Video: Simplifying and Determining the Domain of the Product of Two Rational Functions

Simplify the function 𝑛(π‘₯) = ((π‘₯ + 5)/(π‘₯Β² + 9π‘₯ + 20)) Γ— ((π‘₯Β² + 15π‘₯ + 54)/(7π‘₯Β² + 69π‘₯ + 54)), and determinate its domain.


Video Transcript

Simplify the function 𝑛 of π‘₯ is equal to π‘₯ plus five divided by π‘₯ squared plus nine π‘₯ plus 20 times π‘₯ squared plus 15π‘₯ plus 54 divided by seven π‘₯ squared plus 69π‘₯ plus 54 and determinate its domain.

Essentially, we need to take each expression and simplify it by factoring. The π‘₯ plus five does not factor, so it’s good. However, the one below it, what two numbers multiply to be 20 and add to be nine? That would be five and four. So we can replace that with π‘₯ plus five times π‘₯ plus four. In the top right corner, what are two numbers that multiply to be 54 and add to be 15? That would be nine and six.

Both of those were simple trinomials; however, in the bottom right-hand corner, this is an advanced trinomial because the leading coefficient of seven isn’t one. So we’re going to use the slip and slide method. So let’s go ahead and slip this seven to the back by multiplying it to 54. So now we have π‘₯ squared plus 69π‘₯ plus 378. So what two numbers multiply to be 378 and add to be 69? This might take a little bit, but eventually we will come up with the numbers, and they are 63 and six.

Now, the number that we slipped to the back, we are now going to have two slide underneath and simplify. So 63 divided by seven is nine. Now, six-sevenths, that doesn’t reduce, so I’m going to bring this seven up with the π‘₯. When factoring, we don’t usually leave fractions. So now, our last expression has been replaced.

When simplifying or multiplying two fractions, we can cancel things up and down and diagonally. So the π‘₯ plus fives cancel, the π‘₯ plus nines cancel, and that’s it! So the only thing left on the top is the π‘₯ plus six and on the denominator, there’s an π‘₯ plus four and seven π‘₯ plus six. Now, we have to figure out our domain. So our domain is anything, except for what will make the denominator zero. So we need to set every factor that was on the bottom equal to zero.

And after doing that, if you set π‘₯ plus five equal to zero, you get negative five. And then we get negative four, we get negative nine, and negative six-sevenths. So the domain will be all real numbers minus negative nine, negative five, negative four, and negative six-sevenths. So our function simplified is 𝑛 of π‘₯ equals π‘₯ plus six divided by π‘₯ plus four times seven π‘₯ plus six, with a domain is all real numbers minus negative nine, negative five, negative four, and negative six-sevenths.

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