The degree of the polynomial (x+sqrt{x^4-1})^ | vedclass

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(D) Let $P(x) = (x+\sqrt{x^4-1})^9+(x-\sqrt{x^4-1})^9$. Using the binomial expansion $(a+b)^n + (a-b)^n = 2 \sum_{k=0, 2, 4, \dots} \binom{n}{k} a^{n-

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(D) Let $P(x) = (x+\sqrt{x^4-1})^9+(x-\sqrt{x^4-1})^9$. Using the binomial expansion $(a+b)^n + (a-b)^n = 2 \sum_{k=0, 2, 4, \dots} \binom{n}{k} a^{n-k} b^k$,we have: $P(x) = 2 [ \binom{9}{0} x^9 + \binom{9}{2} x^7 (x^4-1) + \binom{9}{4} x^5 (x^4-1)^2 + \binom{9}{6} x^3 (x^4-1)^3 + \binom{9}{8} x^1 (x^4-1)^4 ]$. Expanding the terms: Term $1$: $x^9$ (degree $9$). Term $2$: $x^7 \cdot x^4 = x^{11}$ (degree $11$). Term $3$: $x^5 \cdot (x^4)^2 = x^{13}$ (degree $13$). Term $4$: $x^3 \cdot (x^4)^3 = x^{15}$ (degree $15$). Term $5$: $x^1 \cdot (x^4)^4 = x^{17}$ (degree $17$). The highest power of $x$ in the expression is $17$. Therefore,the degree of the polynomial is $17$.

The degree of the polynomial $(x+\sqrt{x^4-1})^9+(x-\sqrt{x^4-1})^9$ is

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