Write $f(x) =x^3+x^2-3x-3$ as a product of a linear factor and a quadratic factor

Question

I know that the linear factor is $(x+1) (x^2-3),$ but how would I find the quadratic factor?

Answer 1

It is a polynomial of degree 3, and it is the product of the quadratic factor $x^2-3$ and the linear factor $x+1$. But maybe I miss your point.

Answer 2

Simply calculate the roots of your polynomial (if possible). You have a polynomial of degree 3. Normally you first try to find a root, by "guessing". We have to check the divesors of the constant -3. Which are $\pm 3, \pm 1$.

Easily we see, that x=-1 is a root and then make a long division

$(x^3+x^2-3x-3)\div (x+1)$ giving us the quadratic factor.

On other way is to compare coefficients by:

$(x^3+x^2-3x-3)=(ax+b)(cx^2+dx+e)$

Which should involve more calculation.

Answer 3

You can use Horner's algorithm for the division by $x+1$: \begin{array}{r|rrrr} &1&1&-3&-3\\ \times -1&\downarrow&\nearrow -1&0&3\\ \hline &\color{red}1&\color{red}0&\color{red}{-3}&0 \end{array} so the quotient is $\;x^2+ 0\mkern 1.5mux-3=x^2-3$.