Subtract seven more than twice a number from the square of one-third of the number to get zero?
1 answer:
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Answer:
A) 2 + i
Step-by-step explanation:
F(x) = x^3 - 3x^2 + x + 5
0 = x^3 - 3x^2 + x + 5
0 = (x+1)(x^2 - 4x + 5)
Great, now we can separate these two parenthesis expressions because of the Zero Product Property. Start with the simple one:
0 = x + 1
<u>x = -1</u>
We have our first real root! But it doesn't look like that's one of the answer choices, so move on to the other expression:
0 = (x^2 - 4x +5)
This expression can't be factored, so we will use the quadratic formula (which is x = ).
First solve for the positive part:
= (4 + sqrt(16-20)) / 2
= 4 + sqrt(-4) / 2
= 4 + 2i / 2
<u>= 2 + i</u>
Then for the negative part:
= (4 - sqrt(16-20)) / 2
= 4 - sqrt(-4) / 2
= 4 - 2i / 2
<u>= 2 - i</u>
<u></u>
<u>2 + i</u> is answer choice A! Our other roots, <u>2 - i</u> and <u>-1</u>, aren't answer choices.