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Polynomials are math equations that are multinomial. They have two or more terms with each term separated by a plus (+) or minus (-) sign, and do not have any letters under a radical (√ ) or in a denominator. Their exponents are integer and are always positive:

3x + 5ab + 2y + 4 is a four term polynomial.

6x + 5ab + 4 is a three term polynomial.

√x - 3 is not a polynomial, “x” is under a radical.

3/x + 6 is not a polynomial, “x” is a denominator.

All polynomials are continuous functions. When a polynomial is drawn or is displayed as a graph the resulting line does not break, skip or gap. If you were using pencil and paper the line could be drawn from its beginning to end without lifting the pencil from the paper.

The degree of polynomial that has a single literal factor is the term having the highest degree:

4 x2 - 2x + 4 is a second degree polynomial determined by the “ x2 ” literal factor.

If there are two or more variables in a term the degree of polynomial is the sum of the exponents of the variables:

2ax3 - 4x + 6 is a fourth degree polynomial determined by the sum of the “a” and “ x3 ” factors.

3x + 5ab + 2b + 4 is a second degree polynomial, “a” and “b” each count as one from the “5ab” term.

Solving higher-degree polynomials is often more challenging than linear or quadratic polynomials. Both linear and quadratic polynomials have specific equation formulas, methods and techniques that can be applied to solve these math problems. However, the number of steps or iterations necessary to discover the zeros of higher-degree polynomials can often be described as try and try again.

Let “R” be the set of real numbers. The zero polynomial function is defined to be function “g” as:

g(x) = 0 ; For all “x” in “R”, with a degree not defined.

If some number “R” is a solution of g(x) = 0, then “R” is called a zero of the polynomial. If “R” is a solution of g(x) = 0, then g(R) = 0. By substituting “R” in for “x” results in the value 0 if “R” is a zero of g(x).

Let “R” be the set of real numbers.

For all polynomial functions excluding the zero function:

f(x) = an xn + an -1 xn -1 + …+ a1 x + a0

For all “x” in “R”, and “x” ≠ 0, and “n” is a nonnegative integer. Each “a” is a coefficient of the polynomial function:

a0 is a constant term.

an is the leading coefficient.

fn = 0, 1, 2, respectively we obtain:

f(x) = cx0 = c,

the constant function, is not a polynomial, n = 0

f(x) = ax + b, the linear equation, n = 1

f(x) = ax2 + bx + c,

the quadratic equation, n = 2

n = 3 is a cubic function

n = 4 is a quartic function

If we divide:

(x3 – 6x2 + 3x + 21) / (x – 2)

The result of the division is a quotient, q(x) and a remainder, R, so that:

q(x) = x2 – 4x – 5 and R = 11

R, the remainder, can be stated as a fractional result of the division process:

11 / (x – 2)

Let the polynomial divided be identified as function P(x) so that:

P(x) = x3 – 6x2 + 3x + 21

This allows us to state the P(x) function as:

P(x) / (x – 2) = q(x) + R / (x – 2)

By multiplying both sides of the equation by (x – 2):

P(x) = (x – 2) q(x) + R, or

P(x) = (x – 2) (x2 – 4x – 5) + 11, and

P(x) = x3 – 6x2 + 3x + 21, so that

x3 – 6x2 + 3x + 21 =

(x – 2) (x2 – 4x – 5) + 11

We now state the generalized math of the theorem as:

If P(x) = (x – u) q(x) + R then

P(u) = (u – u) q(x) + R or P(u) = R

Leading us to a formal definition of the Remainder Theorem:

If polynomial P(x) is divided by x – u until a constant remainder is obtained then

P(u) = R

An example of how to find the remainder of a polynomial divided by a binomial polynomial using the Reminder Theorem follows:

P(x) = x4 + 3x3 + x2 – x +3 is divided by

x + 2

P(–2) = 16 – 24 + 4 +2 + 3 = 1, where 1 is the remainder, then

(x4 + 3x3 + x2 – x +3) / (x + 2) = q(x) + 1

This theorem is a method to determine the zeros of a polynomial. It states that if we know a zero of a polynomial it is a factor of the polynomial, and if we know a factor of a polynomial we can find a zero of that polynomial.

