Question

What is the factorization of the polynomial?
-x^2 - 15x - 55

Answer 1

To factor the polynomial \(-x^2 - 15x - 55\), we can first factor out \(-1\) to make the leading coefficient positive. This gives us:

\[
-x^2 - 15x - 55 = -1(x^2 + 15x + 55)
\]

Next, we will focus on factoring the quadratic \(x^2 + 15x + 55\). To factor this quadratic expression, we need to find two numbers that multiply to \(55\) (the constant term) and add up to \(15\) (the coefficient of \(x\)).

The integer pairs that multiply to \(55\) are:
- \(1\) and \(55\)
- \(5\) and \(11\)

Now, we observe that neither of these pairs adds up to \(15\). This indicates that \(x^2 + 15x + 55\) cannot be factored into linear factors with integer coefficients.

Thus, we can conclude that the polynomial \(x^2 + 15x + 55\) is irreducible over the integers. However, we can leave the polynomial in its factored form when considering the factor of \(-1\):

\[
-x^2 - 15x - 55 = -1(x^2 + 15x + 55)
\]

If you are looking for a complete factorization over the real numbers, you can also use the quadratic formula to express it in terms of its roots:

The roots can be calculated using:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where:
- \(a = 1\)
- \(b = 15\)
- \(c = 55\)

Calculating the discriminant:

\[
b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 55 = 225 - 220 = 5
\]

So, the roots are:

\[
x = \frac{-15 \pm \sqrt{5}}{2}
\]

Therefore, the polynomial can be expressed in factored form involving these roots:

\[
-x^2 - 15x - 55 = -1\left(x - \frac{-15 + \sqrt{5}}{2}\right)\left(x - \frac{-15 - \sqrt{5}}{2}\right)
\]

So, the complete factorization is:

\[
-x^2 - 15x - 55 = -1\left(x - \frac{-15 + \sqrt{5}}{2}\right)\left(x - \frac{-15 - \sqrt{5}}{2}\right)
\]

But as mentioned before, over the integers, we have:

\[
-x^2 - 15x - 55 = -1(x^2 + 15x + 55)
\]