Question

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

• A. $1$
• B. $x$
• C. $\frac{x^2+(a+b+c)(ab+bc+ca)}{(a-b)(b-c)(c-a)}$
• D. $0$

Answer 1

The correct option is A $1$

$P\left(x\right)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-z)(c-b)}P\left(a\right)=\frac{(a-b)(a-c)}{(a-b)(a-c)}+\frac{(a-c)(a-a)}{(b-c)(b-a)}+\frac{(a-a)(a-b)}{(c-z)(c-b)}P\left(a\right)=1+0+0=1P\left(b\right)=0+1+0=1P\left(c\right)=0+0+1=1$

Option $\left(B\right),\left(C\right),\left(D\right)$ does not satisfy any one the condtions

$\therefore P\left(x\right)=1$

So option $\left(A\right)$ is correct