Factor Theorem Formal Definition:

Suppose P(x) is a polynomial and suppose u is a real number. Then (x – u) is a factor of P(x) if and only if P(u) = 0. Thus, the Factor Theorem means:

If (x – u) is a factor of P(x) then P(u) = 0

and

If P(u) = 0 then (x – u) is a factor of P(x).

To find the zeros of:

P(x) = (x – 4) (x + 2) (x – 6) therefore,

(x – 4) is a factor of the polynomial, so 4 is a zero of the polynomial,

(x + 2) is a factor of the polynomial, so –2 is a zero of the polynomial, and

(x – 6) is a factor of the polynomial, so 6 is a zero of the polynomial.

Simply, the Factor Theorem states that, P(4), P(–2) or P(6) would result in one of the three factor quantities (x – 4), (x + 2), or (x – 6) to equal zero. The result is that P(x) would equal zero at P(4), P(–2) or P(6). If we were to graph P(x) using Cartesian Coordinates, so that y = P(x), there are 3 coordinates in which y = 0:

(x, y) = (4, 0), (-2, 0) and (6, 0).

We can also ask whether a factor is a factor of a polynomial:

Is (x – 3) a factor of

P(x) = x3 – 3x2 + 2x – 6?

First determine the zero of the factor

(x – 3); x = 3.

By solving for P(3) we find:

33 – 3 (32) + 2 (3) – 6 =

27 – 27 + 6 – 6 = 0

We can now see that (x – 3) is a factor and is also a zero of the polynomial.

The Location Theorem states that if P(x), a polynomial with real coefficients, and two values of x, a and b, are opposite sign (one is positive the other negative) then P(x) has at least one real zero between a and b.

The basic rationalization for this theorem is that polynomials are continuous functions; they do not jump from one point to another when drawn or graphed as Cartesian Coordinates.

As an example suppose that P(x) = x2 – 5x + 6. Observe that P(1) = 2 > 0. P(5/2) = -1/4 < 0. This tells us that P(x) has a zero for the polynomial between 1 and 5/2. The zero is P(2). This does not tell us whether other zeros of the polynomial exist (or do not exist). P(4) = 2 >0, so there must exist a zero for the polynomial between 5/2 and 4. It is x = 3, and P(3) = 0.

As a caution using this theorem; The values chosen for P(a) and P(b) could result in both being a positive or negative value. Suppose P(x) = x2 – 5x + 6, and a = 0, b = 1. There are no zeros for the polynomial between P(0) and P(1). However, if we chose a = 1 and b = 4 we would find that P(1) and P(4) are both positive result, however there are two zeros for the polynomial between P(1) and P(4).

The Upper Bound Theorem is somewhat related to the Location Theorem. It states that if P(x) is a non-constant polynomial function, with real coefficients and c is a nonnegative real number. When x – c is divided into P(x) all the coefficients of the quotient q(x) are nonnegative and that P(c) is non-negative. Then c is an upper bound for the real zeros of P(x). All real zeros of P(x) are less than or equal to c.

To explain farther let P(x) = (x – u) q(x) + R where u is a real number. P(u) is the dividend, (x – u) the divisor, q(u) the quotient and R the remainder of a non-constant polynomial math division result. The Upper Bound Theorem says that if (x – u) > 0, q(x) > 0 and R ≥ 0 then P(x) = (x – u) q(x) + R > 0 (states that at least one of the coefficients of q(x) is positive while the remaining coefficients of q(x) are nonnegative). Its meaning is that u cannot be a zero of the polynomial, all real numbers greater than u are not a zero of the polynomial and all real zeros of the polynomial are less than u. This proves the theorem.

Example of Upper Bound Theorem:

P(x) = 2x3 – 5x2 – x + 7

Divide the polynomial using real number values of u until (x – u) > 0, q(u) > 0 and R ≥ 0:

(The resulting coefficients are provided, not the detail of the division process. I used Synthetic Division to obtain the coefficients)

P(x) = (x – u) q(x) + R

(2x3 – 5x2 – x + 7) / (x – 1) results are

2, -3, -4, 4 coefficients

(2x3 – 5x2 – x + 7) / (x – 2) results are

2, -1, -3, 1 coefficients

(2x3 – 5x2 – x + 7) / (x – 3) results are

2, 1, 2, 13 coefficients

Observe that when x – 3 is divided into the polynomial P(x) = 2x3 – 5x2 – x + 7 all coefficients of the quotient as well as the remainder are positive. Three is an upper bound for the real zeros of the polynomial.

