Chapter 8, Problem 60SGR

To determine

To factorize the trinomial

Answer to Problem 60SGR

  $2(x-\frac{1}{2})(x+\frac{7}{3})$

Explanation of Solution

Given:

Trinomial: $2y^2-9y+3$

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 trinomial $2y^2-9y+3$ ,

Substituting in the above formula where $a=2,b=-9,c=3$

  $\begin{array}{c}\Rightarrow x=\frac{-(-9)\pm \sqrt{-9^2-4(2)(3)}}{2(2)}\\ \Rightarrow x=\frac{9\pm \sqrt{81+24}}{4}\\ \Rightarrow x=\frac{-11\pm \sqrt{289}}{12}\\ \Rightarrow x=\frac{-11\pm 17}{12}\end{array}$

Therefore,

  $\begin{array}{l}\Rightarrow x_1=\frac{-11+17}{12}=\frac{6}{12}=\frac{1}{2}\\ \Rightarrow x_2=\frac{-11-17}{12}=\frac{-28}{12}=\frac{-7}{3}\end{array}$

Thus the expression becomes,

  $\begin{array}{l}\Rightarrow 2(6x^2+11x-7)\\ \Rightarrow 2(x-\frac{1}{2})(x-(-\frac{7}{3}))\\ \Rightarrow 2(x-\frac{1}{2})(x+\frac{7}{3})\end{array}$

Conclusion:

Therefore, the factors of the given trinomial are $2(x-\frac{1}{2})(x+\frac{7}{3}).$