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Learn to factorize the expression of the form $a{x}^{2}+\mathit{bx}+c$.

Procedure to factorize the expression.

Step 1: Determine the coefficient \(a, b\) and \(c\).

Step 2: Calculate the product of \(a\) and \(c\). Product \(= ac\) and sum \(= b\). Thus the middle coefficient is the sum and extreme product is the product value.

Step 3: Express the middle term as sum of two terms such that the sum satisfies the middle term and the product satisfies the extreme product.

Step 4: Now group the expression into two factors by taking the common expression outside.

Example:

**1.**\(x^2+5x+6\)

We have \(a =1\), \(b = 5\) and \(c = 6\).

Here the product \(=\) \(a \times c\) \(=\) \((1 \times 6)\) \(= 6\) and sum \(= b = 5\).

We need to choose two number such that the sum of two numbers is \(5\) and the product of two numbers is \(6\).

\((2+3) = 5\) and \((2 \times 3\) \(= 6\)

We can write as follows.

\(x^2+5x+6\) \(=\) \(x^2+3x+2x+6\)

\(=\)\(x(x+3)+2(x+3)\)

\(=\) \((x+3)(x+2)\)

**2.**\(2x^2-5x-3\)

We have \(a = 2\), \(b = -5\) and \(c = -3\).

Here the product \(=\) \(a \times c\) \(=\) \((2 \times -3)\) \(=\) \(-6\) and sum \(= b = -5\).

We need to choose two number such that the sum of two numbers is \(-5\) and the product of two numbers is \(-6\).

\((-6+1) = -5\) and \((-6\times 1)\)) \(=-6\)

We can rewrite as follows.

\(2x^2-5x-3\) \(=\) \(2x^2-6x+x-3\)

\(= 2x(x-3)+1(x-3)\)

\(= (x-3)(2x+1)\)