Chapter 11.3, Problem 56PPE

To determine

To divide $\left(4a^3{b}^4-6a^2{b}^5+10a^2{b}^4\right)÷2a{b}^2$ .

Answer to Problem 56PPE

  $\left(4a^3{b}^4-6a^2{b}^5+10a^2{b}^4\right)÷2a{b}^2=2a^2{b}^2-3a{b}^3-5a{b}^3$

Explanation of Solution

Given:

  $\left(4a^3{b}^4-6a^2{b}^5+10a^2{b}^4\right)÷2a{b}^2$

Concept Used:

• To conduct a division, $\frac{R\left(x\right)}{Q\left(x\right)}$, cancel out the common factor of numerator and denominator.

Calculation:

In order to divide $\left(4a^3{b}^4-6a^2{b}^5+10a^2{b}^4\right)÷2a{b}^2$ first factorize the numerator(so it would be easy to cancel out denominator) by taking common factor $2a{b}^2$ out, as

  $\begin{array}{c}4a^3{b}^4-6a^2{b}^5+10a^2{b}^4=2a{b}^2\cdot 2a^2{b}^2-2a{b}^2\cdot 3a{b}^3-2a{b}^2\cdot 5a{b}^3\\ =2a{b}^2\left(2a^2{b}^2-3a{b}^3-5a{b}^3\right)\end{array}$

Now, the given divisor can be proceed as:

  $\begin{array}{c}\left(4a^3{b}^4-6a^2{b}^5+10a^2{b}^4\right)÷2a{b}^2=\frac{2a{b}^2\left(2a^2{b}^2-3a{b}^3-5a{b}^3\right)}{2a{b}^2}\\ =\frac{2a^2{b}^2-3a{b}^3-5a{b}^3}{1}\left(\begin{array}{l}\text{dividing common}\\ \text{factor }2a{b}^2\text{.}\end{array}\right)\\ =2a^2{b}^2-3a{b}^3-5a{b}^3\end{array}$

Thus,

  $\left(4a^3{b}^4-6a^2{b}^5+10a^2{b}^4\right)÷2a{b}^2=2a^2{b}^2-3a{b}^3-5a{b}^3$ .