Question

If \[P(x) = \dfrac{{\left( {x - b} \right)\left( {x - c} \right)}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \dfrac{{\left( {x - c} \right)\left( {x - a} \right)}}{{\left( {b - c} \right)\left( {b - a} \right)}} + \dfrac{{\left( {x - a} \right)\left( {x - b} \right)}}{{\left( {c - a} \right)\left( {c - b} \right)}}\], then for all real values of \[x\], \[P(x)\] is equals
A. \[1\]
B. \[0\]
C. \[x\]
D. \[\dfrac{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]

Answer 1

Hint: First we have to know a polynomial function is a function such as a quadratic, a cubic, a quartic, and so on, involving only non-negative integer powers of x. Using the standard algebraic formulas, we simplify the given polynomial function as much as possible. After that we rewrote the simplify function to reduce it.

Complete step by step answer:
Some important algebra formulas:
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
\[{\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)\]
\[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)\]
\[{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\]
\[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\]

Given \[P(x) = \dfrac{{\left( {x - b} \right)\left( {x - c} \right)}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \dfrac{{\left( {x - c} \right)\left( {x - a} \right)}}{{\left( {b - c} \right)\left( {b - a} \right)}} + \dfrac{{\left( {x - a} \right)\left( {x - b} \right)}}{{\left( {c - a} \right)\left( {c - b} \right)}}\]------(1)
The equation (1) can be rewrite as
\[P(x) = - \dfrac{{\left( {x - b} \right)\left( {x - c} \right)}}{{\left( {a - b} \right)\left( {c - a} \right)}} - \dfrac{{\left( {x - c} \right)\left( {x - a} \right)}}{{\left( {b - c} \right)\left( {a - b} \right)}} - \dfrac{{\left( {x - a} \right)\left( {x - b} \right)}}{{\left( {c - a} \right)\left( {b - c} \right)}}\]------(2)
Taking LCM in the RHS of the equation (2) becomes
\[P(x) = - \dfrac{{\left[ {\left( {x - b} \right)\left( {x - c} \right)\left( {b - c} \right) + \left( {x - c} \right)\left( {x - a} \right)\left( {c - a} \right) + \left( {x - a} \right)\left( {x - b} \right)\left( {a - b} \right)} \right]}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]----(3)
We know that \[\left( {x - a} \right)\left( {x - b} \right) = {x^2} - x(a + b) + ab\], then the equation (4) becomes
\[P(x) = - \dfrac{{\left[ {\left( {{x^2} - x(b + c) + bc} \right)\left( {b - c} \right) + \left( {{x^2} - x(a + c) + ac} \right)\left( {c - a} \right) + \left( {{x^2} - x(b + a) + ab} \right)\left( {a - b} \right)} \right]}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]---(4)
Since \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\] then the equation (4) becomes
\[P(x) = - \dfrac{{\left[ {\left( {{x^2}\left( {b - c + c - a + a - b} \right) - x({b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}) + bc\left( {b - c} \right) + ac\left( {c - a} \right) + ab\left( {a - b} \right)} \right)} \right]}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]---(5)
Simplifying the equation (5) we get
\[ \Rightarrow \]\[P(x) = - \dfrac{{\left[ {\left( {{x^2}\left( 0 \right) - x(0) + {b^2}c - b{c^2} + a{c^2} - {a^2}c + {a^2}b - a{b^2}} \right)} \right]}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]
\[ \Rightarrow \]\[P(x) = - \dfrac{{\left[ {{b^2}c - b{c^2} + a{c^2} - {a^2}c + {a^2}b - a{b^2}} \right]}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]
Rewriting the above equation to cancel some terms and reduce it
\[ \Rightarrow \]\[P(x) = - \dfrac{{\left[ {\left( {{b^2} - {a^2}} \right)c + {c^2}\left( {a - b} \right) + ab\left( {a - b} \right)} \right]}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]
\[ \Rightarrow \]\[P(x) = - \dfrac{{\left[ {\left( {a - b} \right)\left( { - \left( {a + b} \right)c + {c^2} + ab} \right)} \right]}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]
\[ \Rightarrow \]\[P(x) = - \dfrac{{\left[ {\left( {a - b} \right)\left( { - ac - bc + {c^2} + ab} \right)} \right]}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]
\[ \Rightarrow \]\[P(x) = - \dfrac{{\left[ {\left( {a - b} \right)\left( {c\left( {c - a} \right) - b\left( {c - a} \right)} \right)} \right]}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]
\[ \Rightarrow \]\[P(x) = \dfrac{{\left( {a - b} \right)\left( {c - a} \right)\left( {b - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\].
\[P(x) = 1\] for all values of \[x\].

So, the correct answer is “Option A”.

Note: Note that a polynomial is defined as an expression which is composed of variables, constants and exponents, that are combined using the mathematical operations such as addition, subtraction, multiplication and division. A polynomial is a function of the form \[f(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + .... + {a_2}{x^2} + {a_1}x + {a_0}\]. The degree of a polynomial is the highest power of \[x\] in its expression.