Harold T. answered 01/14/20
MS in Engineering w/Math Minor and 25 Years Tutoring Experience
We will use the Rational Root Theorem, or p/q test, for both problems.
Given: q*x^n + ... + p
Possible Rational Roots will be ±[factors of p / factors of q]
We will then use synthetic division to see if they are roots.
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Problem #1:
x3 - 10x2 + 23x - 14
p = 14
q = 1
p/q = 14/1
Possible Rational Roots will be ±[(14 * 1, 7 * 2) / 1] = ±{1,2,7,14} = 1, -1, 2, -2, 7, -7, 14, -14
Synthetic Division #1:
1 ⌋ 1 -10 23 -14
1 -9 14
-------------------------
1 -9 14 ⌉ 0
Therefore, x = 1 is a root
Synthetic Division #2:
7 ⌋ 1 -9 14
7 -14
-------------------------
1 -2 ⌉ 0
Therefore, x = 7 is a root
Synthetic Division #3:
2 ⌋ 1 -2
2
-------------------------
1 ⌉ 0
Therefore, x = 2 is a root
The Rational Roots for problem #1 are x = 1, 2, 7
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Problem #2:
x3 + 6x2 + 9x + 4
p = 4
q = 1
p/q = 4/1
Possible Rational Roots will be ±[(4 * 1, 2 * 2) / 1] = ±{1,2,4} = 1, -1, 2, -2, 4, -4
Synthetic Division #1:
-1 ⌋ 1 6 9 4
-1 -5 -4
-------------------------
1 5 4 ⌉ 0
Therefore, x = -1 is a root
Synthetic Division #2:
-4 ⌋ 1 5 4
-4 -4
-------------------------
1 1 ⌉ 0
Therefore, x = -4 is a root
Synthetic Division #3:
-1 ⌋ 1 1
-1
-------------------------
1 ⌉ 0
Therefore, x = -1 is a root
The Rational Roots for problem #2 are x = -1, -1, -4
NOTE: -1 is a double root