Question 1206674: list ALL roots (rational, irrational, and/or complex) of the given polynomial equation by using the methods discussed in this lesson. SHOW YOUR WORK.
x^3 - 5x^2 + 7x - 35 = 0
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
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Since the leading coefficient is 1, we list the factors of the last term to generate all possible rational roots.
List of possible rational roots:
1, -1, 5, -5, 7, -7, 35, -35
List the positive and negative versions of each.
Then plug each of them into the function to see which generates the result of 0.
Let's try x = 1
f(x) = x^3 - 5x^2 + 7x - 35
f(1) = (1)^3 - 5(1)^2 + 7(1) - 35
f(1) = -32
The nonzero result tells us that x = 1 is not a root.
Every other value but x = 5 will also generate nonzero results.
f(x) = x^3 - 5x^2 + 7x - 35
f(5) = (5)^3 - 5(5)^2 + 7(5) - 35
f(5) = 0
Proving that x = 5 is a root and x-5 is a factor.
Let's apply synthetic division with this root.
The last item in the bottom row is 0 which is the remainder. The remainder 0 confirms that x = 5 is indeed a root.
The other items in the bottom row form the quotient.
Quotient = 1x^2 + 0x + 7 = x^2 + 7
We determine x^3-5x^2+7x-35 = (x-5)(x^2+7)
Solving x^2+7 = 0 leads to x = i*sqrt(7) and x = -i*sqrt(7) where i = sqrt(-1)
The three roots are
x = 5, x = i*sqrt(7) and x = -i*sqrt(7)
Answer by greenestamps(13200)  (Show Source):
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