Chapter 8, Problem 16PT

To determine

To factor the given polynomial

Answer to Problem 16PT

Prime polynomial

Explanation of Solution

Given:

Polynomial: $10a^2-50a-a+51$

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,

  $\Rightarrow 10a^2-50a-a+51$

Combining the like terms,

  $\Rightarrow 10a^2-51a+51$

Applying the above formula for $a=10,b=-51,c=51$

  $\begin{array}{l}\Rightarrow x=\frac{-(-51)\pm \sqrt{{(-51)}^2-4(10)(51)}}{2(10)}\\ \Rightarrow x=\frac{51\pm \sqrt{2601-2040}}{20}\\ \Rightarrow x=\frac{51\pm \sqrt{561}}{20}\end{array}$

Therefore, the polynomial cannot be factored further.

Conclusion:

Therefore, the polynomial is prime polynomial.