Suppose a, b, c are three distinct real numbe | vedclass

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(A) Given the expression $P(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$.Evaluate $P(x)$ at $x=

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the expression $P(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$.Evaluate $P(x)$ at $x=a$:$P(a) = \frac{(a-b)(a-c)}{(a-b)(a-c)} + 0 + 0 = 1$.Evaluate $P(x)$ at $x=b$:$P(b) = 0 + \frac{(b-c)(b-a)}{(b-c)(b-a)} + 0 = 1$.Evaluate $P(x)$ at $x=c$:$P(c) = 0 + 0 + \frac{(c-a)(c-b)}{(c-a)(c-b)} = 1$.Since $P(x)$ is a polynomial of degree at most $2$ and it takes the value $1$ for three distinct values $x = a, b, c$,it must be the constant polynomial $P(x) = 1$ for all real $x$.Therefore,the simplified value is $1$.

Suppose $a, b, c$ are three distinct real numbers. Let $P(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$. When simplified,$P(x)$ becomes:

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