Factoring Perfect Square Trinomial and Sum and Difference of Two ... - Free Printable

Problem Breakdown and Solution

The task involves factoring Perfect Square Trinomials and Sum/Difference of Two Cubes. Let's solve each part step by step.

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#### Part 1: Factoring Perfect Square Trinomials

A perfect square trinomial has the form:
1. \( a^2 + 2ab + b^2 = (a + b)^2 \)
2. \( a^2 - 2ab + b^2 = (a - b)^2 \)

We will identify the terms \( a \) and \( b \) for each trinomial and factor accordingly.

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##### Problem 1: \( x^2 + 24x + 144 \)

- The given trinomial is \( x^2 + 24x + 144 \).
- Compare it with the form \( a^2 + 2ab + b^2 \):
- \( a^2 = x^2 \implies a = x \)
- \( b^2 = 144 \implies b = 12 \)
- \( 2ab = 24x \), which matches \( 2(x)(12) \).

Thus, the trinomial factors as:
\[
(x + 12)^2
\]

##### Problem 2: \( x^2 - 12xy + 36y^2 \)

- The given trinomial is \( x^2 - 12xy + 36y^2 \).
- Compare it with the form \( a^2 - 2ab + b^2 \):
- \( a^2 = x^2 \implies a = x \)
- \( b^2 = 36y^2 \implies b = 6y \)
- \( -2ab = -12xy \), which matches \( -2(x)(6y) \).

Thus, the trinomial factors as:
\[
(x - 6y)^2
\]

##### Problem 3: \( 9x^2 - 24x + 16 \)

- The given trinomial is \( 9x^2 - 24x + 16 \).
- Compare it with the form \( a^2 - 2ab + b^2 \):
- \( a^2 = 9x^2 \implies a = 3x \)
- \( b^2 = 16 \implies b = 4 \)
- \( -2ab = -24x \), which matches \( -2(3x)(4) \).

Thus, the trinomial factors as:
\[
(3x - 4)^2
\]

##### Problem 4: \( 2x^2 - 28x + 98 \)

- The given trinomial is \( 2x^2 - 28x + 98 \).
- First, factor out the greatest common factor (GCF), which is 2:
\[
2(x^2 - 14x + 49)
\]
- Now, factor the trinomial inside the parentheses:
- Compare \( x^2 - 14x + 49 \) with \( a^2 - 2ab + b^2 \):
- \( a^2 = x^2 \implies a = x \)
- \( b^2 = 49 \implies b = 7 \)
- \( -2ab = -14x \), which matches \( -2(x)(7) \).

Thus, the trinomial inside the parentheses factors as:
\[
(x - 7)^2
\]
- Including the GCF, the complete factorization is:
\[
2(x - 7)^2
\]

##### Problem 5: \( 121 - 22x + x^2 \)

- The given trinomial is \( 121 - 22x + x^2 \).
- Rearrange the terms to match the standard form:
\[
x^2 - 22x + 121
\]
- Compare it with the form \( a^2 - 2ab + b^2 \):
- \( a^2 = x^2 \implies a = x \)
- \( b^2 = 121 \implies b = 11 \)
- \( -2ab = -22x \), which matches \( -2(x)(11) \).

Thus, the trinomial factors as:
\[
(x - 11)^2
\]

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#### Final Answers for Part 1:
1. \( (x+12)^2 \)
2. \( (x-6y)^2 \)
3. \( (3x-4)^2 \)
4. \( 2(x-7)^2 \)
5. \( (x-11)^2 \)

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#### Part 2: Factoring Sum and Difference of Two Cubes

The formulas for sum and difference of two cubes are:
1. \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)
2. \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)

We will identify the terms \( a \) and \( b \) for each expression and factor accordingly.

