Chapter 6.4, Problem 62STP

To determine

To find: the algebraic equation by simplifying the division.

Answer to Problem 62STP

  $\text{y=-4,-}\frac{1}{4}-\frac{\sqrt{13}}{4}\text{and -}\frac{1}{4}+\frac{\sqrt{13}}{4}$ .

Explanation of Solution

Given:

  $\left({\text{4y}}^{\text{3}}{\text{+18y}}^{\text{2}}\text{+5y-12}\right)\text{÷}\left(\text{y+4}\right)$ .

Concept used:

In division method the term which is dividing or in the second term it will reciprocal and the sign Between them will change to multiplication from division.

Factoring is the decomposition of an algebraic integers or polynomial into the product of its factors which when multiply together gives the original.

Step first is to put the value of $\text{x}$ such that the algebraic sum of the equation will be zero through This it can figure out the first factor of the equation after that the other factors can be easily Deduced by using factoring methods or division polynomial method.

Calculation:

  $\left({\text{4y}}^{\text{3}}{\text{+18y}}^{\text{2}}\text{+5y-12}\right)\text{÷}\left(\text{y+4}\right)$ .

  $\left({\text{4y}}^{\text{3}}{\text{+18y}}^{\text{2}}\text{+5y-12}\right)\times \frac{1}{\left(\text{y+4}\right)}$ .

  $\frac{\left(\text{y+4}\right)\left({\text{4y}}^{\text{2}}\text{+2y-3}\right)}{\left(\text{y+4}\right)}$ .

  ${\text{4y}}^{\text{2}}\text{+2y-3}$ .

Here, $\text{a=4,b=2 and c=-3}$ .

  $\text{y=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}$ .

  $\text{y=}\frac{\text{-2±}\sqrt{{\left(2\right)}^{\text{2}}\text{-4}\left(4\right)\left(-3\right)}}{8}$ .

  $\text{y=-}\frac{1}{4}-\frac{\sqrt{13}}{4},\text{-}\frac{1}{4}+\frac{\sqrt{13}}{4}$ .

Hence, $\text{y=-4,-}\frac{1}{4}-\frac{\sqrt{13}}{4}\text{and -}\frac{1}{4}+\frac{\sqrt{13}}{4}$ .