P(x) is divided by x-c

and the resulting quotient and remainder have no change in sign, the P(x) has no real zero greater than c. Thus c is the upper bound of the zeros of P(x)


• Review (Cont’d)– Location Principle• if a continuous function has opposite signs for two

values of the independent variable, then it is zero for some value of the variable between these two values.

– Lower Bound Theorem• If c is an upper bound of the zeros of P(-x), then –c is a

lower bound of the zeros of P(x).


• Review (Cont’d)– Multiplicity - How many times a particular

number is a zero for a given polynomial. For example, in the polynomial function f(x) = (x – 3)4(x – 5)(x – 8)2, the zero 3 has multiplicity 4, 5 has multiplicity 1, and 8 has multiplicity 2. Although this polynomial has only three zeros, we say that it has seven zeros counting multiplicity.

http://www.mathwords.com/xyz/zero_of_a_function.htm

http://www.mathwords.com/p/polynomial.htm

http://www.mathwords.com/f/function.htm


• Find all the zeros of x4 + 9x3 + 31x2 + 49x + 30 = 0• Step 1 - Use Rational Root Theorem

+ 1, + 2, + 3, + 5, + 6, + 10, + 15, + 30• Step 2 - Use Descartes Rule of Signs

Pos – 0Neg - f(-x) = x4 - 9x3 + 31x2 - 49x + 30 = 0Therefore - 4, 2, or 0 negatives Sometimes it helps to make a chart.


• Find all the zeros of x4 + 9x3 + 31x2 + 49x + 30 = 0• Step 3 - Use Synthetic Division. Since there are only

negative roots, we will start by testing negative possible zeros.

f(-2) -2 1 9 31 49 30-2 -14 -34 -30

1 7 17 15 0


• Find all the zeros of x4 + 9x3 + 31x2 + 49x + 30 = 0• Step 3 - Use Synthetic Division. Since there are only

negative roots, we will start by testing negative possible zeros.

x3 + 7x2 - 17x + 15 = 0

• Follow same process for x3 + 7x2 - 17x + 15 = 0• Try f(-3)

f(-2) -2 1 9 31 49 30-2 -14 -34 -30

1 7 17 15 0


• Find all the zeros of x4 + 9x3 + 31x2 + 49x + 30 = 0Follow same process for x3 + 7x2 - 17x + 15 = 0• Try f(-3) x2 - 4x + 5 = 0

(note: we now have 2 of 4 zeros)• Step 4 - Use Quadratic Formula

These will be imaginary.


Homework - Find all zeros of the following:• x3 -4x2 + x + 2 = 0

• x4 + 3x2 – 4 = 0

• x3 + 3x2 – 2x – 8 = 0

• 2x3 + 7x2 + 7x + 2

using our tools to find the zeros of polynomials 17 november 2014

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