The function below has at least one rational zero. Use this fact to find all zeros of the function h(x) = 7x⁴ - 9x³ - 41x² + 13x + 6. If more than one zero, separate with commas. Write exact values, not decimal approximations.

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 aₙ.

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

q: ±1, ±7 which are all factors of the leading coefficient 7

All possible values are

p/q: ±1, ±2, ±3, ±6, ±1/7, ±2/7, ±3/7, ±6/7

Given, h(x) = 7x⁴ - 9x³ - 41x² + 13x + 6

h(1) = 7(1)⁴ - 9(1)³ - 41(1)² + 13(1) + 6 = 7 - 9 - 41 + 13 + 6 = -24 ≠ 0

h(-1) = 7(-1)⁴ - 9(-1)³ - 41(-1)² + 13(-1) + 6 = 7 + 9 - 41 - 13 + 6 = -32 ≠ 0

h(2) = 7(2)⁴ - 9(2)³ - 41(2)² + 13(2) + 6 = 7(16) - 9(8) - 41(4) + 13(2) + 6 = -92 ≠ 0

h(-2) = 7(-2)⁴ - 9(-2)³ - 41(-2)² + 13(-2) + 6 = 7(16) - 9(-8) - 41(4) + 13(-2) + 6 = 0;

This means (x + 2) is a factor of the polynomial.

Now, by synthetic division,

(7x⁴ - 9x³ - 41x² + 13x + 6) ÷ (x + 2):

The remainder is 0

The quotient is a polynomial of degree 3.

g(x) = 7x³ - 23x² + 5x + 3

Using the Rational Zeros Theorem,

p: ±1, ±3, which are factors of constant term 3

q: ±1, ±7 which are factors of the leading coefficient 7

All possible values are

p/q: ±1, ±3, ±1/7, ±3/7

g(1) = 7(1)³ - 23(1)² + 5(1) + 3 = -8 ≠ 0

g(-1) = 7(-1)³ - 23(-1)² + 5(-1) + 3 = -32 ≠ 0

g(3) = 7(3)³ - 23(3)² + 5(3) + 3 = 189 - 207 + 15 + 3 = 0

By synthetic division:

The remainder is 0

The quotient is a quadratic polynomial.

7x² - 2x - 1 = 0

Using Quadratic Formula = [-b ± √(b² - 4ac)]/2a

x = -(-2) ± √[(-2)² - (4 × 7 × -1)]/ (2 × 7) = (2 ± √32)/14 = (1 ± 2√2)/7