Chapter 8, Problem 76SGR

To determine

To factorize the given polynomial

Answer to Problem 76SGR

  $9(x-\frac{2}{3})(x+\frac{2}{3})$

Explanation of Solution

Given:

Polynomial: $9y^2-12y+4$

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,

  $\Rightarrow 9y^2-12y+4$

Substituting in the above formula where $a=9,b=-12,c=4$

  $\begin{array}{l}\Rightarrow x=\frac{-(-12)\pm \sqrt{{(-12)}^2-4(9)(4)}}{2(9)}\\ \Rightarrow x=\frac{12\pm \sqrt{144-144}}{18}\\ \Rightarrow x=\frac{12\pm 0}{18}\\ \Rightarrow x=\frac{2\pm 0}{3}\\ \Rightarrow x=\frac{2}{3},\frac{2}{3}\end{array}$

The roots are,

  $\Rightarrow x_1=x_2=\frac{2}{3}$

The roots of the quadratic polynomial $9y^2-12y+4=9(x-\frac{2}{3})(x+\frac{2}{3}).$

Conclusion:

Therefore, the factors of the polynomial are $9(x-\frac{2}{3})(x+\frac{2}{3}).$