Question 797714: x^4-14x^3+56x^2-98x+343=0
i have to find the zero of this function, but i don't think i can complete the square or use the quadratic formula. help!?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
In the first place you cannot find "the" zero of a 4th degree polynomial function. That is because, according to the Fundamental Theorem of Algebra, any 4th degree polynomial function has 4 zeros. In the second place, you don't have a function, you have an equation.
\ =\ x^4\ -\ 14x^3\ +\ 56x^2\ -\ 98x\ +\ 343)
is a quartic or 4th degree function. Functions have zeros. nth degree functions have n zeros, counting all multiplicities.
is a quartic equation. Equations have roots, one for each degree of the polynomial.
Use the Rational Roots Theorem. Since the lead coefficient is 1 and 343 is 7 cubed, the only possible rational roots are 7 and -7.
Test with Synthetic Division:
7 | 1 -14 56 -98 343
| 7 -49 49 -343
--------------------------
1 -7 7 -49 0
That worked right off the bat, so now we know that we have
\ =\ (x\ -\ 7)\left(x^3\ -\ 7x^2\ +\ 7x\ -49\right))
Since you don't want to argue with success, let's try the same thing again.
7 | 1 -7 7 -49
| 7 0 49
--------------------------
1 0 7 0
Success again! So now we know:
\ =\ (x\ -\ 7)^2(x^2\ +\ 7))
And since the final factor is a quadratic, it can be solved by any convenient means. If you are only concerned with real number zeros, you can quit now because there are no real roots for  for any positive value of  . However, if you need the complex zeros also, re-write the quadratic factor as an equation thus:
\ =\ 0)
And factor as the difference of two squares:
\left(x\ -\ i\sqrt{7}\right))
In summary, including the complex roots:
For a total of 4 zeros, as advertised.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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