# Factoring Polynomials: Special Cases

**Factoring** is the process of rewriting a sum as a product. It allows us to simplify expressions and solve equations.

For example, the quadratic expression \(x^2+4x+4,\) which is written as a sum, may be expressed as a product \((x+2)(x+2),\) much the way that 14 can be written as a product, \(7\times 2,\) or a sum, \(6+8.\)

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## Factoring Perfect Squares

A perfect square polynomial is one that can be written as the product of two identical factors. The **perfect square** identities below are widely used in algebra.

\[(a+b)^2 = a^2 + 2ab + b^2\] \[(a-b)^2=a^2-2ab-b^2\]

## Is \(4x^2+12x+9\) a perfect square?

Let's begin by looking at the first term in our quadratic, \(4x^2.\) \(4x^2\) is a perfect square because \((2x)(2x)=4x^2.\)

Next, we can look at the last term in our quadratic, \(9.\) \(9\) is a perfect square because \((3)(3)=9.\)

If the square root of the first term multiplied by the square root of the last term multiplied by \(2\) equals the middle term, then our quadratic is a perfect square. Because \((2x)(3)(2)=12x,\) the quadratic \(4x^2+12x+9\) is a perfect square.

\(4x^2+12x+9\) factors into \((2x+3)(2x+3) = (2x+3)^2.\)

## Factor \(9x^2-6x+1.\)

The square root of \(9x^2\) is \(3x.\)

The square root of \(1\) is \(1.\)

\(9x^2-6x+1\) factors into \((3x-1)(3x-1) = (3x-1)^2.\)

## Difference of Squares

The **difference of squares identity** shows how every polynomial that is a difference between two perfect squares can be rewritten in the following factored form:

\[a^2-b^2=(a+b)(a-b).\]

Let's begin with the left side of the expression. We have

\[ \begin{align} (a+b)(a-b) &= a(a-b) + b(a-b) \\ &= a^2 - ab + ab - b^2 \\ & = a^2 - b^2, \end{align} \]

which is equal to the right side of the identity. Hence proved. \(_\square\)

## Factor \(25y^2-49.\)

The square root of \(25y^2\) is \(5y.\)

The square root of \(49\) is \(7.\)

Therefore, \(25y^2-49\) factors into \((5y-7)(5y+7).\)

## Calculate \(299\times 301\).

You can brute force the answer to this problem by using a calculator, but we have a sweeter way. We can apply the difference of two squares identity.

At first we may think about using the long multiplication method, but it wastes time and is, of course, boring. Notice that \(299=300-1\) and \(301=300+1\), so

\[\begin{align*}299\times 301&=(300-1)(300+1)\\&=300^2-1^2\\&=89999. \ _\square \end{align*}\]

## Sums and Differences of Cubes

Every polynomial that is a sum or difference of two perfect cubes can be rewritten in the following factored form: \[\begin{align} x^3 - y^3 &= (x-y) ( x^2 + xy + y^2) \\ x^3 + y^3 &= (x+y) ( x^2 - xy + y^2). \end{align}\]

## Factor \(x^3+8.\)

We recognize that \(x^3+8\) is the sum of \( x^3 \) and \( 2^3 \). Hence, by the sum of cubes factorization, we obtain

\[ x^3 + 8 = ( x + 2) (x ^2 - 2x + 2^2) = (x+2)( x^2 - 2x + 4 ).\ _\square \]

## Factoring Perfect Cubes

A perfect cube polynomial is one that can be written as the product of three identical factors. The **perfect cube** identities below are widely used in algebra.

\[(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\] \[(a-b)^3=a^3-3a^2b + 3ab^2-b^3\]

What is \( (a-2b) ^ 3 \)?

Expanding out, we obtain

\[\begin{align} ( a - 2b)^3 &= a^3 - 3 \times a^2 \times (2b) + 3 \times a \times (2b)^2 - (2b)^3 \\ &= a^3 - 6a^2b + 12a b^2 - 8 b^3.\ _\square \end{align}\]

## See Also

**Cite as:**Factoring Polynomials: Special Cases.

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