Unit 1: Introduction to Polynomial Functions
Activity 4: Factor and Remainder Theorem
In the last activity, you practiced the sketching of a polynomial graph, if you were given the Factored Form of the function statement. In this activity, you will learn a process for developing the Factored Form of a polynomial function, if given the General Form of the function.
Suppose you are given;
f(x) = anxn + an­1xn­1 + an­2xn­2 + ... a2x2 + a1x + a0 (this is a General Form of a polynomial function).
Your task is to sketch a graph of this function. Since you know how to sketch a graph of a polynomial function, IF the function is in Factored Form, it would be handy for you to know how to move easily from General Form to Factored Form.
1. Dividing Polynomials.
2. Finding the first factor.
In the animation you just viewed, the examples illustrated how to divide a function, f(x), by a factor of the form (jx ­ k).
Each of the solutions ended with a Division Statement. The Division Statement was always of the form: f(x) = Quotient . Divisor + Remainder
or
f(x) = q(x) . (x ­ k/j) + r Suppose we were to evaluate f(k/j). Use the Division Statement form of the function to make it easier:
f(k/j) = q(k/j) . (k/j ­ k/j) + r
Because the value of (k/j ­ k/j) = 0, it is easy to see that, regardless of the value of q(k/j), we get this result: f(k/j) = r
This is such a valuable result, that it has been given a name! This is called the Remainder Theorem.
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The Remainder Theorem
When a polynomial function, f(x) is divided by (jx ­ k), the remainder, r, is f(k/j).
This simplifies if j = 1 i.e.: we are dividing by (x ­ k).
When a polynomial function, f(x) is divided by (x ­ k), the remainder is f(k).
Take this idea one step further. We are looking for FACTORS of a polynomial. A factor of a polynomial divides evenly into the polynomial, with a remainder of zero.
So, in the special case of the Remainder Theorem ­ the case when the remainder is zero ­ we know the divisor is a factor of the dividend. This is such an important observation that it has been given a name: The Factor Theorem.
The Factor Theorem
A polynomial function, f(x), has a factor of (x ­ k) if and only if f(k) = 0.
So, if f(k) = 0, then (x­k) is a factor of f(x). Similarly, (jx­k) is a factor of f(x) if and only if f(k/j) = 0.
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Example 1:
Given: f(x) = x³ ­ 4x² + x + 6
Task: Find the fully Factored Form of the function statement Solution:
1. We need a factor to start with. We know the Factor Theorem suggests we would like to find f
(k/j) = 0, which gives us a factor of (x ­ k/j). How should we decide what value of k to begin with? Since the lead coefficient of f(x) is 1, we know that j = 1 in this situation.
That simplifies matters. Now we know that k must be a factor of the final constant in the statement of f(x). So k must be a factor of 6. That leaves us with the task of trying one of {±1, ±2, ±3, ±6} as these are all of the factors of 6.
Now Trial and Error must be used to find the first factor!
f(1) = 1³ ­ 4(1)² + 1 + 6
f(1) = 4 so (x­1) is not a factor. f(­1) = (­1)³ ­4(­1)² + (­1) + 6
f(­1) = 0 so (x+1) is a factor. 2. Now that we know a factor, we can divide the factor into f(x) to find the quotient. This will partially factor the function, and we can look at the quotient to see if it factors further. To divide, use either long division or synthetic division, as described in the animation earlier in this activity. Remember ­ the remainder should be zero, if you chose your factor properly!
ANSWER: f(x) = (x + 1)(x² ­ 5x + 6)
3. Now, look at the quotient. Can it be further factored? If the answer is ‘yes' (as it is in this case) then you have two ways to go:
a) If the quotient is a degree 2 expression, you should be able to factor this using your old methods of trinomial factoring. (Recall, you learned these in Grade 11 Math, and practiced them in the opening Diagnostic Test! This is why you need to know that material!)
b) If the quotient is of higher degree than 2, you will need to repeat steps 1. and 2. Find another factor, and divide again to further reduce the degree of the quotient.
In this case, the quotient is a degree 2 polynomial expression, and can be factored.
The final Factored Form of f(x) for this example:
ANSWER: f(x) = (x + 1)(x ­ 2)(x ­ 3) 3
Example 2:
Given: f(x) = ­2x³ ­ 5x² + 9x + 18
Task: Express f(x) in fully factored form.
Solution:
1. Find a factor, using the Factor Theorem. First of all, we will factor out the negative sign, as it is awkward to factor a polynomial with a lead negative sign. Divide all terms by ­1:
f(x) = ­1(2x³ + 5x² ­ 9x ­ 18)
Now we will briefly ignore the lead negative sign, and factor the bracketed expression, g(x) = 2x³ + 5x² ­ 9x ­ 18
The lead coefficient of our expression is not 1, so we must consider a variety of possible values of k/j for our factor (x ­ k/j):
k must be a factor or ­18 (the final constant term in the function)
j must be a factor of 2 (the lead coefficient) Factors of ­18: {±1, ±2, ±3, ±6, ±9}
Factors of 2: {1, 2} Use Trial and Error to find the first factor!
Choose a factor from the available ones, and test it. Remember, if (x ­ k/j) is a factor of f(x), then f
(k/j) = 0
Try (x­1) first (always try the simple factors first!): g(1) = 2 + 5 ­ 9 ­ 18
g(1) = ­20 so (x ­ 1) is not a factor of g(x).
Try (x ­ 2) next (or your choice, really!) g(2) = 2(2)³ + 5(2)² ­ 9(2) ­ 18
g(2) = 16 + 20 ­ 18 ­ 18
g(2) = 0 so (x ­ 2) is a factor of g(x). 2. Divide to find the quotient. Use either long division or synthetic division.
ANSWER: q(x) = 2x² + 9x + 9
3. Since the quotient is a quadratic expression, you can factor it using your trinomial methods and come up with the Factored Form for g(x). ANSWER: g(x) = ­(2x + 3)(x + 3)(x ­ 2)
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Example 3:
Given: f(x) = x4 ­5x3 + 6x2
Task: Express f(x) in fully factored form.
Solution:
Step 1: ALWAYS look for a COMMON FACTOR first!
This is not something we concerned ourselves with for Examples 1 and 2, but it IS something that should always come first!
Here, there is a common factor: x²
f(x) = x²(x² ­ 5x + 6)
Now, the factor (x² ­ 5x + 6) can indeed be factored, using simple trinomial factoring methods.
What looked like a more complicated expression, now becomes fairly routine to factor, and the final answer is:
f(x) = x² (x ­ 2)(x ­ 3) Steps to Factor a Polynomial Expression
1. Look for a common factor first. 2. Find a factor using the factor theorem. 3. Divide the factor from Step 2, to find a quotient, q(x).
4. If q(x) is a quadratic (degree 2) expression, use your factoring of a trinomial methods to further factor q(x). If q(x) is a polynomial expression of higher degree than 2, go back to step 2, and find another factor using the factor theorem. Then repeat Step 3 to find a simpler quotient. Continue until you DO find a quotient of degree 2 that you can factor.
5. Express your final answer, by collecting all the factors you have discovered, into one statement. HOMEWORK: use "Factor and Remainder Theorem Homework Worksheet"
Quiz tomorrow on Act. 3 & 4
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