2
lesson pretestMultiple Choice

Pretest

Question 2 • Algebra 2

Consider the equation 3 x cubed plus 14 x squared minus 14 x plus 60 is equal to 0 The real solution is negative 6 What are the nonreal solutions?

Answer
A
4 minus 2 i square root of 26 comma 4 plus 2 i square root of 26
B
negative 2 i square root of 26 comma 2 i square root of 26
C
the fraction with numerator negative 2 plus i square root of 26 and denominator 3 comma the fraction with numerator negative 2 minus i square root of 26 and denominator 3
D
the fraction with numerator 2 plus i square root of 26 and denominator 3 comma the fraction with numerator 2 minus i square root of 26 and denominator 3

Explanation

Since −6 is a root, divide the cubic by x + 6. The quotient is 3x² − 4x + 10, so solve 3x² − 4x + 10 = 0. Its discriminant is (−4)² − 4(3)(10) = −104, giving roots (4 ± 2i√26)/6 = (2 ± i√26)/3. These are choice C.

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