Question

Using division method find the square root of the polynomial: $x^4+2x^3-3x^2+2x-1$

Answer 1

To find the square root of the given polynomial $x^4+2x^3-3x^2+2x-1$, we must equate it to the general form of equation that is $(a{x}^2+bx+c)$ as shown below:

$\sqrt{x^4+2x^3-3x^2+2x-1}=a{x}^2+bx+c\Rightarrow {\left(\sqrt{x^4+2x^3-3x^2+2x-1}\right)}^2={(a{x}^2+bx+c)}^2\Rightarrow x^4+2x^3-3x^2+2x-1={(a{x}^2+bx+c)}^2\Rightarrow x^4+2x^3-3x^2+2x-1={\left(a{x}^2\right)}^2+{\left(bx\right)}^2+{\left(c\right)}^2+(2\times a{x}^2\times bx)+(2\times bx\times c)+(2\times c\times a{x}^2)(\because (a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca)\Rightarrow x^4+2x^3-3x^2+2x-1=a^2{x}^4+b^2{x}^2+c^2+2ab{x}^3+2bcx+2ac{x}^2$

Now, comparing the coefficients, we get:

$a^2=1,b^2+2ac=-3,c^2=-1,2ab=2,2bc=2$

$a^2=1\Rightarrow a=1$

$c^2=-1\Rightarrow c=i$

$2ab=2\Rightarrow 2\times 1\times b=2\Rightarrow 2b=2\Rightarrow b=1$

Therefore, $a=1,b=1,c=i$ and substituting the values in the equation $(a{x}^2+bx+c)$, we get $x^2+x+i$.

Hence, the square root of $x^4+2x^3-3x^2+2x-1$ is $\left|\right(x^2+x+i\left)\right|$.