Chapter 1.4, Problem 20E

To determine

To calculate: The rational form of expression, $\frac{y^2-3y-18}{2y^2+5y+3}$

Answer to Problem 20E

The simplified form of the given expression is,

  $\frac{\left(y-6\right)\left(y+3\right)}{\left(2y+3\right)\left(y+1\right)}$

Explanation of Solution

Given information:

The given expression is as, $\frac{y^2-3y-18}{2y^2+5y+3}$

Formula used:

For the algebraic expression the middle term factorization is as, $x^2-x-c$

These are the steps to do this ,

Step 1.multiply the coefficient of $x^2$ and the constant $c$ .

Step 2. Then do factor of that number by doing multiplication or division or subtraction or addition got the coefficient of middle term that means $x$ .

Step 3. Then to take common form first to numbers and then from other two numbers.

Step 4. Got four factors in that two pairs are same and another two are same.

Step 5. Get the final two factors.

Calculation :

Consider the expression, $\frac{y^2-3y-18}{2y^2+5y+3}$

Rewrite the numerator in the form of middle term factorization,

  $\begin{array}{c}\frac{y^2-3y-18}{2y^2+5y+3}=\frac{y^2-\left(6y-3y\right)-18}{2y^2+\left(2y+3y\right)+3}\\ =\frac{y^2-6y+3y-18}{\begin{array}{l}2y^2+2y+3y+3\\ \end{array}}\\ =\frac{y\left(y-6\right)+3\left(x-6\right)}{2y\left(y+1\right)+3\left(y+1\right)}\\ =\frac{\left(y+3\right)\left(y-6\right)}{\left(2y+3\right)\left(y+1\right)}\end{array}$

Rewrite the expression, $\frac{y^2-3y-18}{2y^2+5y+3}=\frac{\left(y-6\right)\left(y+3\right)}{\left(2y+3\right)\left(y+1\right)}$

The above expression can’t be simplified further. $\frac{\left(y-6\right)\left(y+3\right)}{\left(2y+3\right)\left(y+1\right)}$

Thus the simplified form of the expression is .