Factorize the expression

8-5`sqrt(x+2)+x`

Please show your workings clearly.Thanks

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`8-5sqrt(x+2)+x`

`x+8-5sqrt(x+2)`

`x+2-5sqrt(x+2)+6`

`(sqrt(x+2))^2-5sqrt(x+2)+6`

Let `y=sqrt(x+2)`

`y^2=x+2`

`y^2-5y+6`

`=y^2-2y-3y+6`

`=y(y-2)-3(y-2)`

`=(y-2)(y-3)`

`=(sqrt(x+2)-2)(sqrt(x+2)-3)`

You need to change the variable to convert the given radical form of expression into a quadratic, such that:

`x + 2 = t^2 => x = t^2 - 2`

Replacing the variable, yields:

`8 - 5sqrt t^2 + t^2 - 2 = t^2 - 5|t| + 6`

Using the absolute value definition yields:

`|t| = {(t, t>0),(-t, t<0):}`

Considering `t>0` , yields:

`t^2 - 5t + 6`

You may decompose the quadratic in linear factors, using its roots, such that:

`t^2 - 5t + 6 = (t - t_1)(t - t_2)`

You may evaluate the solutions `t_(1,2)` to quadratic equation, using quadratic formula, such that:

`t_(1,2) = (5+-sqrt(25 - 24))/2 => t_(1,2) = (5+-1)/2`

`t_1 = 3; t_2 = 2`

Replacing 3 for `t_1` and 2 for `t_2` yields:

`t^2 - 5t + 6 = (t - 3)(t - 2)`

Replacing back `sqrt(x+2)` for t yields:

`8 - 5sqrt(x+2) + x = (sqrt(x+2) - 3)(sqrt(x+2) - 2)`

Considering `t<0` , yields:

`t^2 + 5t + 6 = (t - t_1)(t - t_2)`

`t_(1,2) = (-5 +- sqrt(25 - 24))/2`

`t_(1,2) = (-5 +- 1)/2 => t_1 = -2, t_2 = -3`

`t^2 + 5t + 6 = (t + 2)(t + 3)`

Replacing back `sqrt(x+2)` for `t` yields:

`8 - 5sqrt(x + 2) + x = (sqrt(x+2) + 2)(sqrt(x+2) + 3)`

**Hence, performing the factorization using substitution and absolute value definition, yields **`8 - 5sqrt(x+2) + x = (sqrt(x+2) - 3)(sqrt(x+2) - 2)` or` 8 - 5sqrt(x + 2) + x = (sqrt(x+2) + 2)(sqrt(x+2) + 3)` .

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