You already have considerable experience with certain polynomials.

• It turns out that linear and quadratic functions are polynomials.

• A linear function is a polynomial of degree 1, since the highest power of x in a linear function f(x) = mx + b is 1.
• A quadratic polynomial is of degree 2, since the highest power of x in a quadratic polynomial g(x) = a x^2 + bx + c is 2.
• It is important to understand that when we multiply two linear polynomials we get a quadratic polynomial (e.g., (x-1) (x+2) = x^2 + x - 2), and we also note that the resulting quadratic polynomial has zeros wherever the linear polynomial does (this will be clarified in the exercises below).

Polynomials as sums of power functions

There are two fundamental ways to look at polynomials.

• The first way is to think of polynomials as sums of power functions of the form a x^p, with p taking only whole-number positive values or zero.
• For example, if we add 4 x^3, 5 x^2, -2x^1 and 7 x^0 we get the polynomial 4 x^3 + 5 x^2 - 2 x + 7.

The table below shows two different ways of representing general polynomials of higher and higher degrees:

degree	with coefficients a, b, c, ...	with coefficients a0, a1, a2, ...	name of class
1	ax+b	a1 x + a0	linear
2	ax^2 + bx + c	a2 x^2 + a1 x + a0	quadratic
3	ax^3 + bx^2 + cx + d	a3 x^3 + a2 x^2 + a1 x + a0	cubic
4	ax^4 + bx^3 + c x^2 + dx + e	a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0	quartic
etc.	etc.	etc.	etc.
.	.	.	.
.	.	.	.
.	.	.	.
.	.	.	.
.	.	.	.
.	.	.	.
n	???	an x^n + a(n-1) x^(n-1)            &nb sp; + ... +  a2 x^2 + a1 x + a0	names become impractical and unnecessary

Using general coefficients a, b, c, ... has an obvious shortcoming if we exceed degree 23 (x is the 24th letter of the alphabet). The a0, a1, a2, ... form is often more convenient, especially when we want to express a polynomial of arbitrary degree n. We can write

an x^n + a(n-1) x^(n-1) + ... + a2 x^2 + x1 x + a0

to represent any degree-n polynomial. The subscript of a will always matched the power of x (e.g., a5 will always go with x^5).

Polynomials as products of linear and quadratic functions

Another way to look at polynomials is in their factored form.

• We saw above that a quadratic polynomial can often be expressed as a product of two linear polynomials (e.g., as before, (x-1) (x+2) = x^2 + x - 2).
• You will see in the exercises below that a quadratic polynomial without zeros cannot be expressed as a product of linear polynomials.

• For example, the expression (x-3) (x^2 + 4x + 12) can be multiplied out to give us x^3 + x^2 - 36, which is a polynomial of degree 3.
• Furthermore the original expression cannot be further factored, since the quadratic factor has no zeros.
• A quadratic polynomial without zeros is called irreducible, since it cannot be factored into two linear factors.

• Looking at the form (x-3)(x^2 + 4x + 12), we see that the only way the polynomial has of attaining the value 0 is for x to equal 3.

• This value will make the linear factor (x-3) zero, so the entire product is 0.
• Any other value of x will make both the linear and quadratic factor (which can never be 0) nonzero, which will make their product nonzero.

• As another example, the expression (x + 2) (x - 1) (x + 3) can be multiplied out to give us x^3 + 4x^2 + x - 6.

• The original expression is as factored as an expression can be, since all its factors are linear.
• If x is equal to -2, or 1, or -3, one of the linear factors must be 0 so the product of the linear factors, and hence the value of the polynomial, must be 0.

• As a final example, the expression (x^2 - 4x + 16) (x^2 + 5) can be multiplied about to give us a degree-4 polynomial.

• Neither of the quadratics in the original expression has zeros, so neither can be factored.
• Since neither factor can ever be 0, the polynomial can never take value 0.

From these examples we see how a series of linear and irreducible quadratic factors gives us a polynomial, with a zero for each linear factor and no zeros corresponding to the irreducible quandratic factors.

