# help with this factoring problem please: `(x^2+1)^(3/2) * (4)*(x+5)^3 + (x+5)^4 (3/2)(x^2+1)^(1/2) * (2x)` is there a specific formula? I really would like a step by step process. thank...

help with this factoring problem please: `(x^2+1)^(3/2) * (4)*(x+5)^3 + (x+5)^4 (3/2)(x^2+1)^(1/2) * (2x)`

is there a specific formula? I really would like a step by step process. thank you in advance.

### 1 Answer | Add Yours

> To start, determine the GCF.To do this, consider the two terms of the expression.

`(x^2+1)^(3/2) * (4)* (x+5)^3`

and

`(x+5)^4 * (3/2) (x^2+1)^(1/2) *(2x) = (x+5)^4 *(3)(x^2+1)^(1/2) *(x)`

> Determine what factors are common to both terms.

These are:

`(x^2+1)` and `(x+5)`

> Then, detemine the smallest exponent for each of these two factors.

For `(x^2+ 1)` , smallest exponent is 1/2.

For `(x+5)` , smallest exponent is 3.

> So, GCF of the above problem is,

`(x^2+1)^(1/2)(x+5)^3`

> Then, factor out GCF.

`(x^2+1)^(3/2)*(4)(x+5)^3 + (x+5)^4 (3/2)(x^2+1)^(1/2)(2x)`

`= (x^2 + 1)^(3/2)*(4)*(x+5)^3 + (x+5)^4 (3)(x^2+1)^(1/2)(x)`

`=(x^2+1)^(1/2)(x+5)^3 ` ( ________ + _______ )

> To determine what goes inside the parenthesis, divide each terms by the GCF.

Note, to divide same base, follow the rules of exponent, which is `a^m/a^n = a^(m-n)`

`((x^2+1)^(3/2) *(4)(x+5)^3 )/((x^2+1)^(1/2)(x+5)^3 ) = (x^2+1)^((3/2)-(1/2)) *(4)*(x+5)^(3-3) = (x^2+1)(4)`

`= 4(x^2+1) = 4x^2 + 4`

`((x+5)^4(3)(x+1)^(1/2)*(x))/((x^2+1)^(1/2)(x+5)^3 )= (x+5)^(4-3)(3)(x^2+1)^((1/2)-(1/2))(x) = (x+5)(3)(x)`

`= 3x(x+5) = 3x^2 + 15`

> Then,

`(x^2+1)^(1/2)(x+5)^3 ( 4x^2 + 4 + 3x^2 + 15x) `

`= (x^2+1)^(1/2)(x+5)^3(7x^2+15x + 4)`

**Hence the factor is** `(x^2+1)^(1/2)(x+5)^3(7x^2+15x+4)` .