Partial Fraction Decomposition Find the partial fraction decomposition of the rational function.

Partial Fraction Decomposition Find the partial fraction...

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is the process of expressing a complex rational function as the sum of simpler rational functions, which are easier to manipulate or integrate. This technique is particularly useful in calculus, algebra, and differential equations to simplify integration and solve problems involving rational expressions.

Rational Functions

Rational functions are expressions represented as the quotient of two polynomials. Understanding their structure, including the degrees of the numerator and the denominator, is essential for techniques such as factoring and partial fraction decomposition. Identifying the types of factors present in the function sets the stage for further analysis.

Irreducible Quadratic Factors

Irreducible quadratic factors are quadratic expressions that do not factor further over the real numbers. In the context of partial fractions, when the denominator of a rational function decomposes into irreducible quadratics, each quadratic factor contributes a term where the numerator is typically a linear expression. Recognizing these factors is key to setting up the correct form of the decomposition.

Equating Coefficients

Equating coefficients is a method used to determine the unknown constants in the partial fraction decomposition. By expanding the assumed decomposed form of the rational function and comparing the coefficients of like powers of the variable to those of the original expression, one can solve for the coefficients systematically. This is an essential algebraic technique within the scope of partial fractions.

Transcript

00:01 In this problem, we're being asked to find the partial fraction decomposition of our given fraction.

00:05 Well, we first have to make sure our denominator is completely factored, which it is.

00:10 So now we can start doing our partial fraction decomposition.

00:14 So let's look at our first factor, which is x squared plus x plus 2.

00:18 Well, that's an irreducible quadratic factor.

00:21 So our first fraction is going to have to be of the form a x plus b all over our factor of x squared plus x plus 2.

00:31 Now, our second factor is x squared plus 1, which is also an irreducible quadratic factor.

00:37 So for this, we're going to have to have plus cx plus d all over x squared plus 1.

00:44 And this is equal to our original fraction, 2x to the 3rd plus 7x plus 5, all over the quantity of x squared plus 2 times x squared plus 1.

00:59 Well, what i'm going to do next is i'm going to multiply both sides of our equation by our common denominator.

01:04 So we're going to multiply by x squared plus x plus 2 times x squared plus 1.

01:11 And we're going to do this to the left hand side too.

01:13 We're going to multiply by x squared plus x plus 2 times x squared plus 1.

01:19 And the reason we do this is because on the right hand side of our equation, our denominator will end up canceling each other route, which is exactly what we want.

01:27 So now let's look on the left hand side of our equation.

01:29 Of our equation.

01:30 Well, our first fraction has a denominator of x squared plus x plus 2.

01:34 So it's missing that factor of x squared plus 1.

01:38 So we need to multiply the numerator a x plus b by x squared plus 1.

01:45 Now for our second fraction, it has the factor x squared plus 1 in the denominator.

01:51 It's missing the x squared plus x plus 2 factor.

01:54 So we're going to have to multiply that second numerator cx plus d by the factor x squared plus x plus two.

02:04 And again, this is all equal to 2x to the third plus 7x plus 5.

02:11 So what we're going to have to do next is we're going to have to go ahead and multiply those binomials and polynomials out.

02:18 Well, as you can see, i've ran out of room.

02:20 So i'm going to start a new sheet here.

02:23 So just as a little reminder as to what we just had, we had a x plus b times the quantity of x squared plus 1 plus cx plus d times to polynomial x squared plus x plus 2 and it's equal to 2x to the 3rd plus 7x plus 5 so now we're going to go ahead and multiply our first to binomial so when we do that we're going to have a x to the 3rd plus a x plus bx squared plus b and then we're going to multiply our binomial by the trinomial.

03:06 So when we do that, we're going to have the term cx to the third plus cx squared plus 2cx plus dx plus dx plus 2d.

03:20 And again, this is all just equal to 2x to the third plus 7x plus 5.

03:26 And now we're going to go ahead and combine our like terms.

03:28 So i'm going to start with my x to the third terms.

03:31 Well, i have ax to the third plus cx to the third.

03:34 So again, we're going to have a x to the third plus cx to the third.

03:38 Next, i'm going to do my quadratic terms.

03:41 Well, i have positive bx squared, positive cx squared, and positive dx squared.

03:47 So again, we're going to have plus bx squared, plus cx squared, plus dx squared.

03:54 Now i'm going to combine our linear terms.

03:57 So i have ax plus two cx plus dx.

04:02 So again, we have plus a x plus a x plus two cx, plus two cx, and plus d x and now we're going to combine our constant terms which in this case are just plus b and plus 2b 2d so we have plus b plus 2d and again this is all just equal to 2x to the 3rd plus 7x plus 5 so now let's look at what we have well our first two terms are a x to the 3rd plus x to the 3rd which means when we combine those coefficients we should get the coefficient of the x to the 3rd term in our answer, which is 2.

04:40 So now we know that a plus c is equal to 2.

04:45 Now let's look at our x squared terms.

04:48 Well, that means when we add b plus c plus d, the coefficient of our x squared terms, we should have the coefficient of our x square term in our answer.

04:57 But notice, we don't have an x squared term because it's zero...

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