Factoring Polynomials Worksheet 2 with algebraic expressions for practice.

Problem Overview:

The worksheet provided involves factoring polynomials using two main techniques:
1. Difference of Squares: \( a^2 - b^2 = (a - b)(a + b) \)
2. Sum of Squares: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)

Additionally, some problems may require other factoring techniques such as:
- Factoring out the greatest common factor (GCF).
- Factoring quadratic trinomials.
- Grouping terms.

Below, I will solve a few representative problems from the worksheet to illustrate the solution process. If you need solutions for specific problems, please let me know!

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Solutions to Selected Problems:

#### Problem 1: \( 3p^2 - 2p - 5 \)

This is a quadratic trinomial. We need to factor it into the form \( (px + q)(rx + s) \).

1. Identify \( a = 3 \), \( b = -2 \), and \( c = -5 \).
2. Find two numbers that multiply to \( ac = 3 \cdot (-5) = -15 \) and add up to \( b = -2 \).
- The numbers are \( -5 \) and \( 3 \) because \( -5 \cdot 3 = -15 \) and \( -5 + 3 = -2 \).
3. Rewrite the middle term using these numbers:
\[
3p^2 - 2p - 5 = 3p^2 - 5p + 3p - 5
\]
4. Factor by grouping:
\[
(3p^2 - 5p) + (3p - 5) = p(3p - 5) + 1(3p - 5)
\]
5. Factor out the common binomial factor \( (3p - 5) \):
\[
(3p - 5)(p + 1)
\]

Answer:
\[
\boxed{(3p - 5)(p + 1)}
\]

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#### Problem 4: \( 4y^2 - 20y + 25 \)

This is a perfect square trinomial. It can be factored as \( (ay + b)^2 \).

1. Notice that \( 4y^2 = (2y)^2 \) and \( 25 = 5^2 \).
2. Check if the middle term \( -20y \) matches \( 2 \cdot (2y) \cdot (-5) \):
\[
2 \cdot 2y \cdot (-5) = -20y
\]
This confirms it is a perfect square trinomial.
3. Factor it as:
\[
(2y - 5)^2
\]

Answer:
\[
\boxed{(2y - 5)^2}
\]

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#### Problem 7: \( 2n^2 + 5mn + 2m^2 \)

This is a quadratic trinomial in two variables. We need to factor it into the form \( (an + bm)(cn + dm) \).

1. Look for two pairs of factors of \( 2n^2 \) and \( 2m^2 \) that combine to give the middle term \( 5mn \).
2. The factors of \( 2n^2 \) are \( (2n, n) \), and the factors of \( 2m^2 \) are \( (2m, m) \).
3. Test the combination:
\[
(2n + m)(n + 2m) = 2n \cdot n + 2n \cdot 2m + m \cdot n + m \cdot 2m = 2n^2 + 4mn + mn + 2m^2 = 2n^2 + 5mn + 2m^2
\]
This works.

Answer:
\[
\boxed{(2n + m)(n + 2m)}
\]

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#### Problem 11: \( z^4 + 6z^2 - 7z^2 \)

Simplify the expression first:
\[
z^4 + 6z^2 - 7z^2 = z^4 - z^2
\]

Now, factor out the GCF \( z^2 \):
\[
z^4 - z^2 = z^2(z^2 - 1)
\]

Notice that \( z^2 - 1 \) is a difference of squares:
\[
z^2 - 1 = (z - 1)(z + 1)
\]

So, the complete factorization is:
\[
z^2(z - 1)(z + 1)
\]

Answer:
\[
\boxed{z^2(z - 1)(z + 1)}
\]

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#### Problem 21: \( 125h^3 - 27k^3 \)

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

Here, \( a = 5h \) and \( b = 3k \) because \( 125h^3 = (5h)^3 \) and \( 27k^3 = (3k)^3 \).

Apply the formula:
\[
125h^3 - 27k^3 = (5h - 3k)((5h)^2 + (5h)(3k) + (3k)^2)
\]
\[
= (5h - 3k)(25h^2 + 15hk + 9k^2)
\]

Answer:
\[
\boxed{(5h - 3k)(25h^2 + 15hk + 9k^2)}
\]

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#### Problem 27: \( 16a^4 - b^4 \)

This is a difference of squares. Use the formula:
\[
a^2 - b^2 = (a - b)(a + b)
\]

First, rewrite \( 16a^4 - b^4 \) as:
\[
(4a^2)^2 - (b^2)^2
\]

Apply the difference of squares formula:
\[
(4a^2)^2 - (b^2)^2 = (4a^2 - b^2)(4a^2 + b^2)
\]

Notice that \( 4a^2 - b^2 \) is also a difference of squares:
\[
4a^2 - b^2 = (2a)^2 - (b)^2 = (2a - b)(2a + b)
\]

So, the complete factorization is:
\[
(4a^2 - b^2)(4a^2 + b^2) = (2a - b)(2a + b)(4a^2 + b^2)
\]

Answer:
\[
\boxed{(2a - b)(2a + b)(4a^2 + b^2)}
\]

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Final Answer:

If you need solutions for specific problems, please specify! For now, the boxed answers for the solved problems are:
\[
\boxed{(3p - 5)(p + 1), (2y - 5)^2, (2n + m)(n + 2m), z^2(z - 1)(z + 1), (5h - 3k)(25h^2 + 15hk + 9k^2), (2a - b)(2a + b)(4a^2 + b^2)}
\]