The cool name is Cool Symmetry. Part II

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 ( a 1 + a 2 ) A 0 ( x ) + ( a 2 + a 0 ) A 1 ( x ) + ( a 0 + a 1 ) A 2 ( x ) = ? (a_1+a_2)A_0(x)+(a_2+a_0)A_1(x)+(a_0+a_1)A_2(x)= ?

A 0 ( x ) = ( x − a 1 ) ( x − a 2 ) ( a 0 − a 1 ) ( a 0 − a 2 ) A_0(x)=\dfrac{(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_1(x)=\dfrac{(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_2(x)=\dfrac{(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(x)=(x-a_0)(x-a_1)(x-a_2)

1 1 a 0 + a 1 + a 2 − x a_0+a_1+a_2-x x − 1 2 ( a 0 + a 1 + a 2 ) x-\frac{1}{2} (a_0+a_1+a_2) x x None of the given.

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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 + a 2 ) , A 1 ( x ) ⋅ ( a 0 + a 2 ) A_0(x)\cdot(a_1+a_2), A_1(x)\cdot(a_0+a_2) and A 2 ( x ) ⋅ ( a 0 + a 1 ) A_2(x)\cdot(a_0+a_1)

A 0 ( x ) ⋅ 0 = 0 A_0(x)\cdot0 = 0

A 1 ( x ) ⋅ ( − 1 ) = − x 2 + x 2 A_1(x)\cdot(-1) = -\dfrac{x^2 + x}{2}

A 2 ( x ) ⋅ ( 1 ) = x 2 − x 2 A_2(x)\cdot(1) = \dfrac{x^2 - x}{2}

Now finding the required expression, we clearly get − x \boxed{-x} . And a 0 + a 1 + a 2 = 0 a_0+a_1+a_2 = 0

Hence, the answer.

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