If f(x) = 2x^3 - 14x^2 + 38x -26 and x-1 is a factor of f(x) find all of the zeros of f(x) algebraically.

Step-by-step 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

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)

                 2

x  = 3 +/- 2i

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

The zeros are

1

3 +/- 2i

answer:

3 bags of apples and 2 bags of oranges.

step-by-step explanation:

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x