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Two zeroes of $x^{4} + 2 x^{3} - 13 x^{2} - 14 x + 24$ are 1 and −2. Find the other two.

Competition practiceAlgebra

A3 and −4
B−3 and 4
C6 and −2
D1 and −12

Step-by-Step Solution

The other two have sum

- 2 - (1 - 2) = - 1

and product

24 \div (1 \times - 2) = - 12

, so they are the roots of

t^{2} + t - 12 = 0

: 3 and −4.

Answer: A) 3 and −4

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