Question

Find the square root of $x^4-2x^3+3x^2-2x+1$ using the division method

• A. $x^2-x+1$
• B. $x^2-x-1$
• C. $x^2+x+1$
• D. $-x^2-x+1$

Answer 1

Answer: (A)

$x^2-x+1$

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:

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

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

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

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

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

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