In this example we discuss the Factor Theorem. First, let us recall the following concept: if \(P(x)\) is a polynomial, then a number \(z\), is a root of \(P(x)\) if \(P(z)=0\). For example, \(x=1\) is a root of the polynomial \(Q(x)=x-1\). Show:
If \(0\) is a root of \(P(x)\), i.e: \(P(0)=0\), then \(P(x)=xQ(x)\) for some polynomial \(Q(x)\).
Use part 1 to show the Factor Theorem: if \(z\) is a root of \(P(x)\), then \(P(x)=(x-z)K(x)\) for some polynomial \(K(x)\).
In this example we discuss one of Vieta’s formulae. Consider the polynomial \(P(x)=x^2+5x-7\). You can take it as a fact that this polynomial has exactly two distinct roots. What is the sum of its roots? What about their product?
The polynomial \(P(x)=x^3+3x^2-7x+1\) has three distinct roots: \(a,b,\) and \(c\). What is the value of \(a^2+b^2+c^2\)?
So far we have discussed polynomials in one variable, i.e: with only an \(x\) as our variable. We can however, include as many as we want. For example, we can talk of a polynomial such as \[P(x,y)=x^2-y^2,\] where both \(x\) and \(y\) are variables. This is an example of an antisymmetric polynomial, which means that \(P(x,y)=-P(y,x)\) (i.e: switching \(x\) for \(y\) gives the original polynomial with a minus sign). Conversely, a polynomial \(Q(x,y)\) such that \(Q(x,y)=Q(y,x)\) is called symmetric. Show that every antisymmetric polynomial \(P(x,y)\) can be factored as \[P(x,y)=(x-y)Q(x,y),\] where \(Q(x,y)\) is a symmetric polynomial.
Find the remainder of dividing \(x^{100}-2x^{51}+1\) by \(x^2-1\). Try not to do a long calculation.
Katie and James take turns replacing the stars with any number of their choice in the following polynomial from left to right: \[x^4 +\star x^3+\star x^2+\star x+\star.\] Katie wins if once the game is over, the resulting polynomial has no integer roots (i.e: no roots which are whole numbers). Does James have a winning strategy?
Show that if a polynomial \(P(x)\) has an integer root (i.e: a root that is a whole number), then it can’t be that \(P(0)\) and \(P(1)\) are both odd.
Solve the equation \[\left(x^2-3x+3\right)^2-3\left(x^2-3x+3\right)+3=x\]
(USO 1974) Let \(a,b,c\) be three distinct integers, and let \(P(x)\) be a polynomial whose coefficients are all integers. Prove that it is not possible that the following three conditions hold at the same time: \(P(a)=b, P(b)=c,\) and \(P(c)=a\).
For a polynomial \(P(x)=ax^2+bx+c\), consider the following two kinds of transformations:
Swap coefficients \(a\) and \(c\). Hence the polynomial \(P(x)\) becomes \(cx^2+bx+a\) after this transformation.
For any number \(t\) of your choice, change the variable \(x\) into \(x+t\). For example, with the choice of \(t=1\), after this transformation, the polynomial \(x^2+x+1\) becomes \((x+1)^2+(x+1)+1=x^2+3x+3\).
Is it possible, using only a sequence of these two transformations, to change the polynomial \(x^2-x-2\) into the polynomial \(x^2-x-1\)?