Factorization of Trinomial

Factorization of Trinomial

Factorization of trinomial expressions of the form x² + px + q:

Let’s consider any two binomial expressions (x + a) and (x + b).

The product of (x + a) and (x + b) = (x + a)(x + b)

                                                 = x(x + b) + a(x + b)

                                                 = x² + bx + ax + ab

                                                 = x² + (a + b)x + ab

If we compare the product x² + (a + b)x + ab with x² + px + q, then

            p = a+b and q = ab

So, while factorizing a trinomial expression of the form x² + px + q, we should find the two value a and b such that a + b = p and a × b = q. Then, the expression is factorized by grouping.

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For example: In x² + 7x + 12, we have 4 + 3 = 7 and 4 × 3 = 12. So the required values of a = 4 and b = 3. Now,

            x² + 7x + 12

            = x² + (4 + 3)x + 12

            = x² + 4x + 3x + 12

            = x(x + 4) + 3(x + 4)

            = (x + 4)(x + 3)

Note: The same process is applied for x² – px + q

For example: In x² - 9x + 20, we have 5 + 4 = 9 and 5 × 4 = 20. So the required values of a = 5 and b = 4. Now,

            x² - 9x + 20

            = x² - (5 + 4)x + 20

            = x² - 5x - 4x + 20

            = x(x - 5) - 4(x - 5)

            = (x - 5)(x - 4)

                                                           

Factorization of trinomial expressions of the form x² + px - q:

While factorizing a trinomial expression of the form x² + px - q, we should find the two value a and b such that a - b = p and a × b = q. Then, the expression is factorized by grouping.

For example: In x² + 3x - 18, we have 6 - 3 = 3 and 6 × 3 = 18. So the required values of a = 6 and b = 3. Now,

            x² + 3x - 18

            = x² + (6 - 3)x - 18

            = x² + 6x - 3x - 18

            = x(x + 6) - 3(x + 6)

            = (x + 6)(x - 3)

Note: The same process is applied for x² – px - q          

For example: In x² - 5x - 14, we have 7 - 2 = 5 and 7 × 2 = 14. So the required values of a = 7 and b = 2. Now,

            x² - 5x - 14

            = x² - (7 - 2)x - 14

            = x² - 7x + 2x - 14

            = x(x - 7) + 2(x - 7)

            = (x - 7)(x + 2)

Factorization of trinomial expressions of the form px² + qx + r:

In the trinomial expressions of the form px² + qx + r, p and q are the numerical coefficients of x² and x respectively and r is the constant term.

To factorize such expressions, we need to find the two values a and b such that a + b = q and a × b = p × r. Then, the expression is expanded to four terms and factorization is performed by grouping.

For example: In 2x² + 7x + 3, we have 6 + 1 = 7 and 6 × 1 = 2 × 3. So the required values of a = 6 and b = 1. Now,

            2x² + 7x + 3

            = 2x² + (6 + 1)x + 3

            = 2x² + 6x + x + 3

            = 2x(x + 3) + 1(x + 3)

            = (x + 3)(2x + 1)

Note: The same process is applied for px² – qx + r        

For example: In 9x² - 12x + 4, we have 6 + 6 = 12 and 6 × 6 = 9 × 4. So the required values of a = 6 and b = 6. Now,

            9x² - 12x + 4

            = 9x² - (6 + 6)x + 4

            = 9x² - 6x - 6x + 4

            = 3x(3x - 2) - 2(3x - 2)

            = (3x - 2)(3x - 2)

 

Factorization of trinomial expressions of the form px² + qx - r:

To factorize the expressions of the form px² + qx + r, we need to find the two values a and b such that a - b = q and a × b = p × r. Then, the expression is expanded to four terms and factorization is performed by grouping.

For example: In 4x² + 4x - 3, we have 6 - 2 = 4 and 6 × 2 = 4 × 3. So the required values of a = 6 and b = 2. Now,

            4x² + 4x - 3

            = 4x² + (6 - 2)x - 3

            = 4x² + 6x - 2x - 3

            = 2x(2x + 3) - 1(2x + 3)

            = (2x + 3)(2x - 1)

Note: The same process is applied for px² – qx - r         

For example: In 15x² - 17x - 4, we have 20 - 3 = 17 and 20 × 3 = 15 × 4. So the required values of a = 20 and b = 3. Now,

            15x² - 17x - 4

            = 15x² - (20 - 3)x - 4

            = 15x² - 20x + 3x - 4

            = 5x(3x - 4) + 1(3x - 4)

            = (3x - 4)(5x + 1)

Factorization of trinomial expressions of the form a² + 2ab + b²:

Let’s find the product of (a + b) and (a + b)

(a + b)(a + b) = a(a+b)+b(a+b)

                      = a² + ab + ab + b²

                      = a² + 2ab + b²

Thus, (a + b)² are the factors of a² + 2ab + b². Similarly, (a – b)² are the factors of a² – 2ab + b². So, the expression of the form a² + 2ab + b² is factorized by making it a perfect square trinomial.

 For example: 4x² + 20x + 25 = (2x)² + 2.2x.5 + 5²

                                                 = (2x + 5)²

And, 4a² – 12a + 9 = (2a)² – 2.2a.3 + 3²

                                = (2a – 3)²

 

You can comment your questions or problems regarding the factorization of trinomial here.