Download Finding Roots of Polynomials

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Math 1314
College Algebra
Finding Roots of Polynomials
SHOW ALL WORK AND JUSTIFY ALL ANSWERS.
Find all zeroes of P(x) = 2x6 − 3x5 − 13x4 + 29x3 − 27x2 + 32x − 12.
1. Make sure all terms of P(x) are written in descending order.
2. Can we factor out a greatest common factor from each term?
0 is a zero if P(0) = 0.
3. Is 0 a zero?
4. Look for sign changes in P(x): P(x) = 2x6 − 3x5 − 13x4 + 29x3 − 27x2 + 32x − 12
sign changes in P(x).
There are
Summary of Descartes’ Rule of Signs: For potential positive real zeroes: Count sign changes in P(x).
Take that number and continue to subtract 2 until you get 0 or 1. The original value and each result of the
subtraction could be the number of positive real zeroes.
Using Descartes’ Rule of Signs, we could have
or
positive real zeroes.
To find P(−x), change signs of all coefficients of xodd power
5. Find P(−x)
P(−x) =
There are
or
sign changes in P(−x).
Summary of Descartes’ Rule of Signs: For potential negative real zeroes: Count sign changes in P(−x).
Take that number and continue to subtract 2 until you get 0 or 1. The original value and each result of the
subtraction could be the number of negative real zeroes.
Using Descartes’ Rule of Signs, we will have
negative real zero.
6. Fill in the values for possible zeroes. I have started for you. Remember that every row must add up to the
degree of the polynomial, which is 6 in this example.
0 is not a zero
0
0
0
positive real zeroes
5
3
1
negative real zeroes
nonreal zeroes
NOTE: If 0 was a zero, we’d
put 1’s in the first column.
7. Factor the leading coefficient. Factor the constant term.
Use the Rational Root Theorem to list all possible rational roots.
Factors of leading coefficient:
Factors of constant term:
1
Summary of Rational Root Theorem:
Potential rational roots look like:
factor of constant term
factor of leading coefficient
1 2 3 4 5 12 1 2 3 4 5 12
Possible rational zeroes: ± , ± , ± , ± , ± , ± , ± , ± , ± , ± , ± , ±
1 1 1 1 1
1
2 2 2 2 2
2
(I have filled in the options for factors of the leading coefficient. You need to fill in all factors of the constant
term for each blank).
1 3
We can reduce the list above to these possible rational zeroes: ±1, ±2, ±3, ±4, ±6, ±12, ± , ± .
2 2
8. We are now going to use synthetic division to find a root. (Remember to write 0 in place of any missing terms).
Which possible rational zero should we try first? We have 16 choices. I hope we choose wisely!
Since we have one negative real root, let’s try a negative possibility first.
How about x = −3?
NOTE: We will have exactly one negative real zero, but this
zero does not have to be a rational number. But it could be!
9. Use synthetic division to see if x = −3 is a zero of P(x).
-3
-3
2
2
-3
-3
-13
-13
29
29
-27
-27
32
32
If the last entry in our synthetic
division is 0, we found a zero.
-12
-12
NOTE: x = −3 is a zero, but we can also conclude something else that is very useful! What do you notice about
the signs of the numbers in the last row? They alternate!
Thus, x = −3 is a lower bound. We do not need to check
x = −4, x = −6 or x = −12.
We have reduced our list of possible rational zeroes to
1 3
±1, ±2, ±3, 4, 6, 12, ± , ± .
2 2
Recall: Lower bounds look like:
-# coefficients of the polynomial P(x)
-3 2 -3 -13 29 -27 32 -12
-3 + +
+
+
OR 0 can take the place of any + or -:
-# coefficients of the polynomial P(x)
-3 2 -3 -13 29 -27 32 -12
-3 0 0 +
+
0
10. The last row from our synthetic division gives us a depressed equation. Thus, we can rewrite P(x) in terms of
the new factor we found and the depressed equation.
P(x) = (x + 3)(2x5 − 9x4 + 14x3 − 13x2 + 12x − 4)
Recall: Factor Theorem: If x = a is a zero of P(x),
then (x − a) is a factor of P(x) and vice versa.
11. Check to see if we can factor out a greatest common factor from the depressed equation. We cannot, so we
continue in our process.
12. Let’s use the Rational Root Theorem on the depressed equation (2x5 − 9x4 + 14x3 − 13x2 + 12x − 4) = 0 to
see if we can eliminate any more rational zeroes.
