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If x is an even positive integer, what is the remainder when \(\frac{9x^3+27x^2- x-3}{3x^2+8x-3}\) is divided by 6?
    (A) 0
    (B) 1
    (C) 2
    (D) 4
    (E) 5

\(\frac{9x^3+27x^2- x-3}{3x^2+8x-3}\) = \(\frac{(9x^2-1)(x+3)}{(3x-1)(x+3)}\)=\(\frac{(3x+1)(3x-1)(x+3)}{(3x-1)(x+3)}\) = \(3x+1\)

Since x is even 3x will be divisible by 6 => Remainder will be 1

Answer - B
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(9x3+27x2−x−3)/(3x2+8x−3 )
= (9x2−1)(x+3)/(3x−1)(x+3)
=(3x+1)(3x−1)(x+3)/(3x−1)(x+3)
= 3x+1
Since x is even 3x will be divisible by 6 => Remainder will be 1
correct answer is B
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Ans: B

(9x3+27x2−x−3)/(3x2+8x−3 )
= (9x2−1)(x+3)/(3x−1)(x+3)
=(3x+1)(3x−1)(x+3)/(3x−1)(x+3)
= 3x+1

No as x is even, so 3x must be divisible by 6. So, remainder is 1.
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If x is an even positive integer, what is the remainder when \(\frac{9x^3+27x^2- x-3}{3x^2+8x-3}\) is divided by 6?
    (A) 0
    (B) 1
    (C) 2
    (D) 4
    (E) 5


Simplify \(\frac{9x^3+27x^2- x-3}{3x^2+8x-3}\)=\(\frac{9x^2(x+3)-1( x+3)}{3x^2+9x-x-3}\)=\(\frac{(9x^2-1)(x+3)}{3x(x+3)-1(x+3)}\)=\(\frac{9x^2-1}{3x-1}=\frac{(3x-1)(3x+1)}{3x-1}=3x+1\)

So the remainder when \(3x+1\) is divided by 6, where x ix even..
So 3x will be divisible by 6 and we are left with only 1

B
we can take x = 2 and get the value.
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The official solution is shared. You can refer to it to check your approach. :)
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chetan2u
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If x is an even positive integer, what is the remainder when \(\frac{9x^3+27x^2- x-3}{3x^2+8x-3}\) is divided by 6?
    (A) 0
    (B) 1
    (C) 2
    (D) 4
    (E) 5


Simplify \(\frac{9x^3+27x^2- x-3}{3x^2+8x-3}\)=\(\frac{9x^2(x+3)-1( x+3)}{3x^2+9x-x-3}\)=\(\frac{(9x^2-1)(x+3)}{3x(x+3)-1(x+3)}\)=\([fraction]9x^2-1/3x-1[/fraction]=\frac{(3x-1)(3x+1)}{3x-1}=3x+1\)

So the remainder when \(3x+1\) is divided by 6, where x ix even..
So 3x will be divisible by 6 and we are left with only 1

B

chetan2u what if you reduced numerator and denominator by 3 in the highlighted part ? how would you arrive at the correct answer in this case?
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dave13
chetan2u
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If x is an even positive integer, what is the remainder when \(\frac{9x^3+27x^2- x-3}{3x^2+8x-3}\) is divided by 6?
    (A) 0
    (B) 1
    (C) 2
    (D) 4
    (E) 5


Simplify \(\frac{9x^3+27x^2- x-3}{3x^2+8x-3}\)=\(\frac{9x^2(x+3)-1( x+3)}{3x^2+9x-x-3}\)=\(\frac{(9x^2-1)(x+3)}{3x(x+3)-1(x+3)}\)=\([fraction]9x^2-1/3x-1[/fraction]=\frac{(3x-1)(3x+1)}{3x-1}=3x+1\)

So the remainder when \(3x+1\) is divided by 6, where x ix even..
So 3x will be divisible by 6 and we are left with only 1

B

chetan2u what if you reduced numerator and denominator by 3 in the highlighted part ? how would you arrive at the correct answer in this case?


You can reduce a fraction only when the entire numerator and entire denominator is a multiple of 3. But her only the coefficients of x^2 are multiple of 3.
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