# Factoring of Special Polynomials

Questions With Solutions

How to use special polynomial forms to factor other polynomials? Questions are presented along with detailed Solutions and explanations. We will study five special polynomial forms.

## 1 - Difference of two squares

*a*

^{ 2}- b^{ 2}= (a - b)(a + b)__Question__Factor the polynomial.

*16 x*

^{ 2}- 9 y^{ 2}Solution

Note that

*16 x*

^{ 2}= (4 x)^{ 2}and 9 y^{ 2}= (3 y)^{ 2}We can write

*16 x*

^{ 2}- 9 y^{ 2}= (4 x)^{ 2}- (3 y)^{ 2}Now that we have written the given polynomial as the the difference of two squares, we use formula above to factor the given polynomial as follows:

*16 x*

^{ 2}- 9 y^{ 2}= (4 x)^{ 2}- (3 y)^{ 2}= (4 x - 3 y)(4 x + 3 y)## 2 - Trinomial Perfect Square

a)*a*

^{ 2}+ 2 a b + b^{ 2}= (a + b)^{ 2}b)

*a*

^{ 2}- 2 a b + b^{ 2}= (a - b)^{ 2}__Question__Factor the polynomials.

*4 x*

^{ 2}+ 20 x y + 25 y^{ 2}Solution

Note that the monomials making the given polynomial may be written as follows:

*4 x*

^{ 2}= (2 x)^{ 2}, 20 x y = 2(2 x)(5 y) and 25 y^{ 2}= (5 y)^{ 2}.

We now write the given polynomial as follows

*4 x*

^{ 2}+ 10 x y + 25 y^{ 2}= (2 x)^{ 2}+ 2(2 x)(5 y) + (5 y)^{ 2}Use the formula

*a*to write the given polynomial as a square as follows:

^{ 2}+ 2 a b + b^{ 2}= (a + b)^{ 2}*4 x*

^{ 2}+ 20 x y + 25 y^{ 2}= (2 x)^{ 2}+ 2(2 x)(5 y) + (5 y)^{ 2}= (2 x + 5 y)^{ 2}__Question__Factor the polynomials.

*1 - 6 x + 9 x*

^{ 2}Solution

Note that the monomials making the given polynomial may be written as follows:

*1 = 1*,

^{ 2}*- 6 x = - 2(3)x*and

*9 x*.

^{ 2}= (3 x)^{ 2}The given polynomial may be written as follows

*1 - 6 x + 9 x*

^{ 2}= 1^{ 2}- 2(3) x + (3 x)^{ 2}Use the formula

*a*to write the given polynomial as a square as follows:

^{ 2}- 2 a b + b^{ 2}= (a - b)^{ 2}*1 - 6 x + 9 x*

^{ 2}= 1^{ 2}- 2(3) x + (3 x)^{ 2}= (1 - 3 x)^{ 2}## 3 - Difference of two cubes

*a*

^{ 3}- b^{ 3}= (a - b)(a^{ 2}+ a b + b^{ 2})__Question__Factor the polynomial.

*8 - 27 x*

^{ 3}Solution

Note that the monomials making the given polynomial may be written as follows:

*8 = (2)*

^{ 3}and 27 x^{ 3}= (3 x)^{ 3}The given polynomial may now be written as follows

*8 - 27 x*

^{ 3}= (2)^{ 3}- (3 x)^{ 3}Use the formula

*a*to write the given polynomial in factored as follows:

^{ 3}- b^{ 3}= (a - b)(a^{ 2}+ ab + b^{ 2})*8 - 27 x*

^{ 3}= (2)^{ 3}- (3 x)^{ 3}= (2 - 3 x)( (2)^{ 2}+ (2)(3x) + (3 x)^{ 2}) = (2 - 3 x)(9 x^{ 2}+ 6x + 4)## 4 - Sum of two cubes

*a*

^{ 3}+ b^{ 3}= (a + b)(a^{ 2}- a b + b^{ 2})__Question__Factor the polynomial.

*8 y*

^{ 3}+ 1Solution

The two monomials making the given polynomial may be written as follows:

*8 y*and

^{ 3}= (2 y)^{ 3}*1 = (1)*

^{ 3}The polynomial to factor may now be written as follows

*8 y*

^{ 3}+ 1 = (2 y)^{ 3}+ (1)^{ 3}Use the formula

*a*to write the given polynomial in factored as follows:

^{ 3}+ b^{ 3}= (a + b)(a^{ 2}- ab + b^{ 2})*8 y*

^{ 3}+ 1 = (2 y)^{ 3}+ (1)^{ 3}= (2 y + 1)( (2 y)^{ 2}- (2 y)(1) + (1)^{ 2}) = (2 y + 1)(4 y^{ 2}- 2 y + 1)

