The expression [x + (x^3 - 1)^{1/2}]^5 + [x - | vedclass

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(C) Let $P(x) = [x + (x^3 - 1)^{1/2}]^5 + [x - (x^3 - 1)^{1/2}]^5$.Using the binomial expansion $(a+b)^n + (a-b)^n = 2 \sum_{k=0, 2, 4, ...} \binom{n}

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$7$

Vedclass · Answer

(C) Let $P(x) = [x + (x^3 - 1)^{1/2}]^5 + [x - (x^3 - 1)^{1/2}]^5$.Using the binomial expansion $(a+b)^n + (a-b)^n = 2 \sum_{k=0, 2, 4, ...} \binom{n}{k} a^{n-k} b^k$,where $a = x$ and $b = (x^3 - 1)^{1/2}$.$P(x) = 2 [ \binom{5}{0} x^5 + \binom{5}{2} x^3 (x^3 - 1) + \binom{5}{4} x (x^3 - 1)^2 ]$.$P(x) = 2 [ 1 \cdot x^5 + 10 (x^6 - x^3) + 5 x (x^6 - 2x^3 + 1) ]$.$P(x) = 2 [ x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x ]$.$P(x) = 10x^7 + 20x^6 + 2x^5 - 20x^4 - 20x^3 + 10x$.The highest power of $x$ in the resulting polynomial is $7$.Therefore,the degree of the polynomial is $7$.

The expression $[x + (x^3 - 1)^{1/2}]^5 + [x - (x^3 - 1)^{1/2}]^5$ is a polynomial of degree :

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