Factoring With Special Forms

Factoring with special forms is a process of using identities to help with different factoring problems.

Factoring binomials of the form \( x^2 - y^2= (x-y)(x+y):\)
This approach applies the difference of two squares identity.

Factor \( x^2 - 16 \).

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We have

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

Factor \( x^4 - 25y^4 \).

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We have

\[ x^4 - 25y^4 = \big(x^2\big)^2 - \big(5y^2\big)^2 = \big(x^2 - 5y^2\big)\big(x^2 + 5y^2\big).\ _\square\]

Factor \( 9x^3y - 36xy \).

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We have

\[ 9x^3y - 36xy = 9xy\big(x^2 - 4\big) = 9xy\big(x^2 - 2^2\big) = 9xy(x-2)(x+2).\ _\square\]

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

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We have

\[ \begin{align} (x+1)^2 - 9(x-2)^2 &= (x+1)^2 - \big(3(x-2)\big)^2 \\ &= \big((x+1) - 3(x-2)\big) \big((x+1) + 3(x-2)\big) \\ &= (x + 1 -3x + 6)(x + 1 + 3x - 6) \\ &= (-2x + 7)(4x - 5). \ _\square \end{align} \]

Factoring binomials of the form \( x^3 + y^3 = (x+y)\big(x^2 - xy + y^2\big) \) and \(x^3 - y^3 = (x-y)\big(x^2 + xy+ y^2\big): \)
This approach applies the sum and difference of cubes identity.

Factor \( 27x^3 - y^3 \).

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We have

\[ \begin{align} 27x^3 - y^3 &= (3x)^3 - y^3 \\ &= (3x - y)\big((3x)^2 + 3xy + y^2\big) \\ &= (3x-y)(9x^2 + 3xy + y^2). \ _\square \end{align} \]

Factor \( 64x^3 + 27y^3 \).

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We have

\[ \begin{align} 64x^3 + 27y^3 &= (4x)^3 + (3y)^3 \\ &= (4x + 3y)\big((4x)^2 - 12xy + (3y)^2\big) \\ &= (4x + 3y)\big(16x^2 -12 xy + 9y^2\big). \ _\square \end{align} \]

Factoring trinomials of the form \( x^2 + 2xy + y^2 = (x+y)^2 \) and \(x^2 - 2xy + y^2 = (x-y)^2: \)

Factor \( x^2 + 6x + 9 \).

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We have

\[ x^2 + 6x + 3 = x^2 + 2(x \cdot 3) + 3^2 = (x+3)^2. \ _\square \]

Factor \( x^2 - 4x + 4 \).

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We have

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

Factor \( 25x^2 - 20x + 4 \).

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We have

\[ 25x^2 - 20x + 4 = (5x)^2 - 2(5x \cdot 2) + 2^2 = (5x-2)^2. \ _\square \]

Factor \( 16x^2 - 24xy + 9y^2 \).

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We have

\[ 16x^2 - 24xy + 9y^2 = (4x)^2 - 2(4x \cdot 3y) + (3y)^2 = (4x-3y)^2. \ _\square \]