b) g(x) = 6x⁴ - 19x³ - 86x² + 304x - 160 \ \

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation a_nxⁿ + a_n – 1x^{n – 1} + ... + a₁x + a₀ = 0, then p is a factor of a₀ and q is a factor if a_n.

If p/q is a rational zero, then p is a factor of 6 and q is a factor of 1.

The possible values of p are   ± 1,  ± 2, ± 3,± 4, ± 8, ± 10, ± 20, ± 40, ± 80 and ± 160. \ \

The possible values for q are ± 1, ± 2, ± 3 and ± 6. \ \

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1,  ± 2, ± 3,± 4, ± 8, ± 10, ± 20, ± 40, ± 80 , ± 60,±1/2, ±3/2, ±5, ±1/3, ±2/3, ±4/3, ±8/3, ±10/3, ±20/3, ±40/3, ±80/3, ±160/3, ± 1/6, ±5/3.

Make a table for the synthetic division and test possible real zeros.

Since f(4) = 0,  x = 4 is a zero. The depressed polynomial is  6x³ + 5x² - 66x + 40 = 0.

If p/q is a rational zero, then p is a factor of 40 and q is a factor of 6.

The possible values of p are   ± 1,  ± 2, ± 4, ± 5, ± 8, ± 10, ± 20 and ± 40.

The possible values of q are   ± 1,  ± 2, ± 3, and ± 6.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1,  ± 2, ± 4, ± 5, ± 8, ± 10, ± 20, ± 40, ±1/2, ±5/2, ±1/3, ±2/3, ±4/3, ±5/3, ±8/3, ±10/3, ± 20/3 , ±40/3, ±1/6,±5/6.

Make a table for the synthetic division and test possible real zeros.

Since f(-4) = 0,  x = -4 is a zero. The depressed polynomial is  6x² - 19x + 10 = 0 \ \

6x² - 15x - 4x + 10 = 0

3x(2x - 5) - 2(2x - 5) = 0

(2x - 5)(3x - 2) = 0

x = 5/2, x = 2/3

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