Question
\begin{tabular}{l} Factorise the following expression \\ completely: \\ \( \qquad 16-y^{8} \) \\ Answer: \\ \hline\end{tabular}
Ask by Harper Davison. in South Africa
Feb 25,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The expression \(16 - y^{8}\) can be factored completely as \((2 - y^{2})(2 + y^{2})(y^{2} - 2y + 2)(y^{2} + 2y + 2)\).
Solution
Factor the expression \( 16-y^{8} \).
Factor the expression by following steps:
- step0: Factor:
\(16-y^{8}\)
- step1: Rewrite the expression:
\(\left(16^{\frac{1}{2}}\right)^{2}-\left(y^{4}\right)^{2}\)
- step2: Factor the expression:
\(\left(16^{\frac{1}{2}}-y^{4}\right)\left(16^{\frac{1}{2}}+y^{4}\right)\)
- step3: Evaluate:
\(\left(4-y^{4}\right)\left(16^{\frac{1}{2}}+y^{4}\right)\)
- step4: Evaluate:
\(\left(4-y^{4}\right)\left(4+y^{4}\right)\)
- step5: Evaluate:
\(\left(2-y^{2}\right)\left(2+y^{2}\right)\left(4+y^{4}\right)\)
- step6: Factor the expression:
\(\left(2-y^{2}\right)\left(2+y^{2}\right)\left(y^{2}-2y+2\right)\left(y^{2}+2y+2\right)\)
The complete factorization of the expression \(16-y^{8}\) is \((2-y^{2})(2+y^{2})(y^{2}-2y+2)(y^{2}+2y+2)\).
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Simplify this solution Bonus Knowledge
To factor the expression \( 16 - y^8 \) completely, we first recognize it as a difference of squares. We can rewrite it as \( (4)^2 - (y^4)^2 \), which factors into \( (4 - y^4)(4 + y^4) \). Next, we see that \( 4 - y^4 \) is also a difference of squares: \( (2)^2 - (y^2)^2 \). Thus, it can be factored further as \( (2 - y^2)(2 + y^2) \). Putting it all together, the complete factorization of \( 16 - y^8 \) is: \[ (2 - y^2)(2 + y^2)(4 + y^4) \] Mathematically, it’s a fun ride through algebraic identities!