Finding the Lower Bound:

The Upper Bound Theorem also provides an explanation to find the Lower Bound for the real zeros of a polynomial. To accomplish this we take the negative of the polynomial P(x); P(-x) and

P(x) = 2x3 – 5x2 – x + 7 and

P(-x) = -2x3 – 5x2 + x + 7

Since the zeros of P(-x) are the roots (or solutions) of the equation P(-x) = 0 and since this equation is equivalent to the equation -P(-x) = 0, the zeros of P(-x) are exactly the zeros of -P(-x) so that

P(-x) = -2x3 – 5x2 + x + 7

-P(-x) = 2x3 + 5x2 – x + 7

Using Synthetic Division we see that

2x3 + 5x2 – x + 7 / (x – 1) has resulting coefficients of 2, 7, 6, -1

2x3 + 5x2 – x + 7 / (x – 2) has resulting coefficients of 2, 9, 17, 27

When x – 2 is divided into -P(-x) all of the coefficients of the quotient, as well as the remainder, are positive. Two is an upper bound for the real zeros of -P(-x) and an upper bound for P(-x). So -2 is a lower bound for P(x) [note the change of positive to negative sign from P(-x) where x = 2 to P(x) where x = -2].

We can now see that all real zeros of the polynomial, P(x) = 2x3 – 5x2 – x + 7, lie between a lower bound -2 and upper bound 3.

The Rational Root Theorem is a method to find the rational zeros of a polynomial.

It states that if P(x) has integer coefficients and a coefficient of 1 on the highest power of x, then all integer solutions of P(x) = 0 are factors of the constant term of P(x).

A coefficient not equal 1 on the highest power of x can be factored out of the polynomial:

P(X) = An Xn + An -1 Xn-1 + … + A1 X + A0

can be written as

P(X) = (An) (P(X) / An), or

P(X) = An (1 Xn + An-1 / An Xn-1 + … + A1 / An X + A0 / An)

This tells us the rational solutions of the polynomial An Xn + An -1 Xn-1 + … + A1 X + A0 = 0, with integer coefficients, must be of the form R / S, where R is a factor of A0 and S is a factor of An. It also tells us that since the coefficient of the highest power is factored to equal 1 then R / S = R / ±1, which, of course, is an integer. Moreover, if R divides A0, it follows that any rational zero of P(X) divides A0. This corollary is also valid when A0 = 0 then R = 0, and when the leading coefficient is -1 (instead of 1).

An example of the Rational Root Theorem follows:

P(x) = x4 – 2x3 – 8x2 +10x +15

By the corollary above all the rational zeros of the polynomial P(x) must be integers that divide 15. The only possible rational zeros of P(x) are: ±1, ±3, ±5, ±15

By substituting these possible values into P(x) leads us to which values are in fact rational zeros of the polynomial. We can see that P(1) = 16 ≠ 0 while P(-1) = 0. Thus -1 is a zero of the polynomial. Hence, by the Factor Theorem we can also validate that x – (-1) is a factor of P(x):

x – (-1) = x + 1

(x + 1) ( x3 – 3x2 – 5x +15 = P(x) = x4 – 2x3 – 8x2 + 10x + 15

The fundamental theorem of algebra states that every polynomial equation P(x) = 0 has at least one real or complex solution. A complex number is any number of the form a + bi, where a and b are both real numbers and i satisfies the equation i2 = -1. If b ≠ 0 the complex number a + bi is called imaginary.

It means that every polynomial P(x) of degree n can be expressed as the product of n linear factors, so P(x) = 0 has exactly n solutions.

If q(x) is the quotient and r1 a solution of P(x) = (x – r1) q(x), when P(x) is divided by (x – r1) we can write:

P(x) = (x – r1) q1(x)

The fundamental theorem also applies to q(x). If we call this solution r2, then x – r2 is a factor of q1(x) so that:

P(x) = (x – r1) (x – r2) q2(x)

Continuing this process until one of the quotients is a constant, c, we have:

P(x) = (x – r1) (x – r2) … (x – rn) c, where c ≠ 0

Note that the degree of a polynomial is reduced one each time it is divided by a linear factor. If a polynomial is of degree n then n divisions are required before the quotient is a constant.

