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rational zeros for f(x)=x^4-4x^3+x^2+16x-20

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2 Answers

Find the rational zeros of f(x)=x^4-4x^3+x^2+16x-20.

This question would come in a unit on polynomial functions after the introduction of the Rational Root Theorem. The student would be expected to solve it this way:

A rational zero is a rational number, say p/q to lowest terms, that makes f(p/q) = 0.

The rational zero theorem says that p is a factor of the constant term of f, and q is a factor of the leading coefficient of f.

So all the possible rational zeros of f are:

factors of -20         ±1, ±2, ±4, ±5, ±10, ±20
--------------- =>  --------------------------------
factors of 1            ±1

=> ±1, ±2, ±4, ±5, ±10, and ±20 are the only possible rational zeros.

Now they need to be tested:

2 |  1  -4   1  16 -20
---       2  -4  -6   20
      ------------------
      1 -2  -3  10  | 0
                         ----

So f(x) = (x - 2) (x^3 - 2 x^2 - 3 x + 10).

2 | 1 -2  -3  10
---     2   0   -6
     -------------
     1  0  -3  | X => no double zero
                  ----

-2 | 1 -2 -3  10
----    -2  8 -10
      ------------
      1 -4  5 | 0
                 ----

So f(x) = (x + 2) (x - 2) (x^2 - 4 x + 5).

For x^2 - 4 x + 5 = 0,
b^2 - 4ac = 16 - 20 = -4,
x = (-(-4) ± sqrt(-4))/2 = 2 ± i, complex conjugates.

So f(x) has only two rational zeros: x = ±2

Notes: Rational implies Real. Real does not imply Rational. The fourth degree polynomial function
g(x) = 5 (x ±√2) (x ±√3) = (x^2 - 2) (x^2 - 3) has 4 real zeros but none of them are rational.
x4-4x3+x2+16x-20 = 0
 
Rational zeros are those values of x for which f(x) = 0, that is to say, the x intercepts.
 
In this case, f(x) factors into
 
(x - 2)(x + 2)(x2 - 4x + 5)
 
from which it is easily seen that x = ±2 are two of the points at which f(x) = 0.
 
Are there others?
 
Well, we are left over with a quadratic equation which cannot be factored into a form (x ± n1)(x ± n2), so let's try the quadratic equation:
 
x = [-b ± (b2 - 4ac)½]/2a
 
In this case a = 1, b = -4, and c = 5
 
We can readily see that the expression (b2 - 4ac) < 0 [16 - 4(1)(5)] = -4 and the square root of -4 is imaginary.
 
Therefore x = ±2 are the only real zeros of f(x).