Question

FID: 43294
Find n if (x^n-27) is divisible by (x-3)

Answer 1

here, given that (x - 3) is a factor of (xⁿ - 27).
f(x) = (xⁿ - 27)
f(3) = 3ⁿ - 27
0 = 3ⁿ - 27
3ⁿ = 27
3ⁿ = 3³
n = 3
hence, we get valus of n = 3

$VERIFICATION:-$

(xⁿ - 27)/(x - 3)

(x³ - 27)/(x - 3)

(x³ - 3³)/(x - 3)

(x - 3)(x² + 9 + 3x)/(x - 3)

(x² + 9 + 3x)
hence, (x - 3) is a factor of (xⁿ - 27) [if n = 3]

Answer 2

Solution:-

given by:-
$p(x) = ( {x}^{n} - 27) \\ if \: (x - 3) \: facor \: of \: ( {x}^{n} - 27) \\ \: then \: x - 3 = 0 = x = 3 \\ p(3) = ({3}^{n} - 27) = 0 \\ > {3}^{n} = 27 \\ > {3}^{n} = {3}^{3} \\ > n = 3$
here (n = 3 ) is right answer is (x^n-27) is divisible by (x-3).


■I HOPE ITS HELP■