Polynomials!

Algebra Level 4

Let there be a cubic polynomial equation P ( x ) P(x) with one root equal to 5, and two other roots, whose average is also 5 5 . If P ( 8 ) = 2016 P(8) = 2016 , find P ( 2 ) P(2) .


The answer is -2016.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Otto Bretscher
Mar 21, 2016

By Viète, the other monic factor must be x 2 10 x + b x^2-10x+b , so f ( x ) = a ( x 5 ) ( x 2 10 x + b ) f(x)=a(x-5)(x^2-10x+b) . We are told that f ( 8 ) = 3 a ( b 16 ) = 2016 f(8)=3a(b-16)=2016 so f ( 2 ) = 3 a ( b 16 ) = 2016. f(2)=-3a(b-16)=\boxed{-2016.}

No fair, no Viete's! I even labelled it as symmetry...

Just kidding + 1 +1 . short, nice solution.

Manuel Kahayon - 5 years, 2 months ago

Log in to reply

I'm using the symmetry f ( x ) = f ( 10 x ) f(x)=-f(10-x) ;)

Otto Bretscher - 5 years, 2 months ago
Manuel Kahayon
Mar 21, 2016

We just have to prove that P ( 5 + x ) = P ( 5 x ) P(5+x) = -P(5-x) , and the answer comes naturally.

Since the average of the two other roots is 5 5 , then the two roots can be expressed as 5 + k 5+k , 5 k 5-k for some number k k .

So, P ( x ) = a ( x 5 ) ( x ( 5 + k ) ) ( x ( 5 k ) ) P(x) = a(x-5)(x-(5+k))(x-(5-k)) for some a a , by the factorization of a polynomial.

So, P ( 5 + x ) = a ( ( x + 5 ) 5 ) ( ( x + 5 ) ( 5 + k ) ) ( ( x + 5 ) ( 5 k ) ) P(5+x) = a((x+5)-5)((x+5)-(5+k))((x+5)-(5-k))

P ( 5 + x ) = a ( x ) ( x + k ) ( x k ) P(5+x) = a(x)(x+k)(x-k)

Similarly, P ( 5 x ) = a ( ( x + 5 ) 5 ) ( ( x + 5 ) ( 5 + k ) ) ( ( x + 5 ) ( 5 k ) ) P(5-x) = a((-x+5)-5)((-x+5)-(5+k))((-x+5)-(5-k))

P ( 5 x ) = a ( x ) ( x + k ) ( x k ) P(5-x) = a(-x)(-x+k)(-x-k)

P ( 5 x ) = a ( x ) ( x + k ) ( x k ) P(5-x) = -a(x)(x+k)(x-k)

But, since P ( 5 + x ) = a ( x ) ( x + k ) ( x k ) P(5+x) = a(x)(x+k)(x-k) ,

P ( 5 x ) = ( a ( x ) ( x + k ) ( x k ) ) P(5-x) = -(a(x)(x+k)(x-k))

P ( 5 + x ) = P ( 5 x ) P(5+x) = -P(5-x)

Which is exactly what we wanted to prove. So,

P ( 8 ) = P ( 5 + 3 ) = ( P ( 5 3 ) = P ( 2 ) P(8) = P(5+3) = -(P(5-3) = -P(2)

2016 = P ( 2 ) , P ( 2 ) = 2016 2016 = -P(2), P(2) = \boxed{-2016} ,

Which is our final answer.

Right .... Same way ( + lim x 0 ln ( 1 + x x ) ) \left(+\displaystyle\lim_{x\to 0}\ln\left(\dfrac{1+x}{x}\right)\right)

Rishabh Jain - 5 years, 2 months ago

Log in to reply

Wow.... Because + 1 +1 is too mainstream XD

Manuel Kahayon - 5 years, 2 months ago

Log in to reply

Very true... ;-P.. I thought of trying something new.

Rishabh Jain - 5 years, 2 months ago

Log in to reply

@Rishabh Jain How did you find this problem...? It's not even in the "New" section yet....

Manuel Kahayon - 5 years, 2 months ago

Log in to reply

@Manuel Kahayon But in Explore section problems appear right at the moment someone posts them.. :-)

Rishabh Jain - 5 years, 2 months ago

Log in to reply

@Rishabh Jain Ohhh...??? Where are those sections...?

Manuel Kahayon - 5 years, 2 months ago

Log in to reply

@Manuel Kahayon On clicking continue after solving a problem we directly go to explore section where on clicking New all the recent new problems appear.

Rishabh Jain - 5 years, 2 months ago
William Isoroku
Mar 23, 2016

Easier solution; by using Vieta's formulae, the sum of the roots is 15 15 and the product let's call it d d , so the polynomial becomes;

P ( x ) = x 3 15 x 2 + ( 50 + d 5 ) x d P(x)=x^3-15x^2+(50+\frac{d}{5})x-d

Substitute 8 8 and 2016 2016 and solve for d d and the answer will come.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...