Question

If f(x) = 2x^3- 2x^2 - 14x + 30 and x + 3 is a factor of f(x), then find all of

the zeros of f(x) algebraically.

Answer 1

Apply the long division to find that there is only 1 real root

Answer 2

Explanation:

First confirm that x = 1 is one of the zeros.

f(1) = 2(1)^3 - 14(1)^2 + 38(1) - 26

f(1) = 2 - 14 + 38 - 26

f(1) = -12 + 38 = + 26

f(1) = 26 - 26

f(1) = 0

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next perform a long division

x -1 || 2x^3 - 14x^2 + 38x - 26 || 2x^2 - 12x + 26

2x^3 - 2x^2

===========

-12x^2 + 28x

-12x^2 +12x

==========

26x -26

26x - 26

========

0

Now you can factor 2x^2 - 12x + 26

2(x^2 - 6x + 13)

The discriminate of the quadratic is negative. (36 - 4*1*13) = - 16

So you are going to get a complex result.

x = -(-6) +/- sqrt(-16)

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2

x = 3 +/- 2i

f(x) = 2*(x - 1)*(x - 3 + 2i)*(x - 3 - 2i)

The zeros are

1

3 +/- 2i