find all the zeros of 2 …
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Sia ? 6 years, 4 months ago
if -2 and -1 are zeros of f(x) = 2x4 + x3 - 14x2 - 19x - 6
x+2 and x+1 are factors of f(x)
So (x + 2)(x + 1) = x2 + x + 2x + 2 = x2 + 3x + 2 is a factor of f(x)
On long division of f(x) by x2 + 3x + 2 we get
f(x) = 2x4 + x3 - 14x2 - 19x - 6 = (2x2 - 5x - 3)(x2 + 3x + 2)
= (2x + 1)(x - 3)(x + 2)(x + 1)
Therefore, zeroes of the polynomial = {tex}\frac{{ - 1}}{2}{/tex}, 3, -2, -1.
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