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Factorization of Expressions of the Form x\(^{2}\) + (a + b)x + ab
Here we will learn the process of Factorization of Expressions of the Form x\(^{2}\) + (a + b)x + ab.
We know, (x + a)(x + b) = x\(^{2}\) + (a + b)x + ab.
Therefore, x\(^{2}\) + (a + b)x + ab = (x + a)(x + b).
1. Factorize: a\(^{2}\) + 7a + 12.
Solution:
Here, constant term = 12 = 3 × 4, and 3 + 4 = 7 (= coefficient of a).
Therefore, a\(^{2}\) + 7a + 12 = a\(^{2}\) + 3a + 4a + 12 (breaking 7a is sum of two terms, 3a + 4a)
= (a\(^{2}\) + 3a) + (4a + 12)
= a(a + 3) + 4(a + 3)
= (a + 3)(a + 4).
2. Factorize: m\(^{2}\) – 5m + 6.
Solution:
Here, constant term = 6 = (-2) × (-3), and (-2) + (-3) = -5 (= coefficient of m).
Therefore, m\(^{2}\) – 5m + 6 = m\(^{2}\) -2m – 3m + 6 (breaking -5m is sum of two terms, -2m - 3m)
= (m\(^{2}\) -2m) +(– 3m + 6)
= m(m - 2) - 3(m - 2)
= (m - 2)(m - 3).
3. Factorize: x\(^{2}\)- x - 6.
Solution:
Here, constant term = -6 = (-3) × 2, and (-3) + 2 = -1 (= coefficient of x).
Therefore, x\(^{2}\) - x - 6 = x\(^{2}\) - 3x + 2x - 6 (breaking -x is sum of two terms, -3x + 2x)
= (x\(^{2}\) - 3x) + (2x - 6)
= x(x - 3)+ 2(x - 3)
= (x - 3)(x + 2).
The method of factorizing x\(^{2}\) + px + q by breaking the middle term, as shown in the above examples, involves the following steps.
Steps:
1. Take the constant term (with the sign) q.
2. Break q into two factors, a, b (with suitable signs) whose sum equals the coefficient of x, i.e., a + b = p.
3. Pair one of these, say, ax with x\(^{2}\), and the other, bx, with the constant term q. Then factorize.
Note: In case step 2 is not possible conveniently, x\(^{2}\) + px + q cannot be factorized as above.
For example, x\(^{2}\) + 3x + 4. Here 4 cannot be broken into two factors whose sum is 3.
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