V=P 3 ={a 3 x 3 +a 2 x 2 +a 1 x+a 0 |a 0 ,a 1 ,a 2, a 3 R } e 1 =1,e 2 =x,e 3 =x

Question

V=P3={a3x3+a2x2+a1x+a0|a0,a1,a2,a3R} e1=1,e2=x,e3=x2,e4=x3 a) Does the function "x3+1,x2-4,x+2,x-5"form a basis in V? b)Present (x+2)3 as a linear combination of thefunction listed in a. V=P3={a3x3+a2x2+a1x+a0|a0,a1,a2,a3R} e1=1,e2=x,e3=x2,e4=x3 a) Does the function "x3+1,x2-4,x+2,x-5"form a basis in V? b)Present (x+2)3 as a linear combination of thefunction listed in a. a) Does the function "x3+1,x2-4,x+2,x-5"form a basis in V? b)Present (x+2)3 as a linear combination of thefunction listed in a.

Explanation / Answer

x3+1,x2-4,x+2,x-5 = ae1 + be2 + ce3 +de4 = (d + a, c - 4a, b + 2a, b - 5a) = (0,0,0,0) if and onlyif d = -a, c = 4a, b = -2a, b = 5a so b = a = 0 thusx3+1,x2-4,x+2,x-5 are linearly independentand form a basis for V. b) (x+2)^3 = (x^2 + 4x + 4)(x + 2) = x^3 + 2x^2 + 4x^2 + 8x +4x + 8 = x^3 + 6x^2 + 12x + 8 = a(x3+1) +b(x2-4) + c(x+2) + d(x-5) = so a = 1 and b = 6and 31 = 2c - 5d, c + d = 12, c = 12-d so 31 = 2(12-d) - 5d = 24 - 7d so d = -1 and c = 13thus (x+2)3 = (x3+1) + 6(x2-4) +13(x+2) + -1(x-5)

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