What are actual roots of FX according to the rational root theorem The following are potential roots of FX 2x 2 2x 24 4 3 2 3 4?

All the rational roots of the polynomial f(x) = 20x4 + x3 + 8x2 + x - 12 are 3/4 and -4/5.

We have the given polynomial:

f(x) = 20x4 + x3 + 8x2 + x - 12

All potential polynomial roots are represented by rational numbers, according to the Rational Root Theorem:

x = a/b

Where a is a factor of a constant term

b is the factor of coefficient of leading term

From the given polynomial:

-12 is the constant term and 20 is the coefficient of leading term

The possible factors of -12 are ±1, ±2, ±3, ±4, ±6, ±12

The possible factors of 20 are ±1, ±2, ±4, ±5, ±10, ±20

So the rational roots = (±1, ±2, ±3, ±4, ±6, ±12)/ (±1, ±2, ±4, ±5, ±10, ±20)

Solve for x when the given function is equal to zero

20x4 + x3 + 8x2 + x - 12 = 0

(4x - 3)(5x + 4)(x2 + 1) = 0

So we get:

4x - 3 = 0 or 5x + 4 = 0 or x2 + 1 = 0

x = 3/4, x = -4/5, x =1, x = -1

Hence, the rational roots are x = 3/4, x = -4/5.

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We have the given polynomial:

f(x) = 20x4 + x3 + 8x2 + x - 12

x = a/b

b is the factor of coefficient of leading term

-12 is the constant term and 20 is the coefficient of leading term

The possible factors of 20 are ±1, ±2, ±4, ±5, ±10, ±20

So the rational roots = (±1, ±2, ±3, ±4, ±6, ±12)/ (±1, ±2, ±4, ±5, ±10, ±20)

Solve for x when the given function is equal to zero

20x4 + x3 + 8x2 + x - 12 = 0

(4x - 3)(5x + 4)(x2 + 1) = 0

4x - 3 = 0 or 5x + 4 = 0 or x2 + 1 = 0

x = 3/4, x = -4/5, x =1, x = -1

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