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ch4aika [34]
1 year ago
7

Precalculus question:

Mathematics
1 answer:
nirvana33 [79]1 year ago
3 0

Considering the polynomial f(x) = x^4  - 16x³ + 70x² + 48x - 219, we have that:

a) The zeros are: x = -1.995, x = 1.995, x = 8 - 3i, x = 8 + 3i.

b) The linear factors are: (x + 1.995)(x - 1.995)(x - 8 + 3i)(x - 8 - 3i).

c) The solutions to f(x) = 0 are: x = -1.995, x = 1.995, x = 8 - 3i, x = 8 + 3i.

<h3>How to obtain the zeros of the polynomial?</h3>

The given zero of the polynomial is:

8 + 3i

Hence it's conjugate is also a zero, that is:

8 - 3i.

Thus the first two factors are given as follows:

(x - 8 - 3i)(x - 8 + 3i) = x² - 16x + 64 + 9i² = x² - 16x + 55.

Then the polynomial can be written as follows:

x^4  - 16x³ + 70x² + 48x - 219 = (ax² + bx + c)(x² - 16x + 55).

As a fourth order polynomial can be the product of two second order polynomials. Applying the distributive property to the right side of the equality, we have that:

x^4  - 16x³ + 70x² + 48x - 219 = ax^4 + x³(b - 16a) + x²(55a + c - 16b) + x(55b - 16c) + 55c.

Then the coefficients are given as follows:

  • a = 1.
  • 55c = -219 -> c = -219/55 -> c = -3.98.
  • b - 16a = -16 -> b = 0.

Hence the other two factors are given as follows:

x² - 3.98 = 0

x = ± sqrt(3.98)

x = ± 1.995.

The linear factors are:

(x - x')(x - x'')(x - x''')(x - x'''').

In which x', x'', x''' and x'''' are the roots.

The solutions to f(x) = 0 are the roots.

More can be learned about the zeros of a polynomial at brainly.com/question/19030198

#SPJ1

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