To find where the value of the expression is positive (or negative), we want to find the values for which it is 0.
For an algebraic approach, start with the rational root theorem which tells us the possible rational roots are 1, -1, 2, -2, 4, and -4.
Checking to see if 1 is a root is easy simply by substituting x=1 in the expression. In this example, we quickly see that 1 is a root: -1+5-8+4=0.
x=1 is a root, so (x-1) is a factor of the expression. Reduce the expression to a quadratic by removing the factor of (x-1) using long division or synthetic division.
1 | -1 5 -8 4
| -1 4 -4
+-------------
-1 4 -4 0
The reduced polynomial is
So the completely factored polynomial is
The zeros of the expression are 1 and 2; those are the only values of x where, as you "walk" along the x axis, the value of the function can change from positive to negative or vice versa.
To answer the question, we need to find the values of x for which the value of the expression is 0 or positive.
To do that, we can "walk" along the x axis. The way I find easiest to do that is to start with a "large" positive value of x and walk along the x-axis to the left.
The factored form of the expression is }.
For "large" values of x (anything greater than the largest root, 2), the signs of the factors are -+++, so the product is negative. So the given inequality is NOT satisfied for large values of x.
Walking left along the x-axis, nothing changes until we reach the largest root, 2. When we pass x=2, the signs of TWO factors change at the same time, so the sign of the expression does not change. The signs of the factors are -+--, so the product is still negative. Since the given inequality is for greater than OR EQUAL TO 0, the inequality is satisfied at the single point x=2.
Continuing our walk to the left, when we pass the next root at x=1, the sign of a single factor changes, so the signs of the factors are now ----, and the product is positive, and the given inequality is satisfied. And x=1 is the smallest root, so the inequality is satisfied for x=1 and for all values of x less than 1.
Our final answer is that the value of the given expression is greater than or equal to zero for and for the single value x=2.
ANSWER (in interval notation): (-infinity,1] U [2,2]
A graph showing that solution set: