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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