Simple Division (not really)

Algebra Level 3

6 x 3 − 8 x − 5 2 x − 4 \large \frac{6x^3-8x-5}{2x-4}

What is the remainder of the expression above?


The answer is 27.

Terry Yu by Terry Yu , 17, USA

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2 solutions

Chew-Seong Cheong
Jul 18, 2017

Let f ( x ) = 6 x 3 − 8 x − 5 f(x)=6x^3-8x-5 and g ( x ) = 2 x − 4 g(x) = 2x-4 . By remainder theorem the remainder of f ( x ) g ( x ) \dfrac {f(x)}{g(x)} is f ( c ) f(c) , where c c is the root of g ( x ) = 0 g(x)=0 . Therefore, we have g ( x ) = 0 ⟹ 2 c − 4 = 0 ⟹ c = 2 g(x) = 0 \implies 2c - 4 = 0 \implies c = 2 . The remainder of f ( x ) g ( x ) \dfrac {f(x)}{g(x)} is f ( c ) = f ( 2 ) = 6 ( 2 3 ) − 8 ( 2 ) − 5 = 27 f(c) = f(2) = 6(2^3)-8(2)-5 = \boxed{27} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Terry Yu
Jun 2, 2017

The fraction can be rewritten as ( 3 x 2 + 6 x + 8 ) ( 2 x − 4 ) + 27 2 x − 4 = ( 3 x 2 + 6 x + 8 ) ( 2 x − 4 ) 2 x − 4 + 27 2 x − 4 \dfrac{(3x^2+6x+8)(2x-4)+27}{2x-4}=\dfrac{(3x^2+6x+8)\xcancel{(2x-4)}}{\xcancel{2x-4}}+\dfrac{\color{#D61F06}\boxed{27}}{2x-4} which the remainder is 27 \color{#D61F06}\boxed{27} .

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I was wondering: how did you come up with that rewriting of the numerator with the 27 27 that better clarifies what the remainder is? It seems like this is a case of solution based on already knowing what the answer is.

Zach Abueg - 3 years, 10 months ago

@Pi Han Goh - Do you know how to make a slash to cross out the two (2x-4)s? I'm not that good with latex but I know you are.

Terry Yu - 4 years ago

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Like this:

( 3 x 2 + 6 x + 8 ) ( 2 x − 4 ) + 27 2 x − 4 = ( 3 x 2 + 6 x + 8 ) ( 2 x − 4 ) 2 x − 4 + 27 2 x − 4 \dfrac{(3x^2+6x+8)(2x-4)+27}{2x-4} = \dfrac{(3x^2+6x+8)\cancel{(2x-4)}}{\cancel{2x-4}} + \dfrac{27}{2x-4}

OR

( 3 x 2 + 6 x + 8 ) ( 2 x − 4 ) + 27 2 x − 4 = ( 3 x 2 + 6 x + 8 ) ( 2 x − 4 ) 2 x − 4 + 27 2 x − 4 \dfrac{(3x^2+6x+8)(2x-4)+27}{2x-4} = \dfrac{(3x^2+6x+8)\bcancel{(2x-4)}}{\bcancel{2x-4}} + \dfrac{27}{2x-4}

OR

( 3 x 2 + 6 x + 8 ) ( 2 x − 4 ) + 27 2 x − 4 = ( 3 x 2 + 6 x + 8 ) ( 2 x − 4 ) 2 x − 4 + 27 2 x − 4 \dfrac{(3x^2+6x+8)(2x-4)+27}{2x-4} = \dfrac{(3x^2+6x+8)\xcancel{(2x-4)}}{\xcancel{2x-4}} + \dfrac{27}{2x-4}

Pi Han Goh - 4 years ago

ahh I see thank you @Pi Han Goh !

Terry Yu - 4 years ago

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