Polynomial and/or Polynomial Functions and Equations Zeros of a polynomial function  Multiplication of polynomials Division of polynomials Factoring polynomials and solving polynomial equations by factoring
Zeros of a polynomial function
The zeros of a polynomial function are the values of x for which the function equals zero.
That is, the solutions of the equation f (x) = 0, that are called roots of the polynomial, are the zeros of the polynomial function or the x-intercepts of its graph in a coordinate plane.
At these points the graph of the polynomial function cuts or touches the x-axis.
If the graph of a polynomial intersects with the x-axis at (a, 0), or  x = a  is a root or zero of a polynomial, then  (x - a)  is a factor of that polynomial.
Every polynomial of degree n has exactly n real or complex zeros.
An nth degree polynomial has at most n real zeros.
Some of the roots may be repeated. The number of times a root is repeated is called multiplicity or order of the root.
The number xi is a root of the polynomial f (x) if and only if f (x) is divisible by (x - xi).
Thus, finding the roots of a polynomial f (x) s equivalent to finding its linear divisors or is equivalent to polynomial factorization into linear factors.
We add or subtract polynomials by combining their like terms. The like terms are terms that have the same variables raised to the same exponents.
Example:   a( - 5x3 + 2x2 - x + 4) + ( - 4x2 + 3x - 7) = - 5x3 + ( 2 - 4) · x2  + ( -1 + 3) · x + 4 - 7 =
= - 5x3 - 2x2  + 2x - 3
b( x4 - 3x3 + 5x - 1) - ( - 2x4 + x3 - 3x2 + 4) = x4 - 3x3 + 5x - 1 + 2x4 - x3 + 3x2 - 4 =
= ( 1 + 2) · x4  + ( -3 - 1) · x3 + 3x2 + 5x - 5 = 3x - 4x3 + 3x2 + 5x - 5
Multiplication of polynomials
When multiplying two polynomials together, multiply every term of one polynomial by every term of the other polynomial using distributive property.
Example:   ( - 2x3 + 5x2 - x + 1) · ( 3x - 2) =
= 3x · ( - 2x3) + 3x · 5x2 + 3x ·  (-x) + 3x · 1 + (- 2) · ( - 2x3) + (- 2) · 5x2 + (- 2) · ( -x) + (- 2) · 1 =
= - 6x4 + 15x3 - 3x2 + 3x + 4x3 - 10x2 + 2x - 2 = - 6x4 + 19x3 - 13x2 + 5x - 2
Division of polynomials
Divide the highest degree term of dividend by the highest degree term of the divisor to get the first term of the quotient.
Take the first term of the quotient and multiply it by every term of divisor. Write this result below the dividend, making sure you line up all the terms with the terms of the dividend that has the same degree.
Subtract the result from the dividend, i.e., reverse all the signs of the terms of the result and add like terms.
Repeat the process of long division until the degree of the new obtained dividend is less than the degree of the divisor.
 Examples: Note, since each second line should be subtracted, the sign of each term is reversed.
b 3x4 + x2 + 5) ¸ ( x2 - x - 1) or like 17 ¸  5 = 3 + 2/5 -15 2
Factoring polynomials and solving polynomial equations by factoring
A polynomial and/or polynomial function with real coefficients can be expressed as a product of its leading coefficient an and n linear factors of the form x - xi, where xi denotes its real roots and/or complex roots,
f (x) = anxn + an-1xn-1 + . . . + a1x + a0 = an(x - x1)(x - x2) . . . (x - xn).
By multiplying the parentheses on the right side and collecting like terms and then comparing the resulting coefficients with the coefficients of the given polynomial obtained are Vieta's formulas that show relations between coefficients and roots of a polynomial.
Thus, for a quadratic or a second degree polynomial
 a2x2 + a1x + a0 = a2(x - x1)(x - x2) = a2[x2 - (x1 + x2)x + x1x2],
and similarly, for a cubic or a third degree polynomial
 a3x3 + a2x2 + a1x + a0 = a3(x - x1)(x - x2)(x - x3) =
 = a3[x3 - (x1 + x2 + x3)x2 + (x1x2 + x1x3 + x2x3)x - x1x2x3].
 Example: Factorize  (2/3)x2 - (2/3)x - 4  using the above theorem. Solution: (2/3)x2 - (2/3)x - 4 = (2/3)(x2 - x - 6) = (2/3)[x2 - (3 + (- 2))x + 3(- 2)] = = (2/3)(x2 - 3x + 2x - 6) = (2/3)[x(x - 3) + 2(x - 3)] = (2/3)(x - 3)(x + 2)
Example:  Given are leading coefficient a2 = -1and the pair of conjugate complex roots, x1 = 1 + and
x2 = 1 - i, of a second degree polynomial, find the polynomial using the above theorem.
 Solution: By plugging the given values into    a2x2 + a1x + a0 = a2(x - x1) (x - x2) a2x2 + a1x + a0 = -1[x - (1 + i)] · [x - (1 - i)] = -[(x - 1) - i] · [(x - 1) + i] = = -[(x - 1)2 - i2] = - (x2 - 2x + 1 + 1)  = - x2 + 2x - 2
 Example: The real root of the polynomial - x3 - x2 + 4x - 6 is x1 = - 3, factorize the polynomial. Solution: We divide given polynomial by one of its known factors, a3x3 + a2x2 + a1x + a0 = a3(x - x1) (x - x2) (x - x3) then we calculate another two roots of given cubic by solving obtained quadratic trinomial, Finally we use the theorem to factorize given polynomial (see the previous example), a3(x - x1)(x - x2)(x - x3) = -1(x + 3)[x - (1 + i)][x - (1 - i)] = -1(x + 3)(x2 - 2x + 2). Notice that given cubic has one real root and the pair of the conjugate complex roots. Odd degree polynomials must have at least one real root.
Example:  Solve polynomial equation  x3  +  2x2 - x - 2 = 0 by factoring.
Solution:          x2(x +  2)  - (x +  2) = 0
(x +  2)·(x2 -  1) = 0
(x +  2)·(x + 1)·(x - 1) = 0
x +  2 = 0      =>     x1 = -2
x +  1 = 0      =>     x2 = -1
x - 1 = 0      =>    x3 = 1
The roots are:    x1 = -2,    x2 = -1   and    x3 = 1.   Intermediate algebra contents 