What are the zeros of (x-2)(x^2-9)
2 answers:
(x - 2)(x² - 9) = 0
(x - 2)(x² + 3x - 3x - 9) = 0
(x - 2)(x(x) + x(3) - 3(x) - 3(3)) = 0
(x - 2)(x - 3)(x + 3) = 0
x - 2 = 0 or x - 3 = 0 or x + 3 = 0
<u> + 2 + 2</u> <u> + 3 + 3</u> <u> - 3 - 3</u>
x = 2 x = 3 x = -3
So because the factors of the equation are (x-2), (x+3) and (x-3), you set them equal to zero and you get x=2, -3, and 3
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