Because the leading coefficient of \(g(x)=-x^3+3x^2+3x-1\) is negative, we’ll never get a row of all non-negative numbers from the synthetic division. But since the set of zeros of \(g ( x )\) is the same as the set of zeros of \(- g ( x )\), we will use \(- g ( x )\) instead of \(g ( x )\) in the synthetic division process.
| \(1\) | \(-3\) | \(-3\) | \(1\) | |
| \(1\) | \(1\) | \(1\) | \(-4\) | \(-6\) |
| \(2\) | \(1\) | \(-1\) | \(-5\) | \(-9\) |
| \(3\) | \(1\) | \(0\) | \(-3\) | \(-8\) |
| \(4\) | \(1\) | \(1\) | \(1\) | \(5\) |
Since \(4\) is the first positive integer to result in a row of non-negative numbers, \(4\) is the smallest positive integer which is an upper bound on all the positive zeros of \(g(x)=-x^3+3x^2+3x-1\).
This is the same result as in problem 2.5.6. In problem 2.5.6 we had \(f(x)=x^3-3x^2-3x+1\) whereas in problem 2.5.7 we had \(g(x)=-x^3+3x^2+3x-1\) which is the negative of \(f(x)\). The zeros of the negative of a function are the same as the zeros of the function. The graph of one is the graph of the other reflected in the \(x\)-axis, therefore they have the same \(x\)-intercepts, and the zeros of a functions are the \(x\)-coordinates of the \(x\)-intercepts.