Solve for remainder

Algebra Level 3

Find the remainder when x 155 x^{155} is divided by x 2 + 3 x + 2. x^2 + 3x +2.

None of the other choices. 2 155 2^{155} 3 154 3^{154} 3 155 3^{155}

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2 solutions

Chew-Seong Cheong
Aug 28, 2016

We note that x 2 + 3 x + 2 = ( x + 1 ) ( x + 2 ) x^2 + 3x +2 = (x+1)(x+2) . Let f ( x ) = x 155 f(x) = x^{155} and by remainder factor theorem ,

  • Remainder of f ( x ) f(x) divided by x + 1 x+1 is = f ( 1 ) = ( 1 ) 155 = 1 = f(-1) = (-1)^{155} = -1 .
  • Similarly, remainder of f ( x ) f(x) divided by x + 1 x+1 is = f ( 2 ) = ( 2 ) 155 = 2 155 = f(-2) = (-2)^{155} = -2^{155} .
  • Therefore, remainder of f ( x ) f(x) divided by ( x + 1 ) ( x + 2 ) (x+1)(x+2) is = ( 1 ) ( x + 2 ) + ( 2 155 ) ( x + 1 ) = (-1)(x+2)+(2^{155})(x+1) = ( 2 155 1 ) x 2 + 2 155 = (2^{155}-1)x - 2 + 2^{155} .

Therefore, the answer is None of the other choices. \boxed{\text{None of the other choices.}}

From where did the 3rd point comes ? can u prove it.

Prince Raj - 4 years, 9 months ago

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Sorry, it should be R ( x ) = ( 1 ) ( x + 2 ) + ( + 2 155 ) ( x + 1 ) R(x) = (-1)(x+2) + (\color{#D61F06}{+}2^{155})(x+1)

It is a combination of statements 1 and 2. R ( x ) = ( 1 ) ( x + 2 ) + ( 2 155 ) ( x + 1 ) R(x) = (-1)(x+2) + (2^{155})(x+1) . When x = 1 x=-1 , R ( 1 ) = ( 1 ) ( 1 + 2 ) + ( 2 155 ) ( 0 ) = 1 R(-1) = (-1)(-1+2) + (-2^{155})(0) = -1 . When x = 2 x=-2 , R ( 2 ) = ( 1 ) ( 0 ) + ( 2 155 ) ( 1 ) = 2 155 R(-2) = (-1)(0) + (2^{155})(-1) = -2^{155} . The other x x also follows the linear R ( x ) R(x) .

Chew-Seong Cheong - 4 years, 9 months ago
Shyam Upadhyay
Aug 28, 2016

Remainder will be linear Remainder=(2^155 -1)x +2^155 -2

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