# `(x-17)/(x^2 -9x+14)` Decompose the above expression into a sum of partial fractions.

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Decompose `(x-17)/(x^2-9x+14)` into partial fractions:

`(x-17)/(x^2-9x+14)=(x-17)/((x-7)(x-2))`

Write the right side as a sum of two fractions; let the numerators be A and B:

`(x-17)/((x-7)(x-2))=A/(x-7)+B/(x-2)` Add the fractions on the right side:

`(x-17)/((x-7)(x-2))=(A(x-2)+B(x-7))/((x-7)(x-2))`

The rational expressions are equivalent with the same denominator, so the numerators must be equal:

x-17=Ax-2A+Bx-7B The linear expressions are equivalent so the coefficients must be the same:

A+B=1 ** Ax+Bx=1x so A+B=1 **

-2A-7B=-17 Solve the system:

-5B=-15 ==> B=3 so A=-2

Therefore `(x-17)/((x-7)(x-2))=(-2)/(x-7)+3/(x-2)`

`(x-17)/(x^2-9x+14)`

To decompose first factorize the denominator to produce a middle term of -9 from the factors o `x^2` and +14:

`x^2-9x+14 = (x-7)(x-2)`

As we do not know the numerator, assign A and B to the unknown values and split the denominator between them:

`therefore (x-17)/(x^2 -9x+14) = A/(x-7) + B/(x-2)`

Now to eliminate the denominator multiply everything by it:

`(x-17)/((x-7)(x-2)) times (x-7)(x-2) = (A/(x-7))times (x-7)(x-2) + (B/(x-2))times (x-7)(x-2)`

`therefore x-17=A(x-2) + B(x-7)`

`x-17= Ax -2A + Bx-7B `

Now rearrange: `x-17 = Ax+Bx -2A - 7B`

`therefore x-17 = x(A+B) -2A -7B`

If x=0 we get: `-17 = -2A -7B` which we will come back to to substitute later. There are various methods to find A and B.

Now go back to `x-17 = Ax-2A +Bx-7B` and factorize differently, grouping As and Bs together:

`x-17 = A(x-2) + B(x-7)` Now assign a value to enable you to eliminