Problem

As an alternative to evaluating polynomials by direct substitution, nested factoring can b...

As an alternative to evaluating polynomials by direct substitution, nested factoring can be used. The method has the advantage of using only products and sums−no powers. For P = x3 + 3x2+ 1x+ 5, we begin by grouping all variable terms and factoring x: P = [x3 + 3x2 + 1x] + 5 = x[x2 + 3x + 1] + 5. Then we group the inner terms with x and factor again: P = x[x2 + 3x+ 1] + 5 = x[x(x+ 3) + 1] + 5.

The expression can now be evaluated using any input and the order of operations. If x= 2, we quickly find that P = 27. Use this method to evaluate H = x3 + 2x2 + 5x− 9 for x= −3.

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