Determinant factorisation using specific method

Question

Consider the determinant $$\begin{vmatrix}x+4&2&3\\-1&x+1&-3\\0&2&x+7\end{vmatrix}$$

I know that $x+3$ and $x+4$ are factors of this determinant. I wish to find the final factor (which is $(x+5)$) specifically by considering the coefficients of $x^2$ and $x^3$. How can I do this? Using this particular method?

Answer 1

Method $1$

If you already know that $(x+5)$ is a factor of the determinant, put $x = -5$ in the determinant

$$\begin{vmatrix}x+4&2&3\\-1&x+1&-3\\0&2&x+7\end{vmatrix}$$

and check if it vanishes.

$$\begin{vmatrix}-1&2&3\\-1&-4&-3\\0&2&2\end{vmatrix} = \begin{vmatrix}0&6&6\\-1&-4&-3\\0&2&2\end{vmatrix} = \begin{vmatrix}0&0&0\\-1&-4&-3\\0&2&2\end{vmatrix}=0$$

Method $2$

Expand the determinant $$\begin{vmatrix}x+4&2&3\\-1&x+1&-3\\0&2&x+7\end{vmatrix} = x^3 + 12x^2 + 47x + 60$$

and consider the equation $$x^3 + 12x^2 + 47x + 60 = 0$$

Sum of the three roots of the above equation is $-12$

If two roots are known to be $-3$ and $-4$, the third root must be $-12 - (-3-4) = -5$

Answer 2

Suppose the determinant is $f(\lambda)=(x+4)(x+3)(x-\lambda)$. (No coefficient in front as is clear from the main diagonal).

Substitute a value of $x$ different from $-3,-4$ into the determinant and solve for it. (Easiest would be expanding along $C_1$ or $R_3$.)

Set that numerical value equal to $f(\lambda)$ which is a linear equation in $\lambda$. You should obtain $\lambda=-5$.

Answer 3

In the development of the determinant by Sarrus' rule, no term is a quadratic polynomial, so only

$$(x+4)(x+1)(x+7)$$ yields cubic and quadratic terms, namely $$x^3+12x^2.$$

Hence by Vieta, the third root must be $$-12+3+4.$$