If \(x=\sqrt{2}-1\), then find the value of \(3x^5+15x^4+30x^3+30x^2+15x+5\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2+12\sqrt{2}\)
The coefficients 3, 15, 30, 30, 15 are 3 times 1, 5, 10, 10, 5, which are the coefficients of \((x+1)^5=x^5+5x^4+10x^3+10x^2+5x+1\). Since \(x+1=\sqrt{2}\), the expression reduces to a power of \(\sqrt{2}\).
\(3(x+1)^5=3x^5+15x^4+30x^3+30x^2+15x+3\)
So the expression \(=3(x+1)^5+2\)
\(x+1=\sqrt{2}\), so \((x+1)^5=(\sqrt{2})^5=4\sqrt{2}\)
Expression \(=3\times4\sqrt{2}+2=2+12\sqrt{2}\)