# Question Video: Converting Improper Rational Expressions to Partial Fractions Mathematics

Mason wants to convert the rational expression (6π₯Β² + 5π₯ β 4)/(5π₯Β² + 6π₯) into partial fractions. His first step is to divide the numerator by the denominator. Complete this division. Mason now converts this expression into partial fractions. Convert the expression into partial fractions.

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### Video Transcript

Mason wants to convert the rational expression six π₯ squared plus five π₯ minus four over five π₯ squared plus six π₯ into partial fractions. His first step is to divide the numerator by the denominator. Complete this division. Mason now converts this expression into partial fractions. Convert the expression into partial fractions.

The first thing weβre going to do is to divide the numerator, thatβs six π₯ squared plus five π₯ minus four, by the denominator. Thatβs five π₯ squared plus six π₯. Now, the reason we can do this is because our fraction is improper. The highest order of π₯ on the numerator is six π₯ squared, whereas on the denominator, itβs five π₯ squared. And so, there are a couple of ways that we can perform this division. Weβre going to use polynomial long division.

To begin with, we divide six π₯ squared by five π₯ squared. Well, thatβs a little bit awkward, but it is six-fifths. Our next job is to multiply this by each term in our divisor. When we do, we get six π₯ squared plus thirty-six fifths π₯. Next, weβre going to subtract all our terms. Six π₯ squared minus six π₯ squared is zero. Then five π₯ minus thirty-six fifths π₯, or 36π₯ over five, is negative 11π₯ over five. And, of course, negative four minus zero is negative four.

Now, at this stage, we canβt divide negative 11π₯ over five by five π₯ squared. So we finished our division. Six π₯ squared plus five π₯ minus four divided by five π₯ squared plus six π₯ is six-fifths with a remainder. And we have a remainder of negative 11π₯ over five minus four. And so, we can write this as six-fifths plus negative 11π₯ over five minus four over five π₯ squared plus six π₯.

This doesnβt look very nice so. So what weβre going to do is take out a constant factor of negative one-fifth. When we do, our numerator becomes 11π₯ plus 20. And it becomes 20 because negative four divided by negative one-fifth is negative four times negative five, which is 20. We can then distribute the parentheses by, now, multiplying the denominator of our faction by five. And so, when we divide the numerator by the denominator, we get six-fifths minus 11π₯ plus 20 over 25π₯ squared plus 30π₯.

The next part of this question asks us to convert the expression into partial fractions. Well, six-fifths is fine. We leave it as it is. But weβre going to need to convert 11π₯ plus 20 over 25π₯ squared plus 30π₯ into partial-fraction form. Now, of course, we can only do this once the denominator is fully factored. So letβs do that first.

25π₯ squared and 30π₯ share a common factor of five π₯. So when we factor the denominator, we get five π₯ times five π₯ plus six. These are both linear terms. So when we split it into partial fractions, we write π΄ over five π₯, where π΄ is some constant, plus π΅ over five π₯ plus six, where π΅ is also a constant. Then our next step is to make the expression on the left look like that on the right.

And, of course, to achieve that, we need to create a common denominator. The common denominator is five π₯ times five π₯ plus six. So weβre going to multiply the numerator and denominator of our first fraction by five π₯ plus six and of our second by five π₯. In doing so, we have our common denominator of five π₯ times five π₯ plus six. So we can add the numerators, which gives us π΄ times five π₯ plus six plus π΅ times five π₯.

Now, this is, of course, still equal to the earlier fraction 11π₯ plus 20 over five π₯ times five π₯ plus six. Now, the denominator of our fractions are now equal. So that must mean, for the fractions themselves to be equal, their numerators must also be. That is, 11π₯ plus 20 must be equal to π΄ times five π₯ plus six plus π΅ times five π₯.

And we have a couple of steps that we can take now. We could distribute the parentheses on the right-hand side and equate coefficients. Alternatively, we can substitute the zeros of five π₯ times five π₯ plus six in. Now, one of these is π₯ equals zero. When we let π₯ be equal to zero on the left-hand side of our equation, we get 11 times zero plus 20, which is 20. Then, on the right, we get π΄ times five times zero plus six, or π΄ times six, plus π΅ times zero. Well, we donβt need to write π΅ times zero. So we get 20 equals π΄ times six. And dividing through by six, we find that π΄ is equal to twenty-sixths or 10 over three.

Weβre now ready to find the value of π΅. Now, this time, we want the expression five π₯ plus six to be equal to zero. The only way for that to occur is if π₯ is equal to negative six-fifths. When we substitute π₯ equals negative six-fifths into the left-hand side, we get thirty-four fifths. Then the coefficient of π΄ becomes zero, as expected. And then we have π΅ times five times negative six-fifths, which is π΅ times negative six. Dividing through by negative six, and we get π΅ equals 34 over negative 30 or negative seventeen fifteenths.

Letβs replace π΄ and π΅ in our expression. Six-fifths minus 11π₯ plus 20 over 25π₯ squared plus 30π₯ becomes six-fifths minus ten-thirds over five π₯ minus 17 over 15 over five π₯ plus six. We simplify a little with this second fraction by dividing through by five. And then we make this look a little bit nicer by distributing the parentheses and rewriting our fraction slightly. So we get six-fifths minus two over three π₯ plus 17 over 15 times five π₯ plus six. Now, of course, it really doesnβt matter which order we write these fractions in. So we can write this as six-fifths plus 17 over 15 times five π₯ plus six minus two over three π₯.