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Find a polynomial f(x) of degree 3 that has the following zeros. -1, 8, 2

User Mtoossi
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Final answer:

A polynomial of degree 3 with zeros at -1, 8, and 2 is constructed by multiplying the factors (x + 1), (x - 8), and (x - 2). The resulting polynomial is f(x) = x^3 - 9x^2 + 6x + 16.

Step-by-step explanation:

The student is asked to find a polynomial f(x) of degree 3 with given zeros: -1, 8, and 2. The zeros of a polynomial are the values of x for which the polynomial equals zero. To use these zeros to construct the polynomial, we place each zero into a binomial (x - zero). For the given zeros, the factors of the polynomial are (x + 1), (x - 8), and (x - 2). Multiplying these factors together gives us the polynomial:

  • (x + 1)
  • (x - 8)
  • (x - 2)

Multiplying (x + 1) and (x - 8) first:

(x + 1)(x - 8) = x² - 8x + x - 8 = x² - 7x - 8

Now, multiply this result by (x - 2):

(x² - 7x - 8) * (x - 2) = x³ - 2x² - 7x² + 14x - 8x + 16 = x³ - 9x² + 6x + 16

Therefore, the polynomial of degree 3 that has the zeros -1, 8, and 2 is f(x) = x³ - 9x² + 6x + 16.

User Fuchsia
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