# Which of the following represents the zeros of f(x) = 3x^{3} - 10x^{2} - 81x + 28?

7, -4, 1/3

7, -4, -1/3

7, 4, 1/3

7, 4, -1/3

**Solution:**

Using the Rational Zeros Theorem, which states that, if the polynomial f(x) = a_{n}x^{n} + a_{n - 1}x^{n - 1} + ... + a_{1}x + a_{0} has integer coefficients,

then every rational zero of f(x) has the form p/q where p is a factor of the constant term a₀ and q is a factor of the leading coefficient a_{n}.

Given:

Function f(x) = 3x^{3} - 10x^{2} - 81x + 28

Here,

p: ±1, ±2, ±4, ±7, ±14, ±28 which are all factors of constant term 28

q: ±1, ±3 which are all factors of the leading coefficient 3

All possible values are

p/q: ±1, ±2, ±4, ±7, ±14, ±28, ±1/3, ±2/3, ±4/3, ±7/3, ±14/3, ±28/3,

From the given options we can select only ±4, ±7, ±1/3 to verify the roots.

f(x) = 3x^{3} - 10x^{2} - 81x + 28

⇒ f(4) = 3(4)^{3} - 10(4)^{2} - 81(4) + 28

= 192 - 160 - 324 + 28

= -264

⇒ f(-4) = 3(-4)^{3} - 10(-4)^{2} - 81(-4) + 28

= -192 - 160 + 324 + 28

= 0

⇒ f(7) = 3(7)^{3} - 10(7)^{2} - 81(7) + 28

= 1029 - 490 - 567 + 28

= 0

⇒ f(1/3) = 3(1/3)^{3} - 10(1/3)^{2} - 81(1/3) + 28

= 0.111 - 1.111 - 27 + 28 = 0

Therefore, the zeros of polynomial are -4, 7, 1/3.

## Which of the following represents the zeros of f(x) = 3x^{3} - 10x^{2} - 81x + 28?

**Summary:**

The zeros of f(x) = 3x^{3} - 10x^{2} - 81x + 28 are -4, 7, 1/3.

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