We just use difference of 2 squares, twice: RULE 4:Denominator Containing a Quadratic FactorĬorresponding to any quadratic factor `(ax^2+īx + c)` in the denominator, there will be a partialĮxpress the following in partial fractions.įirstly, we need to factor the denominator. NOTE: Scientific Notebook can do all this directly for

We now compare the coefficients of `x^3` onīoth sides and then compare the constant values on bothīut since we know 3 values now, we have: `B = 9/4`. Instead, we just useĪppropriate substitutions to find the values of the unknowns `A` to We multiply throughout by `(x-1)^3(x+1)`: (a) Express the following as a sum of partial It's OK to use the ordinary equals sign, too.) Example 3 We normally apply this between 2 expressions when we wish them to be equivalent. (The sign `-=` means 'is identically equal to'. If a linear factor is repeated `n` times in theĭenominator, there will be `n` corresponding partial

Solving this set of simultaneous equations gives: We find the values of `A` and `B` by multiplying both sides by `(2x + 1)(x + 4)`:

1 x 2 2x 8