The zeros of f(x)= 3x^3+16x^2+18x-4
1 answer:
After trial and error, you discover that - 2 works.
-2 || 3 16 18 -4
-6 -20 4
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3 10 - 2 0
What you have left is a quadratic
3x^2 + 10x - 2 You can just use the quadratic formula to solve for the other 2 roots.
x1 = [ - 10 +/- sqrt(10^2 - 4*3*(-2) ) ] / 6
x1 = [ -10 +/- sqrt (100 + 24) ) /6
x1 = [ - 10 +/- sqrt(124) ) / 6
x1 = [ - 10 +/- 11.14 ] / 6 = 0.1893
x2 = [ - 21.14 ] / 6 = - 3.522
Answer
x1 = - 2
x2 = 0.1893
x3 = - 3.522
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