Answer:
[tex]\frac{1}{x+8}[/tex]
Step-by-step explanation:
Here is your original equation:
[tex]\frac{x-5}{x^{2} +3x-40}[/tex]
What we need to focus on is simplification, specifically simplifying the denominator. First, let's isolate the denominator (we'll get back to [tex]x-5[/tex] later):
[tex]x^{2} +3x-40[/tex]
Next, we take a look at -40. Our goal is to find what factors of 40 create a sum of 3 so we can separate the equation for polynomial division.
Let's begin building our expressions. We know that [tex]x^{2}[/tex] has an exponent of 2. When variables are multiplied, the exponents are added, meaning that [tex]x*x=x^{2}[/tex]. (Or [tex]x^{1} + x^{1} =x^{2}[/tex] , the 1s just don't show)
We can also see that -40 is a negative number. A negative multiplied by a positive makes a negative product, so we can also include one positive and one negative in each expression.
[tex]x^{2} +3x-40 = (x+)(x-)[/tex]
Next, let's look at the factors of 40:
1 and 40 2 and 20 4 and 10 5 and 8
Which of these factors can make a sum of 3? Remember that one number has to be positive and the other has to be negative!
[tex]8-5=3[/tex]
In this instance, our positive number is 8 and our negative number is 5. They create the sum of 3 that we're splitting in [tex]3x[/tex]. Therefore, [tex]8(-5)=-40[/tex].
[tex]x^{2} +3x-40 = (x+8)(x-5)[/tex]
Now let's go back to our original equation and substitute our old expression with the new one:
[tex]\frac{x-5}{(x+8)(x-5)}[/tex]
Do you notice that the numerator and part of the denominator are equal? This means that they can cancel each other out! Think of it as a [tex]\frac{1}{1}[/tex].
[tex]\frac{x-5}{(x+8)(x-5)}[/tex] = [tex]\frac{1}{x+8}[/tex]
Therefore, your simplified answer is [tex]\frac{1}{x+8}[/tex] !
Here's the whole process:
[tex]\frac{x-5}{x^{2} +3x-40} \\\\(x+?)(x-?)\\\\40:\\1, 40\\2, 20\\4, 10\\5, 8\\\\8-5=3 \\(8x-5x=3x)\\\\(x+8)(x-5)\\\\\frac{x-5}{(x+8)(x-5)}=\frac{1}{x+8} \\\\\frac{1}{x+8}[/tex]
To check your answer, confirm your expressions using the box method attached.
I hope this helps! If you have any more questions or concerns about the answer or the process, let me know!
