Answer :
Sure, I'd be happy to help you find the Greatest Common Factor (GCF) of [tex]\(28x^3\)[/tex] and [tex]\(16x^2y^2\)[/tex]. Here's a detailed step-by-step solution:
1. Factorize each term into its prime factors and variables:
- For [tex]\(28x^3\)[/tex]:
[tex]\[
28 = 2 \times 2 \times 7
\][/tex]
And we have [tex]\(x^3 = x \times x \times x\)[/tex].
Therefore:
[tex]\[
28x^3 = 2 \times 2 \times 7 \times x \times x \times x
\][/tex]
- For [tex]\(16x^2y^2\)[/tex]:
[tex]\[
16 = 2 \times 2 \times 2 \times 2
\][/tex]
And we have [tex]\(x^2 = x \times x\)[/tex] and [tex]\(y^2 = y \times y\)[/tex].
Therefore:
[tex]\[
16x^2y^2 = 2 \times 2 \times 2 \times 2 \times x \times x \times y \times y
\][/tex]
2. Identify the common factors:
- In both expressions, we need to find the factors that are present in each factorization.
- Common numerical factor:
- The number 2 appears in both factorizations. We take the smallest power of 2 present in both, which is:
[tex]\[
2 \times 2 = 4 \text{ (because there are two 2's in both numbers)}
\][/tex]
- Common variable factors:
- For [tex]\(x\)[/tex]:
- [tex]\(x\)[/tex] appears 3 times in [tex]\(28x^3\)[/tex] and 2 times in [tex]\(16x^2y^2\)[/tex]. We take the smallest power, which is [tex]\(x^2\)[/tex].
- For [tex]\(y\)[/tex]:
- [tex]\(y\)[/tex] is not present in [tex]\(28x^3\)[/tex], so [tex]\(y^2\)[/tex] will not be in the GCF.
3. Combine the common factors:
- Combining these, the GCF is:
[tex]\[
4x^2
\][/tex]
Thus, the Greatest Common Factor (GCF) of [tex]\(28x^3\)[/tex] and [tex]\(16x^2y^2\)[/tex] is [tex]\(4x^2\)[/tex].
1. Factorize each term into its prime factors and variables:
- For [tex]\(28x^3\)[/tex]:
[tex]\[
28 = 2 \times 2 \times 7
\][/tex]
And we have [tex]\(x^3 = x \times x \times x\)[/tex].
Therefore:
[tex]\[
28x^3 = 2 \times 2 \times 7 \times x \times x \times x
\][/tex]
- For [tex]\(16x^2y^2\)[/tex]:
[tex]\[
16 = 2 \times 2 \times 2 \times 2
\][/tex]
And we have [tex]\(x^2 = x \times x\)[/tex] and [tex]\(y^2 = y \times y\)[/tex].
Therefore:
[tex]\[
16x^2y^2 = 2 \times 2 \times 2 \times 2 \times x \times x \times y \times y
\][/tex]
2. Identify the common factors:
- In both expressions, we need to find the factors that are present in each factorization.
- Common numerical factor:
- The number 2 appears in both factorizations. We take the smallest power of 2 present in both, which is:
[tex]\[
2 \times 2 = 4 \text{ (because there are two 2's in both numbers)}
\][/tex]
- Common variable factors:
- For [tex]\(x\)[/tex]:
- [tex]\(x\)[/tex] appears 3 times in [tex]\(28x^3\)[/tex] and 2 times in [tex]\(16x^2y^2\)[/tex]. We take the smallest power, which is [tex]\(x^2\)[/tex].
- For [tex]\(y\)[/tex]:
- [tex]\(y\)[/tex] is not present in [tex]\(28x^3\)[/tex], so [tex]\(y^2\)[/tex] will not be in the GCF.
3. Combine the common factors:
- Combining these, the GCF is:
[tex]\[
4x^2
\][/tex]
Thus, the Greatest Common Factor (GCF) of [tex]\(28x^3\)[/tex] and [tex]\(16x^2y^2\)[/tex] is [tex]\(4x^2\)[/tex].
