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

step 1 Okay, so we're asked to find a partial fraction decomposition. If you look at our denominator an

Partial Fraction Decomposition Find the partial fraction...

Key Concepts

Clearing Denominators and Coefficient Matching

After setting up the partial fraction form, the next step is to multiply through by the common denominator to eliminate fractions. This results in a polynomial equation where the coefficients of corresponding powers of the variable can be equated. Solving these equations yields the unknown constants in the decomposition.

Polynomial Factorization

Polynomial factorization involves expressing a polynomial as a product of its irreducible factors, which could be linear or quadratic. This process is essential in partial fraction decomposition because it allows one to identify the correct form for the decomposed terms, ensuring that each fraction corresponds to a factor of the original denominator.

Decomposition Setup

Setting up the decomposition involves writing the rational function as a sum of fractions where each denominator is one of the factors identified through factorization. The numerators are assigned forms appropriate to the type of factor—constants for linear factors and linear expressions for irreducible quadratic factors.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to rewrite a rational function as the sum of simpler fractions whose denominators are factors of the original denominator. This technique is especially useful when integrating rational functions or solving differential equations.

Rational Functions

Rational functions are expressions defined by the ratio of two polynomials. They are important in many areas of mathematics because understanding their structure and behavior often leads to simplifications that facilitate integration, differentiation, and solving equations.

Transcript

00:01 Okay, so we're asked to find a partial fraction decomposition.

00:04 If you look at our denominator and see that it can factor more, this is a linear factor, so this is okay.

00:12 This term factor into 2x minus 3 times x plus 3.

00:19 You can check this using the quadratic formula.

00:23 And that just turns into, let's see, all our linear factors are distinct.

00:29 So we have a case one.

00:31 So we get a over 2 .0.

00:34 X and that's right start with x plus plus b over 2x minus 3 plus c over x plus 3 okay now we have to multiply by a common denominator which is x plus 2 times 2x minus 3 times x plus 3 and what do we get we get negative 3 x squared minus 3x plus 27 is equal to a times 2x minus 3 times x plus 3 plus b times x plus 2 times x plus 3 and plus c times x plus 2 times 2x minus 3 okay let's expand this we get a is multiplied by 2x squared plus 3 x squared plus 3 x minus 9 plus b which is x squared plus 5 x plus 6 plus c times 2 x squared plus x minus 6 okay now moving on to the next page let's rewrite this we had negative 3 x squared minus 3 x plus 27 and that was all equal to this but i'm going to rewrite this with our a b and c so let's do that.

02:09 We get 2a x squared plus 3a x minus 9a plus bx squared plus 5b x plus 6b plus c up 2 cx squared plus cx minus 6c.

02:32 And now we need to combine our light term.

02:35 So we have x squared x squared x squared x squared x squared x x squared x x and last our constants and let's rewrite that as a system equations or constant here it's negative 3 negative 3 27 okay so we got 2a plus 3 oh that's a sorry about that so i'm going to read it like it like this 2a plus b plus 2c is equal to negative 3 3 plus 3a plus 5 b plus c is equal to negative 3, negative 9, a plus 6b, minus 6c is equal to 27.

03:36 Okay, so let me actually rewrite this equation by dividing it by 3.

03:44 We get negative to 3, a plus 2b, minus 2c is equal to 9.

03:53 Okay, and now i'm going to multiply this equation.

03:56 Or not, i'm just going to actually move this over here and eliminate my a.

04:04 So i get 3a plus 5b plus c is equal to negative 3...

Please give Ace some feedback