Question 402814
{{{P(x) = 3x^5+5x^4+10x^3+14x^2+3x-3}}}
To find the roots of this polynomial we need to factor it. The Greatest Common Factor (GCF) is 1 (which we rarely factor out). There are 6 terms in this poynomial which is too many for the factoring patterns and for factoring trinomials. So we are left with factoring by grouping and factoring by trial and error of the possible rational roots.<br>
I don't immediately see how factoring by grouping will work. And since there are only a few possible rational roots, I am going to use the trial and error method.<br>
The possible rational roots of any polynomial are all the fractions that can be formed, positive and negative, using a factor of the constant term (at the end) over a factor of the leading coefficient. In this polynomial your constant term is -3 and your leading coefficient is a 3. So the possible rational roots of this polynomial are:
3/1 (or 3), -3/1 (or -3), 1/1 (or 1), -1/1 (or -1), 1/3, -1/3
This is a trial and error method. We do not know which, if any, of these numbers are roots. With some polynomials we can be "clever" and figure out how to postpone attempts of roots that are unlikely to work.<br>
With this polynomial, with all its positive coefficients and one small negative constant, it is unlikely that any of the possible positive roots will works. The roots of a polynomial will make it zero. 5 positive numbers a -3 are not likely to add up to zero. (This is not a guarantee that the positive roots will not work. But it makes sense to try the negative ones first.)<br>
The easiest way to check for most roots is synthetic division. We'll start by trying -1:
<pre>
-1 |  3  5  10  14   3  -3
----    -3  -2  -8  -6   3
      ------------------
      3  2   8   6  -3   0
</pre>
The lower right corner is the remainder. It is so 0 so -1 is a root and (x-(-1)) or (x+1) is a factor of P(x). The other factor is found in the rest of the bottom row. The "3 2 8 6 -3" translates into {{{3x^4+2x^3+8x^2+6x-3}}}. So:
{{{P(x) = (x+1)(3x^4+2x^3+8x^2+6x-3)}}}
We will continue to factor but we will be factoring the second factor, {{{(3x^4+2x^3+8x^2+6x-3)}}}. The constant terms is still -3 and the leading coefficient is still 3. So the list of possible rational roots has not changed. And we will continue to ttry the possible negative roots. We'll try -1 again:
<pre>
-1 |  3   2   8   6  -3
----     -3   1  -9   3
     ------------------
      3  -1   9  -3   0
</pre>
Another winner! So (x+1) is a factor once again. And the other factor is {{{3x^3=x^2+9x-3}}}. So
{{{P(x) = (x+1)(x+1)(3x^3-x^2+9x-3)}}}
With the terms of the last factor alternating, + - + -, there will be no further negative roots. (Think about this and try some negative numbers. It should become clear that the terms with the + in front will all be negative because the negative value for x is raised to and odd power. The other terms will be negative, too because the exponent is even which results in a positive but the term as a "-" in front. So when x is negative, <i>all</i> the terms will be negative (which cannot add up to zero.)<br>
Since the positive roots did not look promising earlier, I am going to try the smallest one, 1/3, next:
<pre>
1/3 |   3  -1   9  -3
-----       1   0   3
       --------------
        3   0   9   0
</pre>
Another winner! So {{{(x-1/3)}}} is yet another factor. Now we have:
{{{P(x) = (x+1)(x+1)(x-1/3)(3x^2+9)}}}
The last factor is a quadratic. Other than factoring out a 3:
{{{P(x) = (x+1)(x+1)(x-1/3)3(x^2+3)}}}
Any more factoring is not easy. But the last factor is quadratic. So we can use the Quadratic Formula on it:
{{{x = (-(0) +- sqrt((0)^2-4(1)(3)))/2(1)}}}
which simplifies as follows:
{{{x = (-(0) +- sqrt(0-4(1)(3)))/2(1)}}}
{{{x = (-(0) +- sqrt(-12)))/2(1)}}}
{{{x = (0 +- sqrt(-12)))/2}}}
{{{x = (0 +- sqrt(-1)*sqrt(4)*sqrt(3)))/2}}}
{{{x = (0 +- i*2*sqrt(3)))/2}}}
{{{x = (0 +- i*cross(2)*sqrt(3))/cross(2)}}}
{{{x = 0 +- i*sqrt(3)}}}
In long form this is:
{{{x = 0 + i*sqrt(3)}}} or {{{x = 0 - i*sqrt(3)}}}
or
{{{x = i*sqrt(3)}}} or {{{x = -i*sqrt(3)}}}<br>
So the 5 roots for your fifth degree polynomial are:
-1, -1, 1/3, {{{i*sqrt(3)}}}, {{{-i*sqrt(3)}}}<br>
Mote 1: The -1 counts twice because it was a factor twice.
Note 2: Notice how the complex roots were a conjugate pair. (We'll be using this in the next problem.)<br>
For <i>every</i> polynomial with real coefficients, its complex roots, if any, will come in conjugate pairs. This is important for problems like these. You're asked for a 5th degree polynomial which should have five roots. But you've only been given three. Two of the roots you've been given are complex (because of the "i"). The "missing" two roots are the conjugates of the two you were given. So the five roots are:
-1, 1+2i, 1-2i, -3i and 3i
And if these are the roots then the factors are:
P(x) = (x-(-1))(x-(1+2i))(x-(1-2i))(x-(-3i))(x-(3x))
or
P(x) = (x+1)(x-1-2i)(x-1+2i)(x+3i)(x-3x)
Now all that needs to be done now is to multiply that out. I'll leave that for you to do. Since this is one big product we can use the Commutative and Associative properties in any way we choose. Here's an order I recommend:<ol><li>(x+3i)*(x-3x)</li><li>(x-1-2i)*(x-1+2i)</li><li>(answer from step 1)*(answer from step 2)</li><li>(answer from step 3)(x+1)</li></ol>
Try to use the {{{(a+b)(a-b) = a^2-b^2}}} pattern to multiply the conjugate pairs (steps 1 and 2). If you can see how the pattern fits, the multiplication of the conjugates will go faster.<br>
By using the order I've suggested, you will find that the i's disappear! So after that there are no "i" terms to complicate the rest of the multiplications.