Question 697489: Find the complex zeros of the polynomial function. Write f in factored form.
Use the complex zeros to write f in factored form.
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website!
The problem tells us nothing about any zeros or factors so we will have to find them on our own. Always start factoring by factoring our the greatest common factor (unless it is a 1 which we rarely bother factoring out). The GCF is a 1 so we will not factor it out.
Next we usually try factoring by patterns or trinomial factoring because they are often easier than the other two factoring techniques. However out expression has too many terms for a trinomial or for any of the patterns.
Now we are left with factoring by grouping or with factoring by trial and error of the possible rational roots. Factoring by grouping requires an even number of terms. Since we have 5 terms we would have to split the  into two parts before we try to use factoring by grouping. Since I do not see which particular split ( ,  ,  , etc.) I am going to try to avoid using this method an go on to factoring by trial and error of the possible rational roots.
The possible rational roots of a polynomial are all the ratios, positive and negative, which can be formed by a factor of the constant term (at the end) over a factor of the leading coefficient (at the beginning). Our constant term is 64 and our leading coefficient is 3. The factors of 64 are 1, 2, 4, 8, 16, 32, 64 and the factors of 3 are just 1 and 3. So the possible rational roots of f are:
+1/1 (or +1), +2/1 (or +2), +4/1 (or +4), +8/1 (or +8), +16/1 (or +16), +32/1 (or +32), +64/1 (or +64), +1/3, +2/3, +4/3, +8/3, +16/3, +32/3, +64/3
With so many possible rational roots it could take quite a while to find some roots. It will help if we use our brains to eliminate some of the possible roots logically. With only two terms with negative coefficients, it is unlikely the larger positive roots could make f be zero. So we will not try 4, 8, 16, 32 or 64 (unless nothing else works). Let's try 2 (using synthetic division which I hope you have learned):
2 | 3 -19 54 176 -64
---- 6 -26 56 264
--------------------------
3 -13 28 232 200
The 200 in the lower right corner is the remainder. The remainder is f(2). Since f(2) is not zero, 2 is not a root. (And the fact that it is a large positive number tells us that even 2 was too large of a positive root to work.)
We can try 1 and -1 mentally since powers of 1 and -1 are easy:
f(1) = 3 - 19 + 54 + 176 -64 = 150.
f(-1) = 3 + 19 + 54 - 176 -64 = -164
Neither of them work. But the fact that f(1) is positive and f(-1) is negative tells us that there is a root between them. Since zero is nearly halfway between 150 and -164 and since 150 is a little closer to zero, I'm going to guess that 1/3 might be a root. Let's see:
1/3 | 3 -19 54 176 -64
---- 1 -6 16 64
--------------------------
3 -18 48 192 0
Bingo! The remainder is zero so f(1/3) = 0 and (x - 1/3) is a factor. Not only that, the rest of the bottom line is the other factor of f. "3 -18 48 192" translates into:  . So:
We can simplify this a little if we factor out a 3 from the second factor:
and then multiply (x - 1/3) by the 3:
Now we continue to factor. The last factor,  , has the same constant term, 64, but the leading coefficient is now a 1. So we have the same list of possible rational roots as before except for the fractions. And since a rational root that didn't work before (-1, 1, 2, 4, 8, 16. 32 and 64) cannot magically start working later we only have -2, -4, -8, -16, -32 and -64 left to try. Let's start with -2:
-2 | 1 -6 16 64
---- -2 16 -64
------------------
1 -8 32 0
The remainder is zero. So f(-2) = 0 and (x - (-2)) or (x + 2) is a factor. And the other factor is  :
The last factor is a quadratic. So we can use alternate factoring techniques. But no matter what technique we use, it will not factor. But we can use the quadratic formula to find its roots:
which simplifies as follows:
which is short for:
 or 
With these two roots we can write the remaining factors:
(x - (4 + 4i)) and (x - (4 - 4i))
which simplify to:
(x - 4 - 4i)) and (x - 4 + 4i))
Adding these two factors to our factored f:
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