![]() Now, we can add the three algebraic fractions together by rewriting each to This contradiction means we cannot decompose the algebraic fraction into □ yields □ = 0 however, equating the constant terms gives us ![]() The corresponding coefficients must be equal. We note that for both sides of the equation to be equivalent, If we add these expressions together, then we will Instead, we might try the decomposition □ □ + □ □ − 1 . However, adding these together does not give us a cubic polynomial in the If we tried to decompose this algebraicįraction into into the sum of algebraic fractions with each linear factor as aĭenominator, then we would have an expression of the form ![]() We see that there is a repeated factor of To see why this is the case, let’s consider the decomposition ofģ □ − 1 □ ( □ − 1 ) . However, whenĭealing with repeated linear factors, we have to follow a different process. The sum of algebraic fractions with each factor as a denominator. Is the product of distinct linear factors we split the algebraic fraction into We have seen how to apply this process when the denominator The process of adding algebraic fractions. The decomposition of an algebraic fraction into partial fractions is reversing In this explainer, we will learn how to decompose rational expressions into partialįractions when the denominator has repeated linear factors.
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