Question: User: 64 $m^{4}$ +80 $m^{2}$ $n^{4}$ +25 $n^{8}$
1. **State the problem:**
We need to factor the expression $$64m^{4} + 80m^{2}n^{4} + 25n^{8}$$.
2. **Recognize the form:**
This is a quadratic in terms of $$m^{2}$$ and $$n^{4}$$. Let $$x = m^{2}$$ and $$y = n^{4}$$, then the expression becomes:
$$64x^{2} + 80xy + 25y^{2}$$.
3. **Use the factoring formula for quadratics:**
We look for factors of the form:
$$(ax + by)(cx + dy)$$
such that:
$$ac = 64$$
$$bd = 25$$
$$ad + bc = 80$$
4. **Find factors:**
- Factors of 64: 8 and 8 (since $$8 \times 8 = 64$$)
- Factors of 25: 5 and 5 (since $$5 \times 5 = 25$$)
Try:
$$(8x + 5y)(8x + 5y) = (8x + 5y)^2$$
5. **Verify:**
Expanding:
$$ (8x + 5y)^2 = 64x^2 + 2 \times 8x \times 5y + 25y^2 = 64x^2 + 80xy + 25y^2 $$
which matches the original expression.
6. **Rewrite in original variables:**
Recall $$x = m^{2}$$ and $$y = n^{4}$$, so:
$$ (8m^{2} + 5n^{4})^2 $$
**Final answer:**
$$64m^{4} + 80m^{2}n^{4} + 25n^{8} = (8m^{2} + 5n^{4})^2$$