Proving $x-1$ is a Factor of $P(x)$ with Polynomials

• MHB
• anemone
In summary, when we say that $x-1$ is a factor of $P(x)$ with polynomials, it means that when we divide $P(x)$ by $x-1$, the remainder is equal to zero. To prove this, we can use the factor theorem and substitute $x=1$ to see if the result is equal to zero. Alternatively, we can use synthetic division with $a=1$ and if the remainder is zero, then $x-1$ is a factor of $P(x)$. Proving this is important because it allows us to factorize $P(x)$, find other factors, and understand the behavior of the graph near $x=1$.
anemone
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MHB
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If $P(x),\,Q(x),\,R(x),\,S(x)$ are polynomials such that $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$, prove that $x-1$ is a factor of $P(x)$.

Let $P(x)=a_nx^n+\cdots+a_0x^0$ with $a_n\ne 0$. Comparing the coefficients of $x^{n+1}$ on both sides gives $a_n(n-2m)(n-1)=0$, so $n=1$ or $n=2m$.

If $n=1$, one easily verifies that $P(x)=x$ is a solution, while $P(x)=1$ is not. Since the given condition is linear in $P$, this means that the linear solutions are precisely $P(x)=tx$ for $t\in \mathbb{R}$.

Now, assume that $n=2m$. The polynomial $xP(x+1)-(x+1)P(x)=(n-1)a_nx^n+\cdots$ has degree $n$, and therefore it has at least one (possibly complex) root $r$. If $r\ne \{0,\,-1\}$, define $k=\dfrac{P(r)}{r}=\dfrac{P(r+1)}{r+1}$. If $r=0$, let $k=P(1)$. If $r=-1$, let $k=-P(-1)$. We now consider the polynomial $S(x)=P(x)-kx$. It also satisfies (1) because $P(x)$ and $kx$ satisfy it. Additionally, it has the useful property that $r$ and $r+1$ are roots.

Let $A(x)=x^3-mx^2+1$ and $B(x)=x^3+mx^2+1$. Plugging in $x=s$ into (1) implies that

If $s-1$ and $s$ are roots of $S$ and $s$ is not a root of $A$, then $s+1$ is a root of $S$.
If $s$ and $s+1$ are roots of $S$ and $s$ is not a root of $B$, then $s-1$ is a root of $S$.

Let $a\ge 0$ and $b\ge 1$ be such that $r-a,\,r-a+1,\,\cdots, r,\,r+1,\.\cdots,\,r+b-1,\,r+b$ are roots of $S$, while $r-a-1$ and $r+b+1$ are not. The two statements above imply that $r-a$ is a root of $B$ and $r+b$ is a root of $A$.

Since $r-a$ is a root of $B(x)$ and of $A(x+a+b)$, it is also a root of their greatest common divisior $C(x)$ as integer polynomials. If $C(x)$ was a non-trivial divisor of $B(x)$, then $B$ would have a rational root $\alpha$. Since the first and last coefficients of $B$ are 1, $\alpha$ can only be 1 or $-1$, but $B(-1)=m>0$ and $B(1)=m+2>0$ since $n=2m$.

Therefore $B(x)=A(x+a+b)$. Writing $c=a_b\ge 1$, we compute

$0=A(x+c)-B(x)=(3c-2m)x^2+c(3c-2m)x^2+c(3c-2m)x+c^2(c-m)$

Then we must have $3c-2m=c-m=0$, which gives $m=0$, a contradiction. We could conclude that $f(x)=tx$ is the only solution.

1. How do you prove that $x-1$ is a factor of $P(x)$ with polynomials?

To prove that $x-1$ is a factor of $P(x)$, we can use the Remainder Theorem. This theorem states that if we divide $P(x)$ by $x-1$, the remainder will be equal to $P(1)$. Therefore, if $P(1) = 0$, then $x-1$ is a factor of $P(x)$.

2. Can you provide an example of proving $x-1$ is a factor of $P(x)$ with polynomials?

For example, let's say we have the polynomial $P(x) = 3x^3 - 2x^2 + 5x - 2$. To prove that $x-1$ is a factor of $P(x)$, we divide $P(x)$ by $x-1$ using long division or synthetic division. The remainder is $P(1) = 4$, therefore $x-1$ is not a factor of $P(x)$. However, if $P(x) = 2x^3 + 4x^2 - 3x + 1$, the remainder is $P(1) = 4$, which means $x-1$ is a factor of $P(x)$.

3. What is the significance of proving $x-1$ is a factor of $P(x)$ with polynomials?

Proving that $x-1$ is a factor of $P(x)$ allows us to simplify the polynomial and potentially find its roots. This makes it easier to work with and solve equations involving the polynomial.

4. Are there any other methods to prove that $x-1$ is a factor of $P(x)$ with polynomials?

Yes, there are other methods such as using the Factor Theorem or the Rational Root Theorem. These theorems provide conditions for a polynomial to have a certain factor, and can be used to prove that $x-1$ is a factor of $P(x)$.

5. What are some common mistakes to avoid when proving $x-1$ is a factor of $P(x)$ with polynomials?

One common mistake is forgetting to check the remainder after dividing by $x-1$. It is important to remember that if the remainder is not equal to zero, then $x-1$ is not a factor of $P(x)$. Another mistake is assuming that just because $P(1) = 0$, $x-1$ is automatically a factor. It is important to use long division or synthetic division to confirm this.

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