# Which of the following represent the zeros of f(x) = 6x^{3} - 29x^{2} - 6x + 5.

-5, 1/3, 1/2

5, -1/3, 1/2

5, 1/3, -1/2

5, 1/3, 1/2

**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 a_{0} and q is a factor of the leading coefficient a_{n}.

Here,

p: ±1, ±5 which are all factors of constant term 5

q: ±1, ±2, ±3, ±6 which are all factors of the leading coefficient 6.

All possible values are

p/q: ±1, ±5, ±1/2, ±5/2, ±1/3, ±5/3, ±1/6, ±5/6,

From the given options we can select only ±5, ±1/2, ±1/3 to verify the roots.

Given, f(x) = 6x^{3} - 29x^{2} - 6x + 5

⇒ f(5) = 6(5)^{3} - 29(5)^{2} - 6(5) + 5

f(5) = 750 - 725 - 30 + 5

f(5) = 0

⇒ f(1/2) = 6(1/2)^{3} - 29(1/2)^{2} - 6(1/2) + 5

f(1/2) = 0.75 - 7.25 - 3 + 5

f(1/2) = -4.5 ≠ 0

⇒ f(-1/2) = 6(-1/2)^{3} - 29(-1/2)^{2} - 6(-1/2) + 5

f(-1/2) = - 0.75 - 7.25 + 3 + 5

f(-1/2) = 0

⇒ f(1/3) = 6(1/3)^{3} - 29(1/3)^{2} - 6(1/3) + 5

f(1/3) = 0.222 - 3.222 - 2 + 5

f(1/3) = 0

The roots of the equation are 5, -1/2 and 1/3.

## Which of the following represent the zeros of f(x) = 6x^{3} - 29x^{2} - 6x + 5.

**Summary:**

5, 1/3 , - 1/2 represents the zeros of f(x) = 6x^{3} - 29x^{2} - 6x + 5.

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