FACTORING QUADRATIC POLYNOMIALS EXAMPLES

Example 1 :

Factor :

2a² + 9a + 10

Solution :

Since it is a quadratic expression and coefficient of a² is 2, we have to multiply the constant 10 by 2.

10 ⋅ 2  =  20 

Now we have to find two factors of 20 such that the product is 20 and sum is 9, which is the coefficient of x.

5 ⋅ 4 = 20 and 5 + 4 = 9

Then, 

2a² + 9a + 10  =  2a² + 5a + 4a + 10

  =  a(2a + 5) + 2(2a + 5)

  =  (a + 2)(2a + 5)

So, the factors are (a + 2) and (2a + 5).

Example 2 :

Factor :

5x² - 29xy - 42y²

Solution :

The product of 5 and -42 is -210. Now we have to find two factors of -210 such that the product is -210 and sum is -29, which is the coefficient of x.

-35 ⋅ 6 = -210 and -35 + 6 = -29

Then, 

5x² - 29xy - 42y²  =  5x² - 35xy + 6xy - 42y²

=  5x(x - 7y) + 6y(x - 7y)

=  (5x + 6y)(x - 7y)

So, the factors are (5x + 6y) and (x - 7y).

Example 3 : 

Factor :

9 - 18x + 8x²

Solution :

8x² - 18x + 9

The product of 8 and 9 is 72.

Now we have to find two factors of 72 such that the product is 72 and sum is -18, which is the coefficient of x.

-12 ⋅ (-6) = 72 and -12 - 6 = -18

8x² - 18x + 9  =  8x² - 12x - 6x + 9

=  4x(2x - 3) - 3(2x - 3)

=  (4x - 3)(2x - 3)

So the factors are (4x - 3) and (2x - 3).

Example 4 :

Factor :

6x² + 16xy + 8y²

Solution :

Factor out the greatest common factor.

6x² + 16xy + 8y²  =  2(3x² + 8xy + 4y²)

The product of 3 and 4 is 12.

Now we have to find two factors of 12 such that the product is 12 and sum is 8, which is the coefficient of x.

2 ⋅ 6 = 12 and 2 + 6 = 8

Then, 

  =  2(3x² + 2xy + 6xy + 4y²)

=  2[x(3x + 2y) + 2y(3x + 2y)]

=  2(3x + 2y)(x +  2y)

So, the factors are 2, (3x + 2y) and (x + 2y).

Example 5 : 

Factor :

12x² + 36x²y + 27x²y²

Solution :

Factor out the greatest common factor.

12x² + 36x²y + 27x²y²  =  3x²(4 + 12y + 9y²)

=  3x²(9y² + 12y + 4)

The product of 9 and 4 is 36.

Now we have to find two factors of 36, such that the product is 36 and sum is 12, which is the coefficient of x.

6 ⋅ 6 = 36 and 6 + 6 = 12

Then, 

=  3x²(9y² + 6y + 6y + 4)

=  3x²[(3y(3y + 2) + 2(3y + 2)]

=  3x²(3y + 2)(3y + 2)

=  3x²(3y + 2)²

So, factors are 3, x² and (3y + 2)².

Example 6 :

Factor :

(a + b)² + 9(a + b) + 18

Solution :

Let y = a + b. 

(a + b)² + 9(a + b) + 18  =  y² + 9y + 18

Now we have to find two factors of 18, such that the product is 18 and sum is 9, which is the coefficient of y.

3 ⋅ 6 = 18 and 3 + 6 = 9

Then, 

=  y² + 3y + 6y + 18

=  y(y + 3) + 6(y + 3)

=  (y + 3)(y + 6)

Substitute (a + b) for y. 

=  (a + b + 3)(a + b + 6)

So, the factors are (a + b + 3) and (a + b + 6).

Example 7 :

Factor :

-x² + 16

Solution :

= -x² + 16

Factoring negative from these two, we get

= -(x² - 16)

= -(x² - 4²)

Looks like a² - b² = (a + b)(a - b)

= -(x + 4)(x - 4)

So, the factors of -(x + 4)(x - 4).

Example 8 :

Factor :

36m² - 121

Solution :

= 36m² - 121

Factoring negative from these two, we get

= 6²m² - 11²

= (6m)² - 11²

Looks like a² - b² = (a + b)(a - b)

= (6m + 11)(6m - 11)

So, the factors of (6m + 11)(6m - 11).

Example 9 :

Factor :

12m² - 75

Solution :

= 12m² - 75

Factoring 3 from these two, we get

= 3(4m² - 25)

= 3(2²m² - 5²)

= 3[(2m)² - 5²]

= 3(2m + 5)(2m - 5)

So, the factors of 3(2m + 5)(2m - 5).

Example 10 :

Factor :

5x² - 45y²

Solution :

= 5x² - 45y²

Factoring 5 from these two, we get

= 5(x² - 9y²)

= 5(x² - 3²y²)

= 5[x² - (3y)²]

= 5[(x + 3y)(x - 3y)]

So, the factors of 5[(x + 3y)(x - 3y)]

Example 11 :

Factor :

25a² + 60a + 36

Solution :

= 25a² + 60a + 36

= (5²a²) + 2(5a)(6) + 6²

= (5a)² + 2(5a)(6) + 6²

a² + 2ab + b² = (a + b)²

= (5a + 6)²

= (5a + 6)(5a + 6)

So, the factors are (5a + 6)(5a + 6).

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