Question
Factor. \( 49 s^{4}-64 s^{2} d^{4} \) \( 49 s^{4}-64 s^{2} d^{4}=\square \) \( ( \) Factor completely.)
Ask by Ruiz Lyons. in the United States
Feb 24,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( 49 s^{4} - 64 s^{2} d^{4} = s^{2} (7s - 8d^{2})(7s + 8d^{2}) \)
Solution
To begin, notice that both terms in the expression have a common factor of s². Factor that out:
49s⁴ − 64s²d⁴ = s²(49s² − 64d⁴).
Next, observe that 49s² and 64d⁴ are both perfect squares:
49s² = (7s)² and 64d⁴ = (8d²)².
This sets up a difference of two squares, which factors as:
A² − B² = (A − B)(A + B).
Here, A = 7s and B = 8d², so:
49s² − 64d⁴ = (7s − 8d²)(7s + 8d²).
Thus, the complete factorization is:
49s⁴ − 64s²d⁴ = s²(7s − 8d²)(7s + 8d²).
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Simplify this solution Extra Insights
To factor the expression \( 49 s^{4} - 64 s^{2} d^{4} \), start by recognizing that this is a difference of squares. We can rewrite it as: \[ (7s^{2})^{2} - (8d^{2})^{2} \] Using the difference of squares formula \( a^2 - b^2 = (a - b)(a + b) \), we have: \[ (7s^{2} - 8d^{2})(7s^{2} + 8d^{2}) \] Next, let's see if we can factor any further. The first term \( 7s^{2} - 8d^{2} \) is akin to a difference of squares if we consider it as \( ( \sqrt{7}s - 2\sqrt{2}d)( \sqrt{7}s + 2\sqrt{2}d) \) but not fully factorizable over the integers. Thus, the complete factorization of \( 49 s^{4} - 64 s^{2} d^{4} \) is: \[ (7s^{2} - 8d^{2})(7s^{2} + 8d^{2}) \] There's your answer! 🎉
