Question Video: Converting Improper Rational Expressions to Partial Fractions Mathematics

Convert the rational expression (6π‘₯Β³ βˆ’ 2π‘₯Β² + 5)/(π‘₯Β² + 4π‘₯ + 3) into partial fractions.

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Video Transcript

Convert the rational expression six π‘₯ cubed minus two π‘₯ squared plus five over π‘₯ squared plus four π‘₯ plus three into partial fractions.

Notice how we currently have an improper fraction. We can see that the order of the expression on the numerator is greater than that of the denominator. And this is because the highest power or the highest exponent on the numerator is three. And on the denominator, it’s two. Before we can split it into partial fractions then, we need to divide the numerator by the denominator. There are a number of different ways to do this. Long division with a bus stop or a grid method are the two most common methods.

Let’s look at the bus stop method. Adding zero π‘₯ into this expression isn’t necessary. But it can be helpful and gives us space to complete our working. Six π‘₯ cubed divided by π‘₯ squared is six π‘₯. We then multiply each part of the expression π‘₯ squared plus four π‘₯ plus three by this six π‘₯. And we get six π‘₯ cubed plus 24π‘₯ squared plus 18π‘₯. We’re now going to subtract this from the expression six π‘₯ cubed minus two π‘₯ squared plus zero π‘₯. Six π‘₯ cubed minus six π‘₯ cubed is zero. Negative two π‘₯ squared minus 24π‘₯ squared is negative 26π‘₯ squared. And zero π‘₯ minus 18π‘₯ is negative 18π‘₯. We then bring down this five, since five take away zero is five.

We’re now going to divide negative 26π‘₯ squared by π‘₯ squared. That gives us negative 26. And once again, we multiply negative 26 by each part of the expression π‘₯ squared plus four π‘₯ plus three. And we get negative 26π‘₯ squared minus 104π‘₯ minus 78. And we subtract this from negative 26π‘₯ squared minus 18π‘₯ plus five. Negative 26π‘₯ squared minus negative 26π‘₯ squared is zero. Negative 18π‘₯ minus negative 104π‘₯ is 86π‘₯. And five minus negative 78 is 83. So this tells us that six π‘₯ cubed minus two π‘₯ squared plus five divided by π‘₯ squared plus four π‘₯ plus three is equal to six π‘₯ minus 26 with a remainder of 86π‘₯ plus 83.

Or another way to write this is six π‘₯ minus 26 plus 86π‘₯ plus 83 all over π‘₯ squared plus four π‘₯ plus three. And at this point, we can now convert our expression into partial fractions. We can only do this once the denominator of our fraction 86π‘₯ plus 83 over eight squared plus four π‘₯ plus three is in factorized form. Let’s factor π‘₯ squared plus four π‘₯ plus three. This is a quadratic expression. It’s going to factor into two brackets. We know this because each term in the expression π‘₯ squared plus four π‘₯ plus three has no common factors aside from one.

And we know that the first term in each bracket must be π‘₯. This is because when we expand our brackets using the four method, we begin by multiplying the first term in each bracket. π‘₯ multiplied by π‘₯ gives us the π‘₯ squared as required. To find the numerical parts of our brackets, we need to find two numbers that multiply to make three and add to make four. That’s the coefficient of π‘₯. The only factor pair of the number three is one and three. One and three do indeed add to make four. So we know we factor correctly. But we could check our answer by expanding the brackets. We’ve already seen that multiplying π‘₯ by π‘₯ gives us π‘₯ squared. We then multiply the outer terms π‘₯ multiplied by three is three π‘₯. And we multiply the inner terms one multiplied by π‘₯ is π‘₯. And finally, we multiply the last term in each bracket, one multiplied by three is three.

And if we collect like terms we see that this simplifies to π‘₯ squared plus four π‘₯ plus three which was our original expression. So we must have done this correctly. So forgetting about six π‘₯ minus 26 for the moment, we can see that the fractional part of our expression 86π‘₯ plus 83 over π‘₯ squared plus four π‘₯ plus three can be written as 86π‘₯ plus 83 over π‘₯ plus one multiplied by π‘₯ plus three. And when we write this in partial fraction form, we say that it can be written as some number over π‘₯ plus one plus some other number over π‘₯ plus three.

And then what we do next is we multiply by the denominator of the fraction on the left-hand side. This has the effect of cancelling out each of the denominators. And on the left-hand side, we’re simply left with 86π‘₯ plus 83. On the right, it’s a little trickier. Multiplying this first term by π‘₯ plus one multiplied by π‘₯ plus three and we see that we can divide through by π‘₯ plus one. And we’re just left with 𝐴 multiplied by π‘₯ plus three. Similarly, we multiply the second term by π‘₯ plus one multiplied by π‘₯ plus three. And this time we divide through by π‘₯ plus three. And this time we’re left with 𝐡 multiplied by π‘₯ plus one.

Now at this stage, we can do one of two things. We could expand our two brackets and compare coefficients on both sides of our equation. But that can be quite a long process. Instead, we recognise that we have an identity. This means it’s true for every value of π‘₯. And we can, therefore, substitute any value of π‘₯ into the identity. In fact, we choose to substitute the roots of the denominator π‘₯ plus one π‘₯ plus three equals zero into the identity. And we use that to solve for 𝐴 and 𝐡. The roots are π‘₯ equals negative three and π‘₯ equals negative one.

Why did we choose these? Well, if we substitute π‘₯ equals negative three into the equation, we see that the coefficient of 𝐴 is zero. And we’re left with an equation in terms of 𝐡. Negative three plus one is negative two. And on the left-hand side, we get negative 175. We solve this equation for 𝐡 by dividing through by negative two. And we get that 𝐡 is equal to 175 over two.

The next root is π‘₯ equals negative one. And we choose this because we now see that if we substitute π‘₯ equals negative one into the equation, the coefficient of 𝐡 is now zero. And we’ll have an equation in terms of 𝐴. Negative one plus three is two. And on the left-hand side, we have negative three. So we see that negative three is equal to two 𝐴. And we solve this equation for 𝐴 by dividing through by two. And that tells us that 𝐴 is equal to negative three over two.

We can then replace 𝐴 with negative three over two. And we’re left with negative three over two multiplied by π‘₯ plus one as the first partial fraction. We then replace 𝐡 with 175 over two. And the second partial fraction is 175 over two multiplied by π‘₯ plus three. And our final step is to replace 86π‘₯ plus 83 over π‘₯ squared plus four π‘₯ plus three with these partial fractions. And as its convention to write the constant at the end of our expression, we can now see that our expression can be written as six π‘₯ minus three over two lots of π‘₯ plus one plus 175 over two lots of π‘₯ plus three minus 26.

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