SOLUTION: 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
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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): You can put this solution on YOUR website!
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): You can put this solution on YOUR website!
Looking at the coefficients
1 -5 7 -35
we see the second pair is 7 times the first pair, which means the polynomial will factor by grouping.
The roots are , , and
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