Solutions to Factoring of Special Polynomials

The solutions to the questions on factoring of special polynomials in factoring of special polynomials are presented.

Review the Special Polynomials

Difference of Two Squares

The difference of squares formula is given by: \[ a^2 - b^2 = (a - b)(a + b) \]

Trinomial Perfect Square

This identity comes in two forms:

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

Difference of Two Cubes

The formula for the difference of cubes is: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]

Sum of Two Cubes

The formula for the sum of cubes is: \[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]

Factor the following special polynomials

a) \( -25x^2 + 9 \)

b) \( 16y^4 - x^4 \)

c) \( 36y^2 - 60xy + 25x^2 \)

d) \( \dfrac{1}{2}x^2 + x + \dfrac{1}{2} \)

e) \( -y^3 - 64 \)

f) \( x^6 - 1 \)

Solution to Question a)

Let \(a = 5x\) and \(b = 3\). Then

\[ -25x^2 + 9 = -a^2 + b^2. \]

Using the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\), we get

\[ -25x^2 + 9 = -a^2 + b^2 = (-a + b)(a + b) = (-5x + 3)(5x + 3). \]

Solution to Question b)

Write each term as a square:

\[ 16y^4 - x^4 = (4y^2)^2 - (x^2)^2. \]

Apply the difference of squares:

\[ (4y^2)^2 - (x^2)^2 = (4y^2 - x^2)(4y^2 + x^2). \]

The factor \(4y^2 + x^2\) cannot be factored over the reals, but \(4y^2 - x^2\) is again a difference of squares:

\[ 4y^2 - x^2 = (2y - x)(2y + x). \]

Hence

\[ 16y^4 - x^4 = (2y - x)(2y + x)(4y^2 + x^2). \]

Solution to Question c)

Recognize a perfect square trinomial:

\[ 36y^2 - 60xy + 25x^2 = (6y)^2 - 2\cdot(6y)\cdot(5x) + (5x)^2. \]

Using the square of a binomial formula \(a^2 - 2ab + b^2 = (a - b)^2\), we get

\[ 36y^2 - 60xy + 25x^2 = (6y - 5x)^2. \]

Solution to Question d)

Factor out \(\tfrac12\):

\[ \tfrac12 x^2 + x + \tfrac12 = \tfrac12\bigl(x^2 + 2x + 1\bigr). \]

Notice \(x^2 + 2x + 1 = (x + 1)^2\). Therefore,

\[ \tfrac12 x^2 + x + \tfrac12 = \tfrac12\,(x + 1)^2. \]

Solution to Question e)

Factor out \(-1\) and recognize a sum of cubes:

\[ -y^3 - 64 = -\bigl(y^3 + 4^3\bigr). \]

Apply the sum of cubes formula \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\):

\[ -y^3 - 64 = -\bigl(y + 4\bigr)\bigl(y^2 - 4y + 16\bigr). \]

Solution to Question f)

First write as a difference of squares:

\[ x^6 - 1 = (x^3)^2 - 1^2 = (x^3 - 1)(x^3 + 1). \]

Then factor each as a difference or sum of cubes:

\[ x^3 - 1 = (x - 1)\bigl(x^2 + x + 1\bigr), \quad x^3 + 1 = (x + 1)\bigl(x^2 - x + 1\bigr). \]

Thus

\[ x^6 - 1 = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1). \]