Question

Simplify and state restrictions. Show the factors before you simplify. a 2a2-a-1 a2–1 Х 4a? +40 2a2+7a+3

Answer 1

Given:

`(2a^2-a-1)/(a^2-1)*(4a^2+4a)/(2a^2+7a+3)`

Factorize the given expression.

`\text{ Use -a=-2a+a.}`

`(2a^2-2a+a-1)/(a^2-1)*(4a^2+4a)/(2a^2+7a+3)`

Take out the common terms.

`(2a(a-1)+(a-1))/(a^2-1)*(4a^2+4a)/(2a^2+7a+3)`

`((a-1)(2a+1))/(a^2-1)*(4a^2+4a)/(2a^2+7a+3)`
`Use\text{ }a^2-1=(a-1)(a+1).`

`((a-1)(2a+1))/((a-1)(a+1))*(4a^2+4a)/(2a^2+7a+3)`
`\text{Take out the co}mmon\text{ term }4a^2+4a=4a\mleft(a+1\mright).`

`((a-1)(2a+1))/((a-1)(a+1))*(4a(a+1))/(2a^2+7a+3)`
`\text{Use 7a=6a+a in }2a^2+7a+3.`

`((a-1)(2a+1))/((a-1)(a+1))*(4a(a+1))/(2a^2+6a+a+3)`

Take out common term.

`((a-1)(2a+1))/((a-1)(a+1))*\frac{4a(a+1)}{2a(a^{}+3)+(a+3)}`

`((a-1)(2a+1))/((a-1)(a+1))*\frac{4a(a+1)}{(a^{}+3)(2a+1)}`

The restrictions are

`a\\e1,-1,-3\text{ and }(-1)/(2)\text{.}`

Simplify the given expression

`((a-1)(2a+1))/((a-1)(a+1))*\frac{4a(a+1)}{(a^{}+3)(2a+1)}`

Cancel out the common factors.

`\frac{4a}{(a^{}+3)}`

The answer is :

`(2a^2-2a+a-1)/(a^2-1)*(4a^2+4a)/(2a^2+7a+3),\text{ }a\\e1,-1,-3\text{ and }(-1)/(2)\text{.}`
`\frac{4a}{a^{}+3},\text{ }a\\e-3.`