An algebra problem by Sandeep Bhardwaj

Algebra Level 4

If A 0 ( x ) , A 1 ( x ) , and A 2 ( x ) A_0(x),A_1(x), \text{ and } A_2(x) are the three polynomials and a 0 , a 1 , and a 2 a_0,a_1,\text{ and } a_2 are three distinct real numbers, then compute A 0 ( x ) + A 1 ( x ) + A 2 ( x ) . A_0(x)+A_1(x)+A_2(x).

A 0 ( x ) = ( x − a 1 ) ( x − a 2 ) ( a 0 − a 1 ) ( a 0 − a 2 ) , A 1 ( x ) = ( x − a 0 ) ( x − a 2 ) ( a 1 − a 0 ) ( a 1 − a 2 ) , A 2 ( x ) = ( x − a 0 ) ( x − a 1 ) ( a 2 − a 0 ) ( a 2 − a 1 ) , A ( x ) = ( x − a 0 ) ( x − a 1 ) ( x − a 2 ) A_0(x)=\dfrac{(x-a_1)(x-a_2)}{(a_0-a_1)(a_0-a_2)} ,\quad A_1(x)=\dfrac{(x-a_0)(x-a_2)}{(a_1-a_0)(a_1-a_2)}, \\ A_2(x)=\dfrac{(x-a_0)(x-a_1)}{(a_2-a_0)(a_2-a_1)} , \quad A(x)=(x-a_0)(x-a_1)(x-a_2)

1 x 2 x^2 None of the given choices. 1 + x + x 2 1+x+x^2 x x

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Skanda Prasad
May 6, 2018

Given that a 0 , a 1 , a 2 a_0,a_1,a_2 are distinct real numbers. Hence, let a 0 = 0 a_0=0 , a 1 = 1 a_1=1 and a 2 = − 1 a_2=-1

Obtaining the expressions for A 0 ( x ) , A 1 ( x ) A_0(x), A_1(x) and A 2 ( x ) A_2(x)

A 0 ( x ) = 1 − x 2 A_0(x) = 1-x^2

A 1 ( x ) = x 2 + x 2 A_1(x) = \dfrac{x^2 + x}{2}

A 2 ( x ) = x 2 − x 2 A_2(x) = \dfrac{x^2 - x}{2}

Adding the above equations, we obtain

A 0 ( x ) + A 1 ( x ) + A 2 ( x ) = 1 − x 2 + x 2 2 + x 2 + x 2 2 − x 2 A_0(x) + A_1(x) + A_2(x) = 1-x^2 + \dfrac{x^2}{2} + \dfrac{x}{2} + \dfrac{x^2}{2} - \dfrac{x}{2}

A 0 ( x ) + A 1 ( x ) + A 2 ( x ) = 1 A_0(x) + A_1(x) + A_2(x) = \boxed{1}

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