Find the rational roots if they exist of
x³ - 3x² - 2x - 8
If there are any rational roots, they will be fractions whose
numerator divides evenly into 8, the constant term, and whose
denominator divides evenly into 1, the coefficient of the
highest power of x. They can also be positive or negative.
Possible numerators are the factor of 8, i.e., 1, 2, 4, and 8
Possible denominator is the only factor of 1, namely 1. So
Possible rational roots are ±1/1, ±2/1, ±4/1, and ±8/1 or
±1, ±2, ±4, and ±8
Try x=1, using synthetic division
1| 1 -3 -2 -8
| 1 -2 -4
1 -2 -4 -12
No, for we don't get 0 remainder
So we try x=-1, using synthetic division
-1| 1 -3 -2 -8
| -1 4 -2
1 -4 2 -10
No, for we don't get 0 remainder
So we try x=2, using synthetic division
2| 1 -3 -2 -8
| 2 -2 -8
1 -1 -4 -16
No, for we don't get 0 remainder
So we try x=-2, using synthetic division
-2| 1 -3 -2 -8
| -2 10 -16
1 -5 8 -24
No, for we don't get 0 remainder
So we try x=4, using synthetic division
4| 1 -3 -2 -8
| 4 4 8
1 1 2 0
Hey that gives 0 remainder! So we have now factored
the polynomial as
(x - 4)(1x² + 1x + 2)
or
(x - 4)(x² + x + 2)
Since the trinomial in the second parentheses doesn't
factor, 4 is the only rational root.
------------------------------------------------------------
Determine the number of positive and negative real roots of
2x5 + x4 - 3x3 + x2 - x + 1
To find the number of positive real roots, we use DesCartes'
rule of signs. Count the sign changes going left to right
when the polynomial is in descending order:
2x5 has sign + and x4 also has sign +, so that's NOT a sign change.
x4 has sign + and -3x3 has sign -, that's a change in sign,
so that makes 1 sign change so far.
-3x3 has sign - and +x2 has sign +, that's a change in sign,
so that makes 2 sign changes so far.
+x2 has sign + and -x has sign -, that's a change in sign,
so that makes 3 sign changes so far.
-x has sign - and +1 has sign +, that's a change in sign,
so that makes 4 sign changes so far.
So there are either 4 positive roots, 2 positive roots, or
0 positive root.
To find the possible number of negative real roots, we replace x by
-x, simplify, and find the number of positive roots of the resulting
polynomial.
2(-x)5 + (-x)4 - 3(-x)3 + (-x)2 - (-x) + 1
-2x5 + x4 + 3x3 + x2 + x + 1
We can easily see there is only one sign change here, so
there is exactly one negative root.
So there are either 4 or 2 or 1 positive roots, and exactly 1 negative
root.
----------------------------------------------------------------
Find the other roots of x3+4x2-7x-10 if one root is 2
Use synthetic division:
2|1 4 -7 -10
| 2 12 10
1 6 5 0
So the polynomial is now factored as
(x - 2)(x2 + 6x + 5)
Now factor the trinomial in parentheses
(x - 2)(x + 1)(x + 5)
Set each factor = 0, and solve for x and we get
that the zeros are 2, -1, and -5
The other roots besides 2 are -1 and -5.
Edwin
AnlytcPhil@aol.com