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Definition of terms and symbols when dividing polynomials:

Dividend: | f(x) |

Divisor: | h(x) |

Quotient: | q(x) |

Remainder: | r(x) |

If any of these are ** constants**, for example if r(x) is constant, as in:

r(x) = 5

or:

r(x) = a

then ** variable**, rather than

r = 5

or:

r = a

When f(x) is divided by h(x), the result is the value of q(x) plus r(x)/h(x), as in:

f(x)/h(x) = q(x) + r(x)/h(x)

After multiplying each side by h(x), the above can also be written as:

f(x) = h(x)q(x) + r(x)

The remainder, r(x), will either be equal to 0, or it will be less in degree than the degree of the divisor, h(x).

If h(x) has a degree of 1, then the degree of the remainder must be 0. That is, the remainder must be a constant, as in:

r(x) = cx^{0} = c

Under these conditions variable notation is fine, as in:

r = c

Therefore, if f(x) is divided by the linear polynomial (x - c), the remainder is a constant, r.

Again, consider our basic definition of polynomial division:

Dividend: f(x)

Divisor: h(x)

Quotient: q(x)

Remainder: r(x)

f(x) = h(x)q(x) + r(x)

Make the divisor, h(x), equal to the zero degree polynomial (x - c). This will create a remainder, r, that is a constant.

h(x) = (x - c)

Then this:

f(x) = h(x)q(x) + r(x)

Becomes:

f(x) = (x - c)q(x) + r

Let x = c, so:

f(c) = (c - c)q(x) + r

f(c) = (0)q(x) + r

f(c) = 0 + r

f(c) = r

Therefore, when the polynomial function, f(x), is divided by a linear polynomial function in the form (x - c), the remainder is f(c).

This we will call the ** remainder theorem** for polynomial division.

Using this remainder theorem, if the divisor is the linear function (x - c) as in:

h(x) = (x - c)

Then our basic definition of polynomial division:

f(x) = h(x)q(x) + r(x)

Becomes:

f(x) = (x - c)q(x) + f(c)

Suppose c is a zero of f(x). Then, of course:

f(c) = 0

And the above:

f(x) = (x - c)q(x) + f(c)

Would become:

f(x) = (x - c)q(x) + 0

Or:

f(x) = (x - c)q(x)

That clearly makes (x - c) a factor of f(x).

Therefore, if c is a zero of f(x), then (x - c) is a factor of f(x).

That means that f(x) can be though of as the product of (x - c) times some other polynomial function, q(x). This could be summarized as:

If

f(c) = 0

Then

f(x) = (x - c)q(x)

So, whenever we know a root, or zero, of a function, we know a factor of that function.

Now we are in a position to understand a method for analytically solving a certain group of problems regarding finding roots of polynomial functions.

Suppose you have a polynomial function of degree 3, and you wish to find the real, possibly integer, roots. This function might look like:

f(x) = x^{3} - 9x^{2} + 26x - 24

Now, suppose you knew one root for this function. One happens to be x = 2.

Check:

f(2) = 2^{3} - 9(2^{2}) + 26(2) - 24

f(2) = 8 - 36 + 52 - 24

f(2) = 0

That means:

f(x) = (x - 2)q(x)

Or:

x^{3} - 9x^{2} + 26x - 24 = (x - 2)q(x)

We can find q(x) by dividing f(x) by (x - 2), as in:

q(x) = (x^{3} - 9x^{2} + 26x - 24) / (x - 2)

This yields:

q(x) = x^{2} - 7x + 12

Which means that this:

f(x) = (x - 2)q(x)

Becomes:

f(x) = (x - 2)(x^{2} - 7x + 12)

Now, the second factor, (x^{2} - 7x + 12),
is easily factored, as:

(x^{2} - 7x + 12) = (x - 3)(x - 4)

So, this:

f(x) = (x - 2)(x^{2} - 7x + 12)

Becomes:

f(x) = (x - 2)(x - 3)(x - 4)

Therefore we have factored f(x) into a group of linear factors. The roots of f(x) are clearly demonstrated. They are 2, 3, and 4.

So, if we know one root of a cubic polynomial function, we know a linear factor of that function.

If we know a linear factor of the cubic we can divide the cubic by that factor and get another polynomial factor of one degree less than the cubic. That is, we will get a quadratic polynomial function as a quotient.

That quadratic can be factored by convention means, certainly by the quadratic equation. This factored quadratic equation will yield the other linear factors of the original cubic.

So, a cubic can be factored into a group of linear factors using these methods. When it is so factored, its roots are obvious.

This method can be generalized to situations concerning finding the roots of polynomial functions with degrees past the third. For example, if you had a polynomial function of the fourth degree, and if you could find one zero of that function, then you would have one linear factor of that function. You could divide the fourth degree polynomial by this linear function and find a polynomial of degree three. This third degree polynomial function could be factored as described above.

At the moment there are some things to be concerned with:

The above methods depend upon knowing at least one zero of the polynomial function so as to generate that first linear factor which will subsequently be used as a divisor to reduce the the degree of the original polynomial function. How does one find this zero?

Once the process has gone on long enough, and enough linear functions have been found and used as divisors until the current quotient is of degree two and can easily be factored using the quadratic equation, what does it mean if this quadratic function factors into less than two linear factors, say one or none?

These are questions shortly to be answered.

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