Question

1. the product of four integers is $x^{4}+6x^{3}+11x^{2}+6x$, where $x$ is one of the integers. what are possible expressions for the other three integers?

Explanation:

Step1: Factor out the common factor

First, factor out \(x\) from the given polynomial \(x^{4}+6x^{3}+11x^{2}+6x\). We get \(x(x^{3}+6x^{2}+11x + 6)\).

Step2: Use the rational - root theorem

The rational - root theorem states that if a polynomial \(a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{1}x + a_{0}\) has a rational root \(p/q\), then \(p\) is a factor of \(a_{0}\) and \(q\) is a factor of \(a_{n}\). For the polynomial \(x^{3}+6x^{2}+11x + 6\), \(a_{n}=1\) and \(a_{0}=6\). The possible rational roots are \(\pm1,\pm2,\pm3,\pm6\).

Step3: Test the possible roots

Let's test \(x=-1\): \((-1)^{3}+6(-1)^{2}+11(-1)+6=-1 + 6-11 + 6=0\). So, \((x + 1)\) is a factor of \(x^{3}+6x^{2}+11x + 6\).

Step4: Perform polynomial long - division

Divide \(x^{3}+6x^{2}+11x + 6\) by \((x + 1)\). Using polynomial long - division or synthetic division, we have \(x^{3}+6x^{2}+11x + 6=(x + 1)(x^{2}+5x + 6)\).

Step5: Factor the quadratic polynomial

Factor the quadratic polynomial \(x^{2}+5x + 6\). We find that \(x^{2}+5x + 6=(x + 2)(x+3)\).

Step6: Write the fully factored form

So, \(x^{4}+6x^{3}+11x^{2}+6x=x(x + 1)(x + 2)(x + 3)\).

Answer:

The other three integers are \(x + 1\), \(x + 2\), and \(x + 3\).