To find all zeros of the polynomial function h(x) = 7x4 + 9x3 + 41x2 + 13x + 6, we can start by using the Rational Root Theorem. This theorem states that any rational solution, or zero, of the polynomial equation is of the form ±p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
Here, the constant term (6) has the factors: ±1, ±2, ±3, ±6, and the leading coefficient (7) has the factors: ±1, ±7. Therefore, the possible rational zeros can be:
- ±1, ±2, ±3, ±6, ±1/7, ±2/7, ±3/7, ±6/7
Next, we can test these possible rational zeros by substituting them into the function h(x) to see which values yield h(x) = 0.
After testing, we find that:
- h(-1) = 7(-1)4 + 9(-1)3 + 41(-1)2 + 13(-1) + 6 = 7 – 9 + 41 – 13 + 6 = 32 (not a zero)
- h(-2) = 7(-2)4 + 9(-2)3 + 41(-2)2 + 13(-2) + 6 = 112 – 72 + 164 – 26 + 6 = 184 (not a zero)
- h(-3) = 7(-3)4 + 9(-3)3 + 41(-3)2 + 13(-3) + 6 = 567 – 243 + 369 – 39 + 6 = 660 (not a zero)
- h(-6) = 7(-6)4 + 9(-6)3 + 41(-6)2 + 13(-6) + 6 = 4536 – 1944 + 1476 – 78 + 6 = 3996 (not a zero)
- h(1) = 7(1)4 + 9(1)3 + 41(1)2 + 13(1) + 6 = 7 + 9 + 41 + 13 + 6 = 76 (not a zero)
- h(2) = 7(2)4 + 9(2)3 + 41(2)2 + 13(2) + 6 = 112 (not a zero)
- h(3) = 7(3)4 + 9(3)3 + 41(3)2 + 13(3) + 6 = 413 (not a zero)
- h(6) = 7(6)4 + 9(6)3 + 41(6)2 + 13(6) + 6 = 1536 (not a zero)
- h(1/7) = 7(1/7)4 + 9(1/7)3 + 41(1/7)2 + 13(1/7) + 6 = 6 (not a zero)
- h(-1/7) = 6 (not a zero)
After testing all rational roots, we find rational zeros: 2/7 and -3/7. To find the remaining zeros, we can factor the polynomial:
Using synthetic division or polynomial long division, we can divide by (x – 2/7) and (x + 3/7) to simplify to a quadratic. Solving the quadratic equation will give the exact roots.
After completing the factorization process, you will find all zeros of the function.
In summary, the zeros of the function are approximately given by the rational roots, along with the other roots obtained through factorization.