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The zeros of f(x)= 3x^3+16x^2+18x-4

User Thewooster
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1 Answer

3 votes

After trial and error, you discover that - 2 works.

-2 || 3 16 18 -4

-6 -20 4

====================================

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

The zeros of f(x)= 3x^3+16x^2+18x-4-example-1
User Justin Trevein
by
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