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

Question

i dont understand what rational zeros are.

 

Answer 1

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.

Answer 2

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).