Factor completely.
$x^3+x^2-13 x-13$
We are given the polynomial \(P(x) = x^3 + x^2 - 13x - 13\). Our objective is to factor this polynomial completely. Since the polynomial has four terms, a natural approach is to try factoring by grouping. We look for a pattern where the ratio of coefficients in the first two terms is the same as in the last two terms. Here, the ratio \(1:1\) is consistent across both groups \((x^3, x^2)\) and \((-13x, -13)\).
We can group the terms as \((x^3 + x^2) + (-13x - 13)\). From the first group \((x^3 + x^2)\), we factor out the common term \(x^2\), which gives \(x^2(x + 1)\). From the second group \((-13x - 13)\), we factor out \(-13\), which gives \(-13(x + 1)\). Combining these, the expression becomes \(x^2(x + 1) - 13(x + 1)\).
Now we observe that \((x + 1)\) is a common binomial factor. Factoring out \((x + 1)\), we obtain:
\[x^2(x + 1) - 13(x + 1) = (x + 1)(x^2 - 13).\]
We check if \(x^2 - 13\) can be factored further. Since \(13\) is not a perfect square of a rational number, \(x^2 - 13\) is an irreducible quadratic factor over the rational numbers. Thus, the complete factorization is \((x + 1)(x^2 - 13)\).
Factoring polynomials is a fundamental skill in algebra with applications in solving equations and analyzing functions. For instance, if you were modeling the volume of a shipping crate as \(V(x) = x^3 + x^2 - 13x - 13\), factoring this into \((x+1)(x^2-13)\) helps you easily identify the dimensions of the box or find values of \(x\) for which the volume becomes zero. This helps engineers and designers optimize shapes and sizes efficiently.