Question Video: Simplifying Rational Functions Mathematics

Simplify 𝑛(π‘₯) = ((3π‘₯ + 2)/7π‘₯) + (3π‘₯Β²/(2 βˆ’ π‘₯)).

03:38

Video Transcript

Simplify 𝑛 of π‘₯ equals three π‘₯ plus two over seven π‘₯ plus three π‘₯ squared over two minus π‘₯.

In this problem, we have a function 𝑛 of π‘₯, which is the sum of two algebraic quotients. In order to simplify this function, we need to try and combine the two quotients into a single one. From our knowledge of adding numeric quotients or fractions, we know that we can only do this when we have a common denominator, which is usually the lowest common multiple of the two individual denominators. In the case of adding the fractions one-third and one-quarter, we use a denominator of 12 because this is the lowest common multiple of three and four.

Returning to our algebraic problem, we therefore need to find the lowest common multiple of the two denominators, which are seven π‘₯ and two minus π‘₯. Well, as these are each linear functions of π‘₯ which have no common factors other than one, then their lowest common multiple is actually their product. That’s seven π‘₯ multiplied by two minus π‘₯. This is similar to our numeric example. As the two denominators of three and four had no common factors other than one, their lowest common multiple was their product of 12.

What we want to do now then in order to simplify 𝑛 of π‘₯ is to rewrite each fraction as an equivalent fraction with a denominator of seven π‘₯ multiplied by two minus π‘₯. For the first fraction, we do this by multiplying both the numerator and denominator by two minus π‘₯. And for the second, we have to multiply both the numerator and denominator by seven π‘₯. So 𝑛 of π‘₯ has become three π‘₯ plus two multiplied by two minus π‘₯ over seven π‘₯ multiplied by two minus π‘₯ plus three π‘₯ squared multiplied by seven π‘₯ over seven π‘₯ multiplied by two minus π‘₯.

As we now have the same denominator for both fractions, we can add them by adding the numerators. This gives three π‘₯ plus two multiplied by two minus π‘₯ plus three π‘₯ squared multiplied by seven π‘₯ all over seven π‘₯ multiplied by two minus π‘₯. And we’ve combined our sum into a single fraction.

The next step is to distribute the parentheses in the numerator. We can do this using the FOIL method if we wish. For the first set of parentheses, we have six π‘₯ minus three π‘₯ squared plus four minus two π‘₯, which simplifies to negative three π‘₯ squared plus four π‘₯ plus four. For the second set of parentheses, we just have three π‘₯ squared multiplied by seven π‘₯, which gives 21π‘₯ cubed. So our expression for 𝑛 of π‘₯ simplifies to negative three π‘₯ squared plus four π‘₯ plus four plus 21π‘₯ cubed over seven π‘₯ multiplied by two minus π‘₯.

We could stop here, but it’s more common to write the terms in the numerator in order of descending powers or exponents of π‘₯. Starting with the π‘₯ cubed term then, we have that 𝑛 of π‘₯ is equal to 21π‘₯ cubed minus three π‘₯ squared plus four π‘₯ plus four all over seven π‘₯ multiplied by two minus π‘₯. This can’t be simplified any further as there are no common factors shared by the numerator and denominator of this quotient.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.