Q.If $x = -9$ is a root of $\begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix} = 0$, then the other two roots are ________ .

Concept: Determinant Equality Equation — When a determinant equals zero, the variable $x$ satisfies a polynomial equation. Expanding the determinant gives a cubic in $x$, and one root is given; the other two come from the remaining quadratic factor.

Step 1: Expand the determinant

Step 2: Simplify

So the equation is $x^3 - 67x + 126 = 0$.

Step 3: Use the given root $x = -9$ …

We are given that $x = -9$ satisfies the determinant equation:

This is a determinant equality equation — a polynomial equation in $x$ formed by expanding the determinant. The degree of the polynomial is at most 3 (since the matrix is $3 \times 3$ and each term contains at most one $x$ per row/column). So we expect three roots. One root is given; we need the other two.

1. Expand the determinant

Let’s compute the determinant directly. Using the first row expansion:

Compute each $2 \times 2$ determinant:

• First: $x \cdot x - 2 \cdot 6 = x^2 - 12$.
• Second: $2 \cdot x - 2 \cdot 7 = 2x - 14$.
• Third: $2 \cdot 6 - x \cdot 7 = 12 - 7x$.

So:

Simplify term by term:

Combine like terms:

Thus the equation is:

2. Use the given root to factor

We know $x = -9$ is a root. So $(x + 9)$ is a factor. Divide the cubic by $(x + 9)$.

Perform synthetic division with $-9$:

Coefficients: $1$ (for $x^3$), $0$ (for $x^2$ — note there is no $x^2$ term), $-67$, $126$.

The quotient is $x^2 - 9x + 14$, remainder $0$. So: …

Method: Solving a Determinant Equation Using a Known Root (Factor Theorem)

This method solves "a variable-containing determinant equals zero (or a given value), and one root is given — find the rest" problems, for a $3\times3$ (cubic) determinant.

Steps

Step 1: Expand the determinant into a polynomial

Use cofactor expansion (usually along the row/column with the fewest distinct terms) to write the determinant as an explicit polynomial in the unknown — a cubic in $x$ for a $3\times3$ matrix. Use row/column operations first to create zeros if the entries are messy — this keeps the expansion short.

Step 2: Set the polynomial equal to the given value

Rearrange into the standard polynomial form with all terms on one side.

Step 3: Use the given root to factor

If $x = r$ is a known root, then $(x-r)$ is a factor of the polynomial (Factor Theorem). Divide the polynomial by $(x - r)$ — using synthetic division or long division — to get a reduced polynomial (a quadratic, if you started with a cubic). …

Common Mistakes

Mistake 1: Sign errors while expanding the $3\times3$ determinant

Why it's wrong: the cofactor expansion alternates signs $(+,-,+)$ across a row, and it's easy to drop the minus sign on the middle term — this silently changes the coefficient of $x$ in the cubic and produces a wrong polynomial (and wrong roots). Correct approach: write out each $2\times2$ minor and its sign explicitly before combining, and double-check the middle term's negative sign.

Mistake 2: Forgetting the missing power's zero coefficient during synthetic division

Why it's wrong: when a power of $x$ is missing from the cubic (there is no $x^2$ term here), a student can misalign the coefficients during synthetic division and get a wrong quotient. Correct approach: always write the coefficient as $0$ for a missing power before dividing, so every place value lines up correctly. …