Question 3: Factorize:
\((viii)\) \(\; \) \(x^2-x-2 \)
Solution:
\(\ \ \ \ x^2-x-2\)
\(= x^2+x-2x-2\)
\(= x(x+1)-2(x+1)\)
\( =(x+1)(x-2)\)
Explanation:
Multiply the coefficient of \(x^2\) (which is \(1\)) with the constant term \(2\):
\(1 \times 2 = 2\)
List the factor pairs of \(2\), assigning a negative sign to larger numbers (since the middle term and the constant are negative).
Then, Find the sum of each pair and select the one whose sum equals the coefficient of the middle term, which is \(-1\).
\((1, -2)\) \(\Rightarrow\) \(\text{sum} = \mathbf{-1} \quad \checkmark\)
So, the correct pair is \((1, -2)\)
Use this pair to split the middle term:
\(x^2-x-2\) \(\ =\ \)\( x^2 +x – 2x -2\)
Group the terms and factor: \(= (x^2 + x) + (-2x -2)\)
Take common from first two terms and last terms: \(= x(x +1) -2 (x +1)\)
Again, take common from both terms: \(= (x + 1)(x – 2)\)