Question

How do you evaluate `\frac { x ^ { 2} + 12x + 35} { x + 5}`?

Answer 1

Replace `x` with the desired number and do the arithmetic.

Explanation:

`\frac { x ^ { 2} + 12x + 35} { x + 5}`

To evaluate for `x = 1`, do the arithmetic:

`((1)^2+12(1)+35)/((1)+5) = (1+12+35)/6 = 48/6 = 8`

To evaluate for `x = -3`, do the arithmetic:

`((-3)^2+12(-3)+35)/((-3)+5) = (9-36+35)/2 = 8/2 = 4`

Note that we cannot evaluate the expression for `x=-5` because that would make the denominator `0` and it is not possible to divide by `0`.

Answer 2

Given is `{ x ^ { 2} + 12x + 35} /{ x + 5}`

We see from inspection that numerator can be factorised using split the middle term method as
`x ^ { 2} + 12x + 35`
`=>x ^ { 2} + 7x+5x + 35`
`=>x(x + 7)+5(x + 7)`
`=>(x + 7)(x + 5)`
Now the given expression becomes

`((x + 7)(x + 5))/(x+5)`

`=>(x + 7)`
The expression becomes `0/0` for `x=-5`.
Therefore, to evaluate it applying La Hospitals Rule we find the differential of numerator and of denominator separately.
`lim_(x->-5) { d/dx(x ^ { 2} + 12x + 35)} /{ d/dx(x + 5)}`
`lim_(x->-5) { 2x + 12 } /1`
`=>lim_(x->-5) 2(-5) + 12`
`=2`
We see that this is equal to value of `(x+7)` at `x=-5`

As such the expression can be evaluated for all values of `x` as `=x+7`

Answer 3

See explanation.

I can not help but ask; does the question have something missing?

Explanation:

`color(blue)("Simplifying - this does not really comply with the instruction 'evaluate'")`

Consider `x^2+color(red)(12)x+ color(green)(35)`

Notice that `5xx7=color(green)(35)` and that `5+7=color(red)(12)`

This means that we can write `x^2+12x+35" as "(x+5)(x+7)`

So the whole expression can be written as:

`(cancel((x+5))(x+7))/(cancel((x+5)))`

`x+7`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Evaluate ?")`

You can only 'evaluate' if 'values that work' are substituted

Perhaps the question is really asking you to declare all the values for which this expression works.
................................................................................
`color(brown)("What will not work?")`

The expression becomes 'undefined' if the denominator is 0

Thus `x+5!=0 => x!=-5`

`color(brown)("What will work")`

As what we are given is an expression there is no limiting dependant variable `(y)` that would fix the value of `x`
That is it is `color(red)("not of form")" "y=(x^2+12x+35)/(x+5)`
Where the value of `y` is declared.

`color(white)(.)`

Assuming we limit `x` to `ul(color(red)("not"))` be a complex number' then we have:

The `ul("set")` `x` written as: `{x: x inRR,x!=-5 }`

As an example, Suppose we set `x=1/2`

then we have `((1/2)^2+12(1/2)+35)/(1/2+5) = (color(white)(.)165/4color(white)(.))/(11/2) = 15/2 color(red)(larr" error corrected")`

It has a value and thus is possible to evaluate.