(2).

The inequality is x⁵ + 7x³ + 6x < 5x⁴ + 7x² + 2.

write the inequality in general form with the rational expression on the left and zero on the right.

x⁵ + 7x³ + 6x - 5x⁴ - 7x² - 2 < 0

x⁵ - 5x⁴ + 7x³ - 7x² + 6x - 2 < 0

To find the key numbers, solve the related equation for x.

x⁵ - 5x⁴ + 7x³ - 7x² + 6x - 2 = 0.

The function is p(x) = x⁵ - 5x⁴ + 7x³ - 7x² + 6x - 2. \

If p/q is a rational zero, then p is a factor of constant and q is a factor of leading coefficient 1.

The possible values of p = - 2 are ± 1 and ± 2.

The possible values for q = 1 are ± 1.

By the Rational Roots Theorem, the only possible rational roots are, p / q = ± 1 and ± 2.

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

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

\ p/q \	\ 1 \	\ - 5 \	\ 7 \	\ - 7 \	6	- 2
\ 1 \	\ 1 \	\ - 4 \	\ 3 \	\ - 4 \	2	0

Since, f(1) = 0, x = 1 is a zero. The depressed polynomial is p(x) = x⁴ - 4x³ + 3x² - 4x + 2.

Because, the degree of the polynomial is 4, there are four roots to the polynomial.

The four roots are x₁, x₂, x₃ and x₄.

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The imaginary numbers are not a key numbers, the key numbers are x = 1, x = 3.4 and x = 0.5. So, the polynomial’s test intervals are (-∞, 0.5), (0.5, 1) (1, 3.4) and (3.4, ∞).

In each test interval, choose a representative x-value and evaluate the polynomial.

Test Interval x-value Polynomial Value Conclusion

(-∞, 0.5) x = 0 (0)⁵ - 5(0)⁴ + 7(0)³ - 7(0)² + 6(0) - 2 = - 2 < 0 Negative

(0.5, 1) x = 0.6 (0.6)⁵ - 5(0.6)⁴ + 7(0.6)³ - 7(0.6)² + 6(0.6) - 2 = 0.012 > 0 Positive

(1, 3.4) x = 2 (2)⁵ - 5(2)⁴ + 7(2)³ - 7(2)² + 6(2) - 2 = - 10 < 0 Negative

(3.4, ∞) x = 0.6 (3.5)⁵ - 5(3.5)⁴ + 7(3.5)³ - 7(3.5)² + 6(3.5) - 2 = 8.3 > 0 Positive

From this you can conclude that the inequality is satisfied on the open intervals (-∞, 0.5) and (1, 3.4). So, the solution set is (-∞, -0.5) U (1, 3.4).