An example of the Fundamental Theorem of Algebra follows:

P(x) = x4 + 6x3 + 5x2 – 24x -36 = 0 where 2 and -2 are already known solutions.

Since the highest-degree of P(x) is 4 there are four solutions.

The solutions 2 and -2 must be values in factors of P(x) = 0, we write the factors as

x + 2 and x – 2, and (x + 2) (x – 2) = x2 – 4. Dividing P(x) by x2 – 4 gives us:

P(x) = (x2 – 4) (x2 +6x + 9), and

x2 +6x + 9 = (x + 3) (x + 3)

The solutions of P(x) are 2, -2, -3 and -3.

-3 is a multiple solution, with multiplicity 2, a double solution.

Not all solutions to polynomials are distinct. Non-distinct solutions are called Multiple Solutions. These solutions are complex numbers that might not be real.

If there are two polynomials p(x) and g(x), with real or imaginary coefficients, have the same positive degree n and if there exists n + 1 complex numbers a1, a2 …, an + 1 such that p(ai) = g(ai) for each i = 1, 2, … , n + 1, then p(x) and g(x) are identical polynomials. Thus, p = g.

To prove this corollary suppose that if h(x) = p(x) – g(x). If p(x) ≠ g(x) then h(x) is not the zero polynomial, h(x) has some degree, m. Clearly, m ≤ n since each p(x) and g(x) is of degree n, meaning h(x) has at most n zeros. Then, h(x) must have m complex zeros. However, h(a1) = p(a1) – g(a1) = 0 for i = 1, 2 …, an + 1, contradicts that h(x) must be the zero polynomial. This proves the theorem.

To show the Identity Polynomial Theorem (2 simple examples follow):

Suppose p(x) = 2(x – 1) and g(x) = 2x – 2, then

h(x) = p(x) – g(x) = 2(x – 1) – 2x – 2 = 0

Suppose p = 2 + 2 and g = 4

h = p – g = 2 + 2 – 4 = 0

We conclude that when p(x) = g(x), h(x) = p(x) – g(x) = 0 (or p = g, h = p – g). Then p(x) and g(x) are merely different ways of representing the same expression or number.

To find the solutions of higher-degree equations:

Let P(x) = 2x4 +9x3 +10x2 – 3x – 6 = 0

1. Use the degree of the polynomial to determine how many solutions exist. P(x) is a fourth degree polynomial so there are 4 solutions.

2. Find possible solutions using r/s where r is a factor of the constant and s is a factor of the coefficient of the highest power of x. (r = {6, 3, 2, 1} and s = {2, 1})

r/s combinations are {±6/2, ±6/1, ±3/2, ±3/1, ±2/2, ±2/1, ±1/2, ±1/1}

Possible solutions are {±1, ±1/2, ±2, ±3, ±3/2, ±6}

3. Find P(r/s) to determine if any of the possible solutions is a solution.

P(1) = 2 + 9 + 10 – 3 -6 ≠ 0

P(-1) = 2 – 9 + 10 + 3 – 6 = 0

(-1 is a solution)

4. Use Synthetic Division to find a quotient polynomial which is 1 degree less than the polynomial.

2 + 9 + 10 – 3 – 6 (-1

0 – 2 – 07 – 3 + 6

2 + 7 + 03 – 6

The quotient is 2x3 + 7x2 +3x – 6.

5. Repeat steps 2 to 4 until the quotient polynomial is a quadratic.

Possible solutions are (from step #2):

±1, ±1/2, ±2, ±3, ±3/2, ±6

P(2) = 32 + 72 + 40 – 6 – 6 ≠ 0

P(-2) = 32 – 72 + 30 + 6 – 6 = 0

(-2 is a solution)

2 + 7 + 3 – 6 (-2

0 – 4 – 6 + 6

2 + 3 – 3

The quotient polynomial is 2x2 + 3x – 3, a quadratic equation.

6. Use the quadratic formula or factoring to find the remaining two solutions.

(The following values of x were obtained by using the quadratic formula)

x = ( -3 ± √(9 + 24) ) / 4 = ( -3 ± √33 ) / 4

7. The solution set is:

{-1, -2, ( -3 + √33 ) / 4, ( -3 – √33 ) / 4}