# The function below has at least one rational zero. Use this fact to find all zeros of the function h(x) = 7x^{4} - 9x^{3} - 41x^{2} + 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_{n}x^{n} + a_{n - 1}x^{n - 1} +...+ a_{1}x + a_{0} 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) = 7x^{4} - 9x^{3} - 41x^{2} + 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,

(7x^{4} - 9x^{3} - 41x^{2} + 13x + 6) ÷ (x + 2):

The remainder is 0

The quotient is a polynomial of degree 3.

g(x) = 7x^{3} - 23x^{2} + 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.

7x^{2} - 2x - 1 = 0

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

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

## The function below has at least one rational zero. Use this fact to find all zeros of the function h(x) = 7x^{4} - 9x^{3} - 41x^{2} + 13x + 6. If more than one zero, separate with commas. Write exact values, not decimal approximations.

**Summary:**

The function h(x) = 7x^{4} - 9x^{3} - 41x^{2} + 13x + 6 has four zeroes, they are, -2, 3, (1 + 2√2)/7, (1 - 2√2)/7.

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