Question

Y= x^3-19x+30 State the number of complex zeros for each function

Answer 1

Answer:

  0

Step-by-step explanation:

Using Descartes' rule of signs, we observe that the signs of the coefficients, + - +, have two changes. Thus there will be 0 or 2 positive real roots, hence 0 or 2 complex roots.

We can do further work to determine if it is 0 or 2. A graphing calculator provides an easy answer.

This function of x has no complex zeros. They are all real.

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For a cubic, it isn't always easy to find the zeros. The rational root theorem tells you any rational zeros will be factors of 30. Possibilities are ...

  ±1, ±2, ±3, ±5, ±6 . . . . . we're pretty sure no roots have magnitude > 6

For x=0, y = 30 . . . . the constant

For x=1, y = 12 . . . . . a smaller value, so we're going in the right direction

For x=2, y = 0 . . . . . one of the real roots

Dividing out this factor*, we get ...

  y = (x -2)(x^2 +2x -15)

Factoring the quadratic gives ...

  y = (x -2)(x -3)(x +5) . . . . . . all real zeros

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* For dividing x^3 -19x +30, synthetic division works well. The work for that is shown in the second attachment.