Long Term Division
Find
The problem is written like this:
The quotient and remainder can then be determined as follows:
- Divide the first term of the numerator by the highest term of the denominator (meaning the one with the highest power of x, which in this case is x). Place the result above the bar (x3 ÷ x = x2).
- Multiply the denominator by the result just obtained (the first term of the eventual quotient). Write the result under the first two terms of the numerator (x2 · (x − 3) = x3 − 3x2).
- Subtract the product just obtained from the appropriate terms of the original numerator (being careful that subtracting something having a minus sign is equivalent to adding something having a plus sign), and write the result underneath ((x3 − 12x2) − (x3 − 3x2) = −12x2 + 3x2 = −9x2) Then, "bring down" the next term from the numerator.
- Repeat the previous three steps, except this time use the two terms that have just been written as the numerator.
- Repeat step 4. This time, there is nothing to "pull down".
The polynomial above the bar is the quotient, and the number left over (−123) is the remainder.
The log division algorithm for arithmetic can be viewed as a special case of the above algorithm, in which the variable x is replaced by the specific number 10.
Remainder Theorem
When you divide a polynomial P(x) by x-b you get:
p(x) = (x-b)·q(x) + r(x)
But r(x) is simply the constant r (remember? when you divide by (x-b) the remainder is a constant) .... so we get this:
f(x) = (x-b)·q(x) + r
Now see what happens when you have x equal to b:
p(b) = (b-b)·q(b) + r
p(b) = (0)·q(b) + r
p(b) = r
So we get this:
The Remainder Theorem:
When you divide a polynomial p(b) by x-b the remainder r will be p(b)
So if you want to know the remainder after dividing by x-b you don't need to do any division:
Just calculate p(b).
Let us see that in practice:
Example: 2x2-5x-1 divided by x-3
(Continuing our example from above)
We don't need to divide by (x-3) ... just calculate f(3):
2(3)2-5(3)-1 = 2x9-5x3-1 = 18-15-1 = 2
And that is the remainder we got from our calculations above.
We didn't need to do Long Division at all!
Example: Dividing by x-4
(Continuing our examplee)
What would the remainder be if we divided by "x-4" ?
"c" is 4, so let us check f(4):
2(4)2-5(4)-1 = 2x16-5x4-1 = 32-20-1 = 11
Once again ... We didn't need to do Long Division to find it.
Factor Theorem
Now ...
What if you calculated p(b) and it was 0?
... that means the remainder is 0, and ...
... (x-b) must be a factor of the polynomial!
Example: x2-3x-4
f(4) = (4)2-3(4)-4 = 16-12-4 = 0
so (x-4) must be a factor of x2-3x-4
And so we have:
The Factor Theorem:
When p(b)=0 then x-b is a factor of the polynomial
And the other way around, too:
When x-b is a factor of the polynomial then p(b)=0
Polynomial Equations
- Real roots of a polynomial equation P(x) = 0 correspond to the x-intercepts of the graph of the polynomial function P(x).
- X-intercepts of the graph of a polynomial function correspond to the real roots of the related polynomial equation.
- If a polynomial equation is factorable, factoring the polynomial, setting its factors equal to zero, and solving each factor will determine the roots.
Families of a Polynomial Function
- A family of functions is a set of functions with the same characteristics.
- Polynomial functions with graphs that have the same x-intercepts belong to the same family.
- A family of polynomial functions with zeros a1,a2,a3,…an can be represented by an equation of the form:
y = k(x – a1)(x – a2)(x – a3) . . . (x – an), where k is a real number not equal to zero
- An equation for a particular member of a family of polynomial functions can be determined if a point on the graph is known.
Solving Inequalities or Inequations
- A polynomial inequality results when the equal sign in a polynomial equation is replaced with an inequality symbol.
- The real zeros of a polynomial function, or x-intercepts of the corresponding graph, divide the x-axis into intervals that can be used to solve a polynomial inequality.
- Polynomial inequalities may be solved graphically by determining the x-intercepts and then using the graph to determine the intervals that satisfy the inequality.
- Factoring inequalities can be solved algebraically by:
- Considering all cases
- Using intervals and testing values in each interval
- Table and number lines can help organize intervals and to provide a visual clue to solutions.
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