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Algebra Level 1

( 5 x 2 + 14 x + 2 ) 2 − ( 4 x 2 − 5 x + 7 ) 2 x 2 + x + 1 = a ( x 2 + 19 x − 5 ) + b \dfrac{(5x^2 + 14x + 2)^2 - (4x^2-5x+7) ^2}{x^2+x+1} = a(x^2+19x-5) + b

What is a + b a+b ?


The answer is 9.

Jason Chrysoprase by Jason Chrysoprase , 18, Indonesia

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1 solution

Jason Chrysoprase
Jan 22, 2016

Assume a a = 5 x 2 + 14 x + 2 5x^2 + 14x +2 , b b = 4 x 2 − 5 x + 7 4x^2 - 5x +7

By using a 2 − b 2 = ( a + b ) ( a − b ) a^2 - b^2 = (a+b)(a-b) equality,

and then our expression become :

( 5 x 2 + 14 x + 2 + 4 x 2 − 5 x + 7 ) ( 5 x 2 + 14 x + 2 − ( 4 x 2 − 5 x + 7 ) ) x 2 + x + 1 \frac {(5x^2 + 14x +2 +4x^2 - 5x +7 ) ( 5x^2 + 14x +2 - (4x^2 - 5x +7))}{x^2+x+1}

= ( 9 x 2 + 9 x + 9 ) ( x 2 + 19 x − 5 ) x 2 + x + 1 = \frac {(9x^2 + 9x +9 )(x^2 + 19x -5)}{x^2+x+1}

= 9 ( x 2 + x + 1 ) ( x 2 + 19 x − 5 ) x 2 + x + 1 = \frac {9(x^2 + x + 1 )(x^2 + 19x -5)}{x^2+x+1}

= 9 ( x 2 + 19 x − 5 ) = a ( x 2 + 19 x − 5 ) + b = 9(x^2 + 19x -5) = a(x^2+19x-5)+b

So, a = 9 , b = 0 a = 9, b = 0

a + b = 9 + 0 = 9 a + b = 9 + 0 = 9

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