Please solve as polynomial to third degree: the product of three consecutive odd numbers is 6,783. What are the numbers? The solution being 17, 19, and 21, what are those numbers when considering:
x*(x+2)*(x+4)=6,783?
Multiplied out:
x^3+6*x^2+8*x-6,783=0
What are the roots? 17 is a root, but 19 and 21 are not. What am I missing?
The way you have written the equation, x is the smallest of the odd numbers.
x+2 and x+4 are not roots of the equation. Only x=17 is.
The check is that 17*19*21 = 6783
kill me
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To solve the given polynomial equation x^3 + 6x^2 + 8x - 6,783 = 0, there are a few possible approaches. One common method is to use factorization or synthetic division if a root is known. In this case, you mentioned that 17 is a root, so we can start by using synthetic division to divide the polynomial by (x - 17):
```
17│ 1 6 8 -6783
────────────────
1 23 391 666
```
The result of the synthetic division is a new polynomial with coefficients 1, 23, 391, and 666. The quotient represents the remaining polynomial after factoring out (x - 17). Now we have a new cubic polynomial: x^2 + 23x^ + 391x + 666 = 0.
To further find the other two roots, we can try to factorize this new polynomial or use another method such as the Rational Root Theorem. In this case, since we know that the roots are consecutive odd numbers, we can test the remaining odd numbers to find the other roots. We can try x = 19:
```
19│ 1 23 391 666
────────────────
1 42 803 6237
```
The synthetic division gives us a remainder of 6237, which means 19 is not a root. Doing the same process with x = 21:
```
21│ 1 23 391 666
────────────────
1 44 935 7656
```
The synthetic division gives us a remainder of 7656, which means 21 is also not a root. Since none of the remaining odd numbers satisfy the equation, there are no other consecutive odd number roots for this polynomial.
From this analysis, we can conclude that the roots of the polynomial x^3 + 6x^2 + 8x - 6,783 = 0 are 17, 19 (not a root), and 21 (not a root).