If f ( x ) = 2 x 3 + 18 x 2 + 56 x + 40 f(x)=2x 3 +18x 2 +56x+40 and f ( − 1 ) = 0 f(−1)=0, then find all of the zeros of f ( x ) f(x) algebraically

Question

If f(x)=2x3+18x2+56x+40f(x)=2x3+18x2+56x+40 and f(−1)=0f(−1)=0, then find all of the zeros of f(x)f(x) algebraically

Answer 1

You're given one real root. From that you can do synthetic division to determine the other roots.

This is a synthetic division shortcut. Write down the real root, then under a division symbol, the coefficients of the variables, 2, 18, 56, and 40. Drop down the 2 first, multiply by the root (-1) and add it to the next coefficient (18). 2*(-1) = -2, -2 + 18 = 16. Drop that one down. Next multiply 16 by the root, (-1) and add that to the next coefficient, 56 + (-16) = 40, drop down the 40, then lastly repeat this, multiply 40 by the root (-1) and add to the last coefficient 40 + (-40) = 0. If the remainder was not 0, then -1 would not have been a real root.

-1 | 2 18 56 40

......0 -2 -16 -40

--------------------

......2 16 40 0

Put these into another equation, (2x² + 16x + 40)

Factoring out a 2 we get 2(x² + 8x + 20)

The factors of 20 are 2*10 and 4*5. 2+10 = 12, 4+5 = 9, so we must use the quadratic equation to find our roots.

-8 ± √(64 - 4(20))/2 = -4 ± √(-16)/2 = -4 ± 4i/2 = -4 ± 2i

So the other two roots are x = (-4 + 2i) and x = (-4 - 2i)