Time remaining:
simplify the following polynomial function

Algebra
Tutor: None Selected Time limit: 1 Day

p(x) = a (x-b)(x-c) / (a-b)(a-c) +b (x-a)(x-c)/(b-a)(b-c) + c (x-a)(x-b)/(c-a)(c-b)

Dec 2nd, 2014

The answer is p(x) = x. The simplest (while not most elegant) way to solve this task is to compute directly.

First, lead it to a common denominator (a-b)(b-c)(c-a), denote this denominator as D. We must consider a sign - at each summand:
p(x) = [-a(x-b)(x-c)(b-c) - b(x-a)(x-c)(c-a) - c(x-a)(x-b)(a-b)]/[D]


We see that D didn`t depend on the x, let`s calculate the numerator:

-a(x^2 - bx - cx + bc)(b-c) - b(x^2 - ax - cx + ac)(c-a) - c(x^2 - ax - bx + ab)(a-b).


Let`s gather the summands with x^2, x^1 and x^0 (without x):

x^2*[-a(b-c) - b(c-a) - c(a-b)] + x^1*[a(b+c)(b-c) + b(a+c)(c-a) + c(a+b)(a-b)] +[-abc(b-c) - abc(c-a) - abc(a-b)] =

= x^2*[-ab + ac - bc + ba - ca + cb] + x^1*[a(b^2-c^2) + b(c^2-a^2) + c(a^2-b^2)] + [-abc(b - c + c - a + a - b)].


x^2 and x^0 obviously have zero factors, so our numerator is

x^1 * [ab^2 - ac^2 + bc^2 - ba^2 + ca^2 - cb^2].


And finally, open the brackets in D:

D = (a-b)(b-c)(c-a) = (ab - b^2 - ac + bc)(c-a) = abc - cb^2 - ac^2 + ca^2 - ba^2 + ab^2 + ca^2 -  abc.

+- abc vanishes and we see that the numerator equals x*D. So, p(x) = x.

Feb 5th, 2015

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