Question

Simplify and state restrictions. Show the factors before cancelling. 2x2-6x x2-9 . 4x3+28x2 x2+8x+15

Answer 1

Given:

`(2x^2-6x)/(x^2-9)/(4x^3+28x^2)/(x^2+8x+15)`
`\begin{gathered} (2x^2-6x)/(x^2-9)*(x^2+8x+15)/(4x^3+28x^2) \\ \\ To\text{ simplify the expression, we n}eed\text{ to factorize each expression and cancel them out.} \\ \text{Hence,} \\ (2x(x-3))/((x-3)(x+3))*((x+3)(x+5))/(4x^2(x+7)) \end{gathered}`

Canceling out the common factors,

`\begin{gathered} (2x(x-3))/((x-3)(x+3))*((x+3)(x+5))/(4x^2(x+7)) \\ ((x+5))/(2x(x+7)) \end{gathered}`

Thus,

`(2x^2-6x)/(x^2-9)/(4x^3+28x^2)/(x^2+8x+15)=((x+5))/(2x(x+7))`

The restrictions will occur for the expression to be undefined.

The expression will be undefined when the denominator is zero.

Therefore,

`\begin{gathered} ((x+5))/(2x(x+7))\text{ will have the following restrictions,} \\ x\\e0\text{ and x }\\e-7 \end{gathered}`