Chapter 8, Problem 56SGR

To determine

To factorize the given polynomial

Answer to Problem 56SGR

  $(x-6)(x+8)$

Explanation of Solution

Given:

Polynomial: $x^2-2x-48$

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,

  $\Rightarrow x^2-2x-48$

Substituting in the above formula where $a=1,b=-2,c=-48$

  $\begin{array}{l}\Rightarrow x=\frac{-(-2)\pm \sqrt{{(-2)}^2-4(1)(-48)}}{2(1)}\\ \Rightarrow x=\frac{2\pm \sqrt{4+192}}{2}\\ \Rightarrow x=\frac{2\pm \sqrt{196}}{2}\\ \Rightarrow x=\frac{-2\pm 14}{2}\end{array}$

The roots are,

  $\begin{array}{l}\Rightarrow x_1=\frac{-2+14}{2}=\frac{12}{2}=6\\ \Rightarrow x_2=\frac{-2-14}{2}=\frac{-16}{2}=-8\end{array}$

The roots of the quadratic polynomial $x^2-2x-48=(x-6)(x+8).$

Conclusion:

Therefore, the factors of the polynomial are $(x-6)(x+8).$