Question

Factorization of \( x^2 - x - 6 \)

• A. \( (x-3)(x-2) \)
• B. \( (x-3)(x+2) \)
• C. \( (x+3)(x-2) \)
• D. \( (x+3)(x+2) \)

Hint:

To factorize a quadratic expression \( ax^2 + bx + c \): - If \(a=1\) (i.e., \(x^2+bx+c\)), find two numbers that multiply to \(c\) and add to \(b\). Let these numbers be \(p\) and \(q\). The factorization is \( (x+p)(x+q) \). - If \(a \ne 1\), find two numbers that multiply to \(ac\) and add to \(b\). Let these be \(p\) and \(q\). Rewrite the middle term \(bx\) as \(px+qx\), then factor by grouping. - Alternatively, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) to find roots \(x_1, x_2\). Then factorization is \( a(x-x_1)(x-x_2) \).

Answer 1

The Correct Option is B

Step 1: Analyze the quadratic expression \( x^2 - x - 6 \).
We need to find two numbers that: - Multiply to the constant term (-6).
- Add up to the coefficient of the x term (-1).

Step 2: Find the two numbers.
Pairs of factors for -6: - (1, -6) or (-1, 6) → Sums are -5 or 5.
- (2, -3) or (-2, 3) → Sums are -1 or 1.
The pair (2, -3) adds up to \(2 + (-3) = -1\).

Step 3: Rewrite the middle term using these numbers and factor by grouping.
\( x^2 - x - 6 = x^2 + 2x - 3x - 6 \) Group terms: \( = (x^2 + 2x) + (-3x - 6) \) Factor out common factors from each group: \( = x(x+2) - 3(x+2) \) Factor out the common binomial factor \( (x+2) \): \( = (x-3)(x+2) \)
Step 4: Verify by expanding the factored form.
\( (x-3)(x+2) = x(x+2) - 3(x+2) = x^2 + 2x - 3x - 6 = x^2 - x - 6 \).
The factorization is correct.
This matches option (2).