Using the distributive law to multiply polynomials

When we multiply polynomials we repeatedly use the distributive law of multiplication over addition, as in the following examples:

Example 1: To multiply the two linear factors (x-3) and (x+4) we proceed as follows:

(x-3)(x-4) = x(x+4) - 3(x-4) (a+b) (c) = a(c) + b(c); let a=x and b=-3

= x^2 + 4x - (3x - 12) a(b+c) = ab + ac; apply to each term and be careful to group the second

= x^2 + 4x - 3x + 12 distribute the - (really a -1) through the 3x-12

= x^2 + x + 12

Example 2: To multiply the two irreducible quadratic factors (x^2 + x + 4) and (x^2 - 3x + 14) we proceed in a similar manner:

(x^2 + x + 4) (x^2 - 3x + 14) = x^2(x^2 - 3x + 14) + x(x^2 - 3x + 14) +4(x^2 - 3x + 14)

= x^4 - 12x^3 + 14x^2 + x^3 - 3x^2 + 14x + 4x^2 - 12x + 56

= x^4 + (-12 x^3 + x^3) + (14 x^2 - 3 x^2 + 4 x^2) + (14x - 12x) + 56

= x^2 - 11 x^3 + 15 x^2 + 2x + 56

Example 3: To multiply the three linear factors (x-2), (x+4) and (x-1) we first multiply two of the factors to get a quadratic then multiply the remaining linear factor by this quadratic:

(x-2)(x+4)(x-1) = (x-2) [ (x+4)(x-1) ]

= (x-2) [ x(x-1) + 4(x-1) ]

= (x-2) [x^2-x + 4x - 4]

= (x-2) (x^2 + 3x - 4)

= x (x^2 + 3x - 4) - 2 (x^2 + 3x - 4)

= x^3 + 3x^2 - 4x - (2x^2 + 6x - 8)

= x^3 + 3x^2 - 4x - 2x^2 - 6x + 8

= x^3 + x^2 - 10x + 8

(x-3)(x^2 + 2x + 7)

(x - 5) (2x + 7) (x^2 - 3x + 9)

(x^2+x+3) (x^2 - 2x + 10)

(x-2) (x+3) (x+5).

What are the zeros of the linear polynomials f(x) = 2x - 6 and g(x) = x + 2?

What quadratic polynomial q(x) do you get if you multiply f(x) by g(x)?

Using the quadratic formula find the zeros of the quadratic polynomial q(x). Are the zeros identical to those of f(x) and g(x)?

2.  Using the quadratic formula determine the zeros of x^2 - x - 6. Call the zeros z1 and z2.

Show that if we multiply (x - z1) by (x - z2), for the values of z1 and z2 obtained above, we get the quadratic polynomial x^2 - x - 6.

3.  Explain why the following statement must be true: If the quadratic polynomial f(x) = a x^2 + bx + c has no zeros, then that polynomial is not the product of two linear polynomials.

4.  Explain why the following statement must be true: No polynomial of degree 2 can be the product of three or more polynomials of degree 1.

The table and graph of (x-a)(x-b)

The graph of y = (x-a)(x-b) is clearly a parabola, since (x-a)(x-b) always multiplies out to a quadratic. However, it is instructive to see how the (x-a)(x-b) form can be used to obtain the overall shape and some important details of the graph.

A table for y = f(x) = (x-2) (x+1) might look like the following:

This table shows not only the end result, the y value, but the values of the two factors for each value of x. It is clear that the value of y will be zero whenever one of the factors (x-2)(x+1) is 0. It is also clear that unless one of these factors is 0, y cannot be 0, since we can't get 0 by multiplying two nonzero numbers.

We obtain the following graph:

This graph shows how the function is 0 at x = -1 and x = 2, whenever one of the linear factors is 0, and only when one of the factors is zero.

The table can be extended to include a large positive and a large negative value of x:

A 'large' value of x is a value whose absolute value is considerably greater than that of any of the zeros. The absolute values of the zeros in this example are 2 and 1; the absolute values of the 'large' x's are both 100, considerably greater than either 2 or 1.

It is clear that when x is large in this sense, each of the linear factors will have a value pretty close to the value of x. When two large factors are multiplied, of course, the result is even larger. So we see that for large x, the values of y rapidly become extremely large. We see also that when there are two such factors, we must always get a positive result for large absolute values of x, since both factors will have the same sign.

The following graph shows how the y values rapidly increase as x increases its absolute value in either the positive or the negative direction.

The table and graph of (x-a)(x-b)(x-c)

A table for the function y = f(x) = (x-2)(x+1)(x-3) is shown below:

Again we see that the value of f(x) is 0 whenever the value of one of its factors is 0.

We see that for large positive x all three factors are positive and y is a very large positive number, while for large negative x all three factors are negative and y is a very large negative number.

The corresponding graph is shown below, first for a range near the zeros then for a range extending slightly beyond the zeros:

Note that between the zeros at x = 2 and x = 3 the graph doesn't just flatten out along the x axis, despite the fact that the table doesn't show any values between x = 2 and x = 3. This couldn't happen because none of the factors (x-2), (x+1) or (x-3) is 0 anywhere between x = 2 and x = 3. Rather than flattering out the curve follows a smooth path, as indicated on the graph.

Exercises 5-6