FACTORING POLYNOMIALS USING ALGEBRAIC IDENTITIES

Trinomials—Look for Perfect-Square Trinomials :

a² + 2ab + b²  =  (a + b)²

a² - 2ab + b²  =  (a - b)²

Examples : 

x² + 4x + 4  =  x² + 2(x)(2) + 2²  =  (x + 2)²

m² - 6m + 9  =  m² - 2(m)(3) + 3²  =  (m - 3)²

Binomials—Look for a Difference of Two Squares :

a² - b²  =  (a + b)(a - b)

Example : 

4x² - 9y²  =  2²x² - 3²y²

=  (2x)² - (3y)²

=  (2x + 3y)(2x - 3y)

Solved Problems

Problem 1 :

Factor :

x² + 6xy + 9y²

Solution : 

x² + 6xy + 9y²  =  x² + 2(x)(3y) + (3y)²

=  (x + 3y)²

=  (x + 3y)(x + 3y)

Problem 2 :

Factor :

4a² - 20ab + 25b²

Solution : 

4a² - 20ab + 25b²  =  2²a² - 20ab + 5²b²

=  (2a)² - 20ab + (5b)²

=  (2a)² - 2(2a)(5b) + (5b)²

=  (2a - 5b)²

=  (2a - 5b)(2a - 5b)

Problem 3 :

Factor :

2x² + 12xy + 18y²

Solution : 

Factor out the GCF.

=  2(x² + 6xy + 9y²)

=  2[x² + 2(x)(3y) + (3y)²]

=  2(x + 3y)²

=  2(x + 3y)(x + 3y) 

Problem 4 :

Factor :

2ab² - 16ab + 32a

Solution : 

Factor out the GCF.

=  2a(b² - 8b + 16)

=  2a[b² - 2(b)(4) + 4²]

=  2a(b - 4)²

Problem 5 :

Factor :

x² - 25y²

Solution : 

x² - 25y²  =  x² - 5²y²

=  x² - (5y²)

=  (x + 5y)(x - 5y)

Problem 6 :

Factor :

9m² - 16n²

Solution : 

9m² - 16y²  =  3²m² - 4²n²

=  (3m)² - (4n²)

=  (3m + 4n)(3m - 4n)

Problem 7 :

Factor :

x⁴ - y⁴

Solution : 

x⁴ - y⁴  =  (x²)² - (y²)²

=  (x² + y²)(x² - y²)

=  (x² + y²)(x + y)(x - y)

Problem 8 :

Factor :

8p⁴ - 18q⁴

Solution : 

Factor out the GCF.

32p⁴ - 162q⁴  =  2(16p⁴ - 81q⁴)

=  2[4²(p²)² - 9²(q²)²]

=  2[(4p²)² - (9q²)²]

=  2(4p² + 9q²)(4p² - 9q²)

=  2(4p² + 9q²)[(2p)² - (3q)²]

=  2(4p² + 9q²)(2p + 3q)(2p - 3q)

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.