Which of the following represents the zeros of f(x) = 5x³ - 6x² - 59x + 12?
4, 3, 1/5
4, 3, -1/5
4, -3, 1/5
4, -3, -1/5

Solution:

Using the Rational Zeros Theorem, which states that, if the polynomial f(x) = a_nxⁿ + a_{n - 1}x^{n - 1} + ... + a₁x + a₀ 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) = 5x³ - 6x² - 59x + 12

Here,

p: ±1, ±2, ±3, ±4, ±6, ±12 which are all factors of constant term 12

q: ±1, ±5 which are all factors of the leading coefficient 5

All possible values are 

p/q: ±1, ±2, ±3, ±4, ±6, ±12, ±1/5, ±2/5, ±3/5, ±4/5, ±6/5, 12/5,

From the given options we can select only ±4/5, ±3/5, ±12/5 to verify the roots.

f(x) = 5x³ - 6x² - 59x + 12

⇒ f(4/5) = 5(4/5)³ - 6(4/5)² - 59(4/5) + 12

= 2.56 - 3.84 - 47.2 + 12

= -36.48

⇒ f(-4/5) = 5(-4/5)³ - 6(-4/5)² - 59(-4/5) + 12

= -2.56 - 3.84 +47.2 + 12

= 52.8

⇒ f(3/5) = 5(3/5)³ - 6(3/5)² - 59(3/5) + 12

= 1.08 - 2.16 -35.4 +12

= -24.48

⇒ f(-3/5) = 5(-3/5)³ - 6(-3/5)² - 59(-3/5) + 12

= -1.08 - 2.16 +35.4 +12

= 44.16

⇒ f(12/5) = 5(12/5)³ - 6(12/5)² - 59(12/5) + 12

= 69.12 - 34.56 - 141.6 +12

= -95.04

⇒ f(-12/5) = 5(-12/5)³ - 6(-12/5)² - 59(-12/5) + 12

= -69.12 -34.56 +141.6 +12

= 49.92

The given options are not the zeros of the given polynomial. However on verification it is identified that -3, 4 and 1/5.

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Which of the following represents the zeros of f(x) = 5x³ - 6x² - 59x + 12?

Summary: 

The zeros of f(x) = 5x³ - 6x² - 59x + 12 are -3, 4, 1/5.