Question

1. $3m^4 - 48n^2$
2. $x^3y^2 - 343y^5$

Explanation:

Problem 8: \(3m^4 - 48n^2\)

Step 1: Factor out the GCF

First, we find the greatest common factor (GCF) of the two terms \(3m^4\) and \(-48n^2\). The GCF of 3 and 48 is 3. So we factor out 3:
\(3m^4 - 48n^2 = 3(m^4 - 16n^2)\)

Step 2: Apply the difference of squares

Notice that \(m^4 - 16n^2\) is a difference of squares, since \(m^4=(m^2)^2\) and \(16n^2=(4n)^2\). The formula for the difference of squares is \(a^2 - b^2=(a + b)(a - b)\). Here, \(a = m^2\) and \(b = 4n\):
\(m^4 - 16n^2=(m^2 + 4n)(m^2 - 4n)\)

Step 3: Check for further factoring

Now, we look at \(m^2 - 4n\). It is not a difference of squares (since \(4n\) is not a perfect square of a monomial with integer coefficients for all \(n\)) and has no common factors. However, \(m^2 + 4n\) also cannot be factored further with real coefficients in a simple way. So putting it all together:
\(3m^4 - 48n^2 = 3(m^2 + 4n)(m^2 - 4n)\)

Step 1: Factor out the GCF

First, we find the greatest common factor (GCF) of the two terms \(x^3y^2\) and \(-343y^5\). The GCF of \(y^2\) and \(y^5\) is \(y^2\). So we factor out \(y^2\):
\(x^3y^2 - 343y^5 = y^2(x^3 - 343y^3)\)

Step 2: Apply the difference of cubes

Notice that \(x^3 - 343y^3\) is a difference of cubes, since \(x^3=x^3\) and \(343y^3=(7y)^3\). The formula for the difference of cubes is \(a^3 - b^3=(a - b)(a^2 + ab + b^2)\). Here, \(a = x\) and \(b = 7y\):
\(x^3 - 343y^3=(x - 7y)(x^2 + 7xy + 49y^2)\)

Step 3: Combine the factors

Putting it all together with the factored out \(y^2\):
\(x^3y^2 - 343y^5 = y^2(x - 7y)(x^2 + 7xy + 49y^2)\)

Answer:

\(3(m^2 + 4n)(m^2 - 4n)\)

Problem 10: \(x^3y^2 - 343y^5\)