+0  
 
0
93
1
avatar+45 

1. Let $a,$ $b,$ $c$ be the roots of $x^3 - x + 4 = 0.$ Compute \[(a^2 - 1)(b^2 - 1)(c^2 - 1).\]

 

2. The fourth degree polynomial $P(x)$ satisfies $P(1) = 1,$ $P(2) = 2,$ $P(3) = 3,$ $P(4) = 4,$ and $P(5) = 125.$ What is $P(6)?$

 

3. Let $a > 0$, and let $P(x)$ be a polynomial with integer coefficients such that

 

\[P(1) = P(3) = P(5) = P(7) = a,\]and\[P(2) = P(4) = P(6) = P(8) = -a.\]

 

What is the smallest possible value of $a$?

 Aug 15, 2021
 #1
avatar+436 
+1

pls try to have one question per post :)

1.

Let the polynomial be $f(x).$

It is useful to find a polynomial with roots of \(a^2,b^2,c^2\) so we can quickly solve it using Vieta's.

Notice that \(f(\sqrt{x})\) has roots of \(a^2, b^2, c^2\), but only if that root is positive (specifically if the real part is positive). That is because \(\sqrt{x}=k\) for some constant where the real part is negative has no solutions.

Also, notice that \(f(-\sqrt{x})\) has roots of \(a^2, b^2, c^2\), but only of the real part of that root is negative. That means that \(f(\sqrt{x})\cdot f(-\sqrt{x})\) will ensure that it has roots of \(a^2, b^2, c^2\), regardless of if the roots are positive or negative. 

Also note that since the polynomial \(f(\sqrt{x})\cdot f(-\sqrt{x})\) is even, it can be expressed as a polynomial in terms of \(\sqrt{x}^2=x\).

Let g(x) be the polynomial with roots of \(a^2,b^2,c^2\). We can see that the polynomial, after some simplifying, is equal to:

\(-x^3 + 2 x^2 - x + 16\)

Note that \((a^2 - 1)(b^2 - 1)(c^2 - 1)=a^2 b^2 c^2 - a^2b^2 - a^2 c^2 - b^2 c^2 + a^2 + b^2 + c^2 - 1\). By Vieta's, \(a^2b^2c^2=-16, a^2b^2+a^2c^2+b^2c^2=-1, a^2+b^2+c^2=-2\), and the rest can be solved easily.

2.

This problem can be solved quickly using polynomial interpolation using remainder theorem as follows:

Notice that \(P(x)-x=0\) for \(x=1, 2, 3, 4\), which means that \(P(x)-x=A(x-1)(x-2)(x-3)(x-4)\) for some constant A. Rearranging, we get that \(P(x)=x+A(x-1)(x-2)(x-3)(x-4)\). We can solve for A by substituting 5 for x:

\(5+A(5-1)(5-2)(5-3)(5-4)=125\\ A(4)(3)(2)=120\\ A=5\)

So the polynomial is equal to \(5+5(x-1)(x-2)(x-3)(x-4)\).

Substitute x=6 to get the answer. 

 Aug 16, 2021

35 Online Users

avatar