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 $1$

$a,b,c$ are $3$ distinct real numbers

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

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

$=\frac{(c-b+a-c+b-a)x^2-(c^2-b^2+a^2-c^2+b^2-a^2)x+(b{c}^2-b^{2}c+c{a}^2-c^{2}a+a{b}^2-a^{2}b)}{(a-b)(b-c)(c-a)}$

$=\frac{c^2(b-a)+c(a^2-b^2)-ab(a-b)}{(a-b)(b-c)(c-a)}$

$=\frac{(a-b)(-ab+bc+ac-c^2)}{(a-b)(b-c)(c-a)}$

$=\frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}$

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