Question 3: Factorize:
\((i)\) \(\;\) \( x^2+x-12\)
Solution:
\(\ \ \ \ x^2+x-12\)
\(= x^2+4x-3x-12\)
\(= x(x+4)-3(x+4)\)
\( =(x+4)(x-3)\)
Explanation:
Multiply the coefficient of \(x^2\) (which is \(1\)) with the constant term \(12\):
\(1 \times 12 = 12\)
List the factor pairs of \(12\), assigning a negative sign to the smaller number (since the middle term is positive and the constant is negative).
Then, Find the sum of each pair and select the one whose sum equals the coefficient of the middle term, which is \(1\).
\((-1, 12)\) \(\Rightarrow\) \(\text{sum} = 11\)
\((-2, 6)\) \(\Rightarrow\) \(\text{sum} = 4\)
\((-3, 4)\) \(\Rightarrow\) \(\text{sum} = \mathbf{1} \quad \checkmark\)
So, the correct pair is \((-3, 4)\)
Use this pair to split the middle term:
\(x^2 + x – 12\) \(\ =\ \)\( x^2 – 3x + 4x – 12\)
Group the terms and factor: \(= (x^2 – 3x) + (4x – 12)\)
Take common from first two terms and last terms: \(= x(x – 3) + 4(x – 3)\)
Again, take common from both terms: \(= (x + 4)(x – 3)\)