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##### Problem 1: \( x^3 - 8 \)

- The given expression is \( x^3 - 8 \).
- Recognize it as a difference of cubes:
\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]
- Here, \( a^3 = x^3 \implies a = x \)
- \( b^3 = 8 \implies b = 2 \)

Thus, the factorization is:
\[
(x - 2)(x^2 + 2x + 4)
\]

##### Problem 2: \( y^3 + 64 \)

- The given expression is \( y^3 + 64 \).
- Recognize it as a sum of cubes:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
- Here, \( a^3 = y^3 \implies a = y \)
- \( b^3 = 64 \implies b = 4 \)

Thus, the factorization is:
\[
(y + 4)(y^2 - 4y + 16)
\]

##### Problem 3: \( 2y^3 + 2 \)

- The given expression is \( 2y^3 + 2 \).
- Factor out the GCF, which is 2:
\[
2(y^3 + 1)
\]
- Recognize \( y^3 + 1 \) as a sum of cubes:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
- Here, \( a^3 = y^3 \implies a = y \)
- \( b^3 = 1 \implies b = 1 \)

Thus, \( y^3 + 1 \) factors as:
\[
(y + 1)(y^2 - y + 1)
\]
- Including the GCF, the complete factorization is:
\[
2(y + 1)(y^2 - y + 1)
\]

##### Problem 4: \( 8x^9y^6 + z^{15} \)

- The given expression is \( 8x^9y^6 + z^{15} \).
- Recognize it as a sum of cubes:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
- Here, \( a^3 = 8x^9y^6 \implies a = 2x^3y^2 \)
- \( b^3 = z^{15} \implies b = z^5 \)

Thus, the factorization is:
\[
(2x^3y^2 + z^5)((2x^3y^2)^2 - (2x^3y^2)(z^5) + (z^5)^2)
\]
- Simplify the terms:
- \( (2x^3y^2)^2 = 4x^6y^4 \)
- \( (2x^3y^2)(z^5) = 2x^3y^2z^5 \)
- \( (z^5)^2 = z^{10} \)

So, the complete factorization is:
\[
(2x^3y^2 + z^5)(4x^6y^4 - 2x^3y^2z^5 + z^{10})
\]

##### Problem 5: \( 2m^4 - 128m \)

- The given expression is \( 2m^4 - 128m \).
- Factor out the GCF, which is \( 2m \):
\[
2m(m^3 - 64)
\]
- Recognize \( m^3 - 64 \) as a difference of cubes:
\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]
- Here, \( a^3 = m^3 \implies a = m \)
- \( b^3 = 64 \implies b = 4 \)

Thus, \( m^3 - 64 \) factors as:
\[
(m - 4)(m^2 + 4m + 16)
\]
- Including the GCF, the complete factorization is:
\[
2m(m - 4)(m^2 + 4m + 16)
\]

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#### Final Answers for Part 2:
1. \( (x-2)(x^2+2x+4) \)
2. \( (y+4)(y^2-4y+16) \)
3. \( 2(y+1)(y^2-y+1) \)
4. \( (2x^3y^2+z^5)(4x^6y^4-2x^3y^2z^5+z^{10}) \)
5. \( 2m(m-4)(m^2+4m+16) \)

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#### Boxed Final Answers:

Part 1:
1. \( (x+12)^2 \)
2. \( (x-6y)^2 \)
3. \( (3x-4)^2 \)
4. \( 2(x-7)^2 \)
5. \( (x-11)^2 \)

Part 2:
1. \( (x-2)(x^2+2x+4) \)
2. \( (y+4)(y^2-4y+16) \)
3. \( 2(y+1)(y^2-y+1) \)
4. \( (2x^3y^2+z^5)(4x^6y^4-2x^3y^2z^5+z^{10}) \)
5. \( 2m(m-4)(m^2+4m+16) \)

\boxed{
\text{Part 1: } (x+12)^2, (x-6y)^2, (3x-4)^2, 2(x-7)^2, (x-11)^2 \\
\text{Part 2: } (x-2)(x^2+2x+4), (y+4)(y^2-4y+16), 2(y+1)(y^2-y+1), (2x^3y^2+z^5)(4x^6y^4-2x^3y^2z^5+z^{10}), 2m(m-4)(m^2+4m+16)
}