Factors of leading coefficient:
Factors of constant term:
2
Summary of Rational Root Theorem:
Potential rational roots look like:
factor of constant term
factor of leading coefficient
1 2 4 1 2 4
Possible rational zeroes: ± , ± , ± , ± , ± , ±
1 1 1 2 2 2
(I have filled in the options for factors of the leading coefficient. You need to fill in all factors of the constant
term for each blank).
1
We can reduce the list above to these possible rational zeroes: ±1, ±2, ±4, ± .
2
Now, go back to our chart from Descartes’ Rule of Signs. We had only one negative real zero and we found
it, so we do not have to check any more negatives!
1
Thus, our list is reduced further to 1, 2, 4, .
2
13.
14.
What should we try next? How about x = 1?
Use synthetic division to see if x = 1 is a zero of P(x).
1 2 -9 14 -13 12 -4
-3 2 -3 -13 29 -27 32
NOTE: Once you find a zero, it is a good idea
to test it again, in case it is a zero of
multiplicity > 1. However, we do not need to
test x = −3 again since P(x) has only one
negative real zero.
Drat! x = 1 is not a zero.
Also, we cannot conclude that it is an upper bound.
Recall: Upper bounds look like:
+# coefficients of the polynomial P(x)
-3 2 -3 -13 29 -27 32 -12
-3 + + +
+
+
+
+
OR 0 can take the place of any + (the entries
must be nonnegative):
+# coefficients of the polynomial P(x)
-3 2 -3 -13 29 -27 32 -12
-3 0 0 +
+
+
+
0
1 is not a zero, so our possible rational roots are 2, 4,
Let’s try another possible rational root on our list:
What about x = 2?
15. Use synthetic division to see if x = 2 is a zero of P(x).
2 2 -9 14 -13 12 -4
-3 2 -3 -13 29 -27 32
1
2
16. x = 2 is a zero. Is x = 2 an upper bound? No.
17. Rewrite P(x): P(x) = (x + 3)(x − 2)(2x4 − 5x3 + 4x2 − 5x + 2).
Check to see if we can factor the depressed equation. We cannot, so we continue in our process.
18. Using the Rational Root Theorem on our depressed equation does not eliminate any possible rational zeroes.
19. Let’s use synthetic division to see if x = 2 is a zero of multiplicity > 1.
2 2 -5 4 -5 2
3 2 -3 -3 9 7
3
20. x = 2 is a zero of multiplicity because synthetic division shows us it is a zero again.
Rewrite P(x): P(x) = (x + 3)(x − 2)(x − 2)(2x3 − x2 + 2x − 1)
Check to see if we can factor the depressed equation. If we want to continue testing zeroes with syn1
We can factor by grouping.
thetic division, we can. We would test x = , see
2
that it is a zero and continue from there.
21. Factor 2x3 − x2 + 2x − 1: Factor the greatest common factor from the first and second terms. Then factor the
greatest common factor from the third and fourth terms.
2x3 − x2 + 2x − 1 = x2 (
) + 1(
)
Thus, 2x3 − x2 + 2x − 1 = (x2 + 1)(2x − 1)
22. Rewrite P(x): P(x) = (x + 3)(x − 2)(x − 2)(2x − 1)(x2 + 1)
Now, we have written P(x) as a product of linear factors and one quadratic factor. We can solve x2 + 1 = 0 by
factoring (if possible), completing the square, or quadratic formula.
We cannot factor (x2 + 1) over the real numbers.
Solve x2 + 1 = 0.
Did you get a complex number? If so, did you get
its conjugate?
Recall: Conjugate Pairs Theorem: If a + bi is a
root of P(x), a − bi is also a root of P(x).
List all zeroes of P(x):
23. Strategy to find all zeroes of a polynomial P(x):
(a) Make sure all terms of P(x) are written in descending order.
(b) Factor P(x) if possible.
(c) Check to see if x = 0 is a zero of P(x).
(d) Use Descartes’ Rule of signs to determine the number of positive, negative, and nonreal roots.
(e) Use the Rational Root Theorem to list all possible rational roots.
(f) Use synthetic division to find a root. (Remember to write 0 in place of any missing terms).
(g) Rewrite P(x), using the root to write as a product of factors.
(h) Look at the depressed equation to see if some rational roots can be eliminated. Use this and results from
Descartes’ Rule to choose the next potential zero to try.
(i) Continue until P(x) is written as a product of linear factors and one quadratic factor (i.e. depressed
equation is quadratic). Solve the quadratic equation by factoring, completing the square, or quadratic
formula.
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