## More Question of Factoring Special Polynomials

Factor the following special polynomialsa)

*- 25 x*

^{ 2}+ 9b)

*16 y*

^{ 4}- x^{ 4}c)

*36 y*

^{ 2}- 60 x y + 25 x^{ 2}d)

*(1/2) x*

^{ 2}+ x + (1/2)e)

*- y*

^{ 3}- 64f)

*x*

^{ 6}- 1

## Solutions to Above Questions

a) If we let a = 5 x and b = 3, the given polynomial may be written as:*- 25 x*

^{ 2}+ 9 = - a^{ 2}+ b^{ 2}Use the special polynomial

*a*and factor the given polynomial as follows:

^{ 2}- b^{ 2}= (a - b)(a + b)*- 25 x*

^{ 2}+ 9 = - a^{ 2}+ b^{ 2}= (- a + b)(a + b) = (-5 x + 3)(5 x + 3)b) The given polynomial has the form of the difference of two squares and may be written as:

*16 y*

^{ 4}- x^{ 4}= (4 y^{ 2})^{ 2}- (x^{ 2})^{ 2}Use the special polynomial

*a*and factor the given polynomial as follows:

^{ 2}- b^{ 2}= (a - b)(a + b)*16 y*

^{ 4}- x^{ 4}= (4 y^{ 2})^{ 2}- (x^{ 2})^{ 2}= (4y^{ 2}- x^{ 2})(4y^{ 2}+ x^{ 2})The term

*(4y*in the above is the sum of two squares and cannot be factored using real numbers. However the term (4y

^{ 2}+ x^{ 2})^{ 2}- x

^{ 2}) is the difference of two squares and can be further factored. Hence the given polynomial is factored as follows:

*16 y*

^{ 4}- x^{ 4}= (2 y - x)(2 y + x)(4y^{ 2}+ x^{ 2})c) The given polynomial may be written as:

*36 y*

^{ 2}- 60 x y + 25 x^{ 2}= (6 y)^{ 2}- 2(6 y)(5 x) + (5 x)^{ 2}Use the special trinomial

*a*to factor the given polynomial as follows:

^{ 2}- 2 a b + b^{ 2}= (a - b)^{ 2}*36 y*

^{ 2}- 60 x y + 25 x^{ 2}= (6 y)^{ 2}- 2(6 y)(5 x) + (5 x)^{ 2}= (6 y - 5 x)^{ 2}d) Factor (1/2) out and rewrite the given polynomial as:

*(1/2) x*

^{ 2}+ x + (1/2) = (1/2) x^{ 2}+ 2 (1/2) x + (1/2) = (1/2)( x^{ 2}+ 2 x + 1)Use the special trinomial

*a*to factor

^{ 2}+ 2 a b + b^{ 2}= (a + b)^{ 2}*x*and the given polynomial as follows:

^{ 2}+ 2 x + 1 = x^{ 2}+ 2(x)(1) + 1^{ 2}*(1/2) x*

^{ 2}+ x + (1/2) = (1/2)( x^{ 2}+ 2 x + 1) = (1/2)(x + 1)^{ 2}e) Factor - 1 out and rewrite the given polynomial as:

*- y*

^{ 3}- 64 = - (y^{ 3}+ 64) = - ( y^{ 3}+ 4^{ 3})Use

*a*to factor the given polynomial as follows:

^{ 3}+ b^{ 3}= (a + b)(a^{ 2}- a b + b^{ 2})*- y*

^{ 3}- 64 = - (y^{ 3}+ 64) = - ( y^{ 3}+ 4^{ 3})*= -(y + 4)(y*

^{ 2}- (y)(4) + 4^{ 2}) = -(y + 4)(y^{ 2}- 4 y + 16)f) Let us write the given polynomial as the difference of two squares as follows:

*x*

^{ 6}- 1 = (x^{ 3})^{ 2}- (1)^{ 2}Use the special difference of squares polynomial

*a*and factor the given polynomial as follows:

^{ 2}- b^{ 2}= (a - b)(a + b)*x*

^{ 6}- 1 = (x^{ 3})^{ 2}- (1)^{ 2}= (x^{ 3}- 1)(x^{ 3}+ 1)In the above we have the product of the sum and difference of two cubes. Hence

*x*

^{ 6}- 1 = (x^{ 3})^{ 2}- (1)^{ 2}= (x^{ 3}- 1)(x^{ 3}+ 1)*= (x - 1)(x*

^{ 2}+ x + 1)(x + 1)(x^{ 2}- x + 1)

### More References and links

Factor PolynomialsFactor Polynomials by Common Factor

Factor polynomial by Grouping

High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers

Home Page