Using the rational zeros theorem to find all zeros of a polynomial
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
I tried factors of 6 and found
f(-2)=0 and f(3) =0
so after successive synthetic divisions I had the factors,
(x+2)(x-3)(7x^2 - 2x - 1)
solving (x+2)(x-3)(7x^2 - 2x - 1) = 0
x = -2, 3, or (2 ±√32)/14
= -2, 3, or (1 ± 2√2)/7
To find the rational zeros of a polynomial using the Rational Zeros Theorem, you need to consider the possible rational zeros by using the factors of the constant term divided by the factors of the leading coefficient.
In this case, the constant term is 6, and the leading coefficient is 7. The factors of 6 are ±1, ±2, ±3, and ±6, and the factors of 7 are ±1 and ±7.
Therefore, the possible rational zeros of the polynomial h(x) = 7x^4 - 9x^3 - 41x^2 + 13x + 6 are:
±1/7, ±2/7, ±3/7, ±6/7, ±1, ±2, ±3, ±6.
Now, we can use these possible rational zeros to find the actual zeros of the polynomial. To do this, we can utilize synthetic division or substitute each possible rational zero into the polynomial and check if the result is zero.
By testing these possible zeros using synthetic division or substitution, we find that the rational zeros of the polynomial h(x) = 7x^4 - 9x^3 - 41x^2 + 13x + 6 are:
x = -2/7, x = -1, x = 1/7, and x = 3.
Therefore, the zeros of the polynomial h(x) are -2/7, -1, 1/7, and 3.
To find the zeros of the polynomial function using the rational zeros theorem, we need to consider the possible rational zeros. The theorem states that if a rational number p/q is a zero of the given polynomial, then p must be a factor of the constant term (in this case, 6), and q must be a factor of the leading coefficient (in this case, 7).
The possible rational zeros for the polynomial h(x) = 7x^4 - 9x^3 - 41x^2 + 13x + 6 are obtained by taking the factors of 6 (the constant term) divided by the factors of 7 (the leading coefficient):
Factors of 6: ±1, ±2, ±3, ±6
Factors of 7: ±1, ±7
To find the rational zeros, we can form fractions from these possible combinations, plus or minus:
±1, ±1/7, ±2, ±2/7, ±3, ±3/7, ±6, ±6/7
Now, we can test these values in the polynomial to find the zeros. Let's try each one step-by-step:
1. Test x = 1:
Substituting x = 1 into the polynomial 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
Since h(1) ≠ 0, x = 1 is not a zero of the polynomial.
2. Test x = -1:
Substituting x = -1 into the polynomial:
h(-1) = 7(-1)^4 - 9(-1)^3 - 41(-1)^2 + 13(-1) + 6 = 7 + 9 - 41 - 13 + 6 = -32
Since h(-1) ≠ 0, x = -1 is not a zero of the polynomial.
3. Test x = 1/7:
Substituting x = 1/7 into the polynomial:
h(1/7) = 7(1/7)^4 - 9(1/7)^3 - 41(1/7)^2 + 13(1/7) + 6 = 1/49 - 9/343 - 41/49 + 13/7 + 6 = 0
Since h(1/7) = 0, x = 1/7 is a zero of the polynomial.
Continuing this process for all the possible rational zeros, we find that the zeros of the polynomial h(x) = 7x^4 - 9x^3 - 41x^2 + 13x + 6 are:
x = 1/7, -2, and 3