Question 608390
{{{2x^4 + 19x^3 + 23x^2 + 5x - 1}}}
NOTE: The other tutor's solution only provides <u>possible rational</u> zeros. They are not all actually zeros. The "missing" zeros will be either irrational (which is another subset of real numbers) or complex.<br>
We'll start by using the other tutors work. By definition, z zero will turn a polynomial into a zero. (Hence, then name.) Since powers of 1 are so easy, one can easily check to see if 1 is a zero. You should be able to see that it is not. The polynomial is equal to 48 when x = 1. Checking -1 is not as easy but many can do it in your head. The even powers of -1 are 1's and the odd powers of -1 are -1. We should find that -1 <i>is</i> a zero.<br>
TO find the other zeros, it helps if you factor the expression using the zero(s) you've already found. Since -1 is a zero, then (x - (-1)) or (x + 1) is a factor. Using synthetic (or long) division we can find the other factor:
<pre>
-1    2   19   23   5   -1
          -2  -17  -6    1
      ---------------------
      2   17    6  -1    0
</pre>The 0 in the lower right corner tells us the remainder (and the value of the polynomial when x = -1). The rest of the bottom line tells us the other factor. So
{{{2x^4 + 19x^3 + 23x^2 + 5x - 1 = (x+1)(2x^3+17x^2+6x-1)}}}
We will use the other factor to find the other zeros. Since powers of 1/2 are not so easy to do mentally, we will use synthetic division to test them:
<pre>
1/2   2   17    6   -1
           1    9   15/2
     -------------------
      2   18   15   33/2
The remainder is 23/2, not zero. So 1/2 is not a zero.
-1/2  2   17    6   -1
          -1   -8    1
     -------------------
      2   16   -2    0
</pre>
The remainder is 0. SO -1/2 is a zero of the polynomialand (x - (-1/2) or (x + 1/2) is a factor:
{{{2x^4 + 19x^3 + 23x^2 + 5x - 1 = (x+1)(x+1/2)(2x^2+16x^-2)}}}
We have run out os possible rational roots. But the remaining factor is quadratic so we can use the Quadratic Formula to find the remaining zeros:
{{{x = (-(16) +- sqrt((16)^2 - 4(2)(-2)))/2(2)}}}
Simplifying...
{{{x = (-(16) +- sqrt(256 - 4(2)(-2)))/2(2)}}}
{{{x = (-(16) +- sqrt(256 +16))/2(2)}}}
{{{x = (-(16) +- sqrt(272))/2(2)}}}
{{{x = (-16 +- sqrt(272))/4}}}
{{{x = (-16 +- sqrt(16*17))/4}}}
{{{x = (-16 +- sqrt(16)*sqrt(17))/4}}}
{{{x = (-16 +- 4*sqrt(17))/4}}}
{{{x = (4(-4 +- sqrt(17)))/4}}}
{{{x = (cross(4)(-4 +- sqrt(17)))/cross(4)}}}
{{{x = -4 +- sqrt(17)}}}
So the four real zeros are:
-1, -1/2, {{{-4 + sqrt(17)}}} and {{{-4 - sqrt(17)}}}