Question

The number of positive integral roots of ${{x}^{4}}+{{x}^{3}}-4{{x}^{2}}+x+1=0$ is/are

Answer 1

Hint: Use polynomial factorization method to factorize the given polynomial and find out all the roots or alternatively you can use Descartes’ sign rule to solve.

Complete step by step answer:
Let us denote the given polynomial equation as $p\left( x \right)$. So we can write
\[p\left( x \right)={{x}^{4}}+{{x}^{3}}-4{{x}^{2}}+x+1=0..(1)\]
Let us separate the term $-4{{x}^{2}}$ to convert the polynomial equation into a sum quadratic equation in ${{x}^{2}}$ and cubic equation in $x$
\[p\left( x \right)=\left( {{x}^{4}}-2{{x}^{2}}+1 \right)+\left( {{x}^{3}}-2{{x}^{2}}+x \right)=0\]
Let us use the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and proceed
\[\begin{align}
  & p\left( x \right)=\left( {{x}^{4}}-2{{x}^{2}}+1 \right)+\left( {{x}^{3}}-2{{x}^{2}}+x \right)=0 \\
 & \Rightarrow \left( {{\left( {{x}^{2}} \right)}^{2}}-2{{x}^{2}}+{{\left( 1 \right)}^{2}} \right)+x\left( {{x}^{2}}-2x+1 \right)=0 \\
 & \Rightarrow {{\left( {{x}^{2}}-1 \right)}^{2}}+x{{\left( x-1 \right)}^{2}}=0 \\
\end{align}\]
Let us use the algebraic identity of $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ and proceed
\[\begin{align}
  & \Rightarrow {{\left( x-1 \right)}^{2}}{{\left( x+1 \right)}^{2}}+x{{\left( x-1 \right)}^{2}}=0 \\
 & \Rightarrow {{\left( x-1 \right)}^{2}}\left( {{x}^{2}}+3x+1 \right)=0 \\
\end{align}\]
 We put $p\left( x \right)=0$ to find out the roots of the equation . So
\[{{\left( x-1 \right)}^{2}}=0,{{x}^{2}}+3x+1=0...(2)\]
  We get the roots $x=1,1$ are roots from the first equation. So we got two roots who are equal.\[\]
We know that any quadratic equation in one variable of the form $a{{x}^{2}}+bx+c=0,a\ne 0$ will have integral roots if the discriminant $D={{b}^{2}}-4ac$ satisfies the following condition\[\]
1. $D\ge 0$\[\]
2. $D$ is a perfect square.\[\]
3. $-b\pm \sqrt{D}$ must be divisible by $2a$\[\]
Let us calculate discriminant for the quadratic equation $a{{x}^{2}}+3x+1=0.$ from in equation(2)\[\]
$D={{3}^{2}}-4\cdot 1\cdot 1=5$ \[\]
$D$ though greater than 0 ,is not a perfect square. So we will not get any integral root from the equation $a{{x}^{2}}+3x+1=0.$
So the total number of positive integral roots is 2 which are equal. \[\]
Alternatively,
We know from Descartes' rule of sign that a polynomial has as many as positive integral roots as same or less than by an even number, the coefficients of polynomial change sign when written in descending order of the variable. The given polynomial is written in descending order of power,
\[p\left( x \right)={{x}^{4}}+{{x}^{3}}-4{{x}^{2}}+x+1=0..(1)\]
We can see coefficients are 1,1,-4,1,1. The coefficients change their sign 2 times(1 to -4 and -4 to 1) which is an even number .

So the number of real roots is 2.

Note: The question tests your knowledge of polynomial factorization and Descarate’s sign rule. Careful solving of simultaneous equations, substitution and use of rules will lead us to the correct result. The question can also be framed as whether another polynomial satisfies or or more than one of the roots.