x=√3+1/2 find 4x cube +2x square-8x+7 Please Reply asap exam today at 7:45 AM
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4x^3 + 2x^2 − 8x + 7
given x = (√3 +1)/2
4((√3 +1)/2)^3 + 2((√3 +1)/2)^2 − 8((√3 +1)/2) + 7
given (a/b)^n = a^n/b^n
=> 4(√3 +1)^3/8 + 2(√3 +1)^2/4 − 8(√3 +1)/2 + 7
=> (√3 +1)^3/2 + (√3 +1)^2/2 − 4(√3 +1) + 7
given (a + b)^n = ∑[k=0,n] C(n,k) * a^(n−k) * b^k
hence (a + b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3
and (a + b)^2 = 1a^2 + 2ab + 1b^2
=> (3√3 + 9 + 3√3 +1)/2 + (3 + 2√3 +1)/2 − 4(√3 +1) + 7
=> (6√3 + 10)/2 + (2√3 +4)/2 − 4(√3 +1) + 7
=> 3√3 + 5 + √3 + 2 − 4√3 − 4 + 7
=> 10
given x = (√3 +1)/2
4((√3 +1)/2)^3 + 2((√3 +1)/2)^2 − 8((√3 +1)/2) + 7
given (a/b)^n = a^n/b^n
=> 4(√3 +1)^3/8 + 2(√3 +1)^2/4 − 8(√3 +1)/2 + 7
=> (√3 +1)^3/2 + (√3 +1)^2/2 − 4(√3 +1) + 7
given (a + b)^n = ∑[k=0,n] C(n,k) * a^(n−k) * b^k
hence (a + b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3
and (a + b)^2 = 1a^2 + 2ab + 1b^2
=> (3√3 + 9 + 3√3 +1)/2 + (3 + 2√3 +1)/2 − 4(√3 +1) + 7
=> (6√3 + 10)/2 + (2√3 +4)/2 − 4(√3 +1) + 7
=> 3√3 + 5 + √3 + 2 − 4√3 − 4 + 7
=> 10
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