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A polynomial \( f(x) \) and two of its zeros are given.

\[ f(x) = 2x^5 + 11x^4 + 44x^3 + 31x^2 - 148x + 60 \]

Two of its zeros are \(-2 - 4i\) and \(\frac{11}{13}\).

(a) Find all the zeros. Write the answer in exact form.

Answer :

Given that f(x) = 2x³ + 11x² + 44x³ + 31x² - 148x + 60; -2 - 4i and 11/13 are the zeros. The zeros of the given polynomial are -2 - 4i, 11/13, and -2 + 4i.

The given polynomial is f(x) = 2x³ + 11x² + 44x³ + 31x² - 148x + 60.

Thus, f(x) can be written as 2x³ + 11x² + 44x³ + 31x² - 148x + 60 = 0

We are given that -2 - 4i and 11/13 are the zeros. Let's find out the third one. Using the factor theorem,

we know that if (x - α) is a factor of f(x), then f(α) = 0.

Let's consider -2 + 4i as the third zero. Therefore,(x - (-2 - 4i)) = (x + 2 + 4i) and (x - (-2 + 4i)) = (x + 2 - 4i) are the factors of the polynomial.

So, the polynomial can be written as,f(x) = (x + 2 + 4i)(x + 2 - 4i)(x - 11/13) = 0

Now, let's expand the above equation and simplify it.

We get, (x + 2 + 4i)(x + 2 - 4i)(x - 11/13) = 0

⇒ (x + 2)² - (4i)²(x - 11/13) = 0 (a² - b² = (a+b)(a-b))

⇒ (x + 2)² + 16(x - 11/13) = 0 (∵ 4i² = -16)

⇒ x² + 4x + 4 + (16x - 176/13) = 0

⇒ 13x² + 52x + 52 - 176 = 0 (multiply both sides by 13)

⇒ 13x² + 52x - 124 = 0

⇒ 13x² + 26x + 26x - 124 = 0

⇒ 13x(x + 2) + 26(x + 2) = 0

⇒ (13x + 26)(x + 2) = 0

⇒ 13(x + 2)(x + 2i - 2i - 4i²) + 26(x + 2i - 2i - 4i²) = 0 (adding and subtracting 4i²)

⇒ (x + 2)(13x + 26 + 52i) = 0⇒ x = -2, -2i + 1/2 (11/13)

Therefore, the zeros of the given polynomial are -2 - 4i, 11/13, and -2 + 4i.

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