Condition for a general third degree polyomial to a have a particular second degree polynomial as a factor

Question

Hi I have a question that I just cannot answer so I would be very great full for some help :)

If $x^2 +ax-1$ is a factor of $x^3+px^2+qx+r$ show that $q=-(ar+1)$

Welcome to MSE. In order for MathJax commands to be effective, they must be surrounded by $ signs. For example, $x^2$ shows up as $x^2$. — saulspatz, Apr 09 '21 at 17:09
What happens if you perform long division of the cubic polynomial by the quadratic polynomial? — Greg Martin, Apr 09 '21 at 17:13
I tried long division but it didn’t work out so I then used unknown coefficients and that worked out in the end .Thanks so much for your help ! — user912758, Apr 11 '21 at 16:28

score 1 · Answer 1 · answered Apr 09 '21 at 17:19

1

Let $\alpha, \beta $ be the roots of the first polynomial, and $\gamma$ the remaining root of the second polynomial. We need to compute

$q = \alpha\beta+\beta\gamma +\gamma\alpha = \alpha\beta + \gamma(\alpha + \beta)$.

We know that

$\alpha + \beta = -a$

$\alpha\beta = -1$

$\gamma = \frac {-r}{\alpha \beta} = r$

Thus $q = \alpha\beta + \gamma(\alpha + \beta) = -1 - ar=-(ar+1)$

answered Apr 09 '21 at 17:19

Maffred

1

Thank you I finally got it after trying a couple of methods !! – user912758 Apr 11 '21 at 16:25

score 0 · Accepted Answer · answered Apr 09 '21 at 18:10

0

Since both leading coefficients equal $1$ we know that for some number $s$ it is the case that

$$ (x^2+ax-1)(x-s)=x^3+px^2+qx+r $$

Multiplying the polynomials on the right gives

$$ x^3+(a-s)x^2-(1+as)x+s=x^3+px^2+qx+r$$

The coefficients of equal powers must be equal. Therefore

answered Apr 09 '21 at 18:10

2 Answers2