Chapter 8, Problem 31PT

To determine

To factor the given polynomial

Answer to Problem 31PT

  $(x-1)(x-1)$

Explanation of Solution

Given:

Polynomial: $a^2-2a+1$

Formula used:

Roots of the polynomial of the form $a{x}^2+bx+c=a(x-x_1)(x-x_2)$ are

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Consider the given polynomial,

  $\Rightarrow a^2-2a+1$

Applying the above formula for $a=1,b=-2,c=1$

  $\begin{array}{l}\Rightarrow x=\frac{-(-2)\pm \sqrt{{(-2)}^2-4(1)(1)}}{2(1)}\\ \Rightarrow x=\frac{2\pm \sqrt{4-4}}{2}\\ \Rightarrow x=\frac{2\pm 0}{2}\\ \Rightarrow x=\frac{2}{2},\frac{2}{2}\\ \Rightarrow x=1,1\end{array}$

Therefore, the roots are

  $\Rightarrow x_1=x_2=1$

Therefore, the polynomial can be factorised as,

  $\Rightarrow (x-1)(x-1)$

Conclusion:

Therefore, the polynomial is factored as $(x-1)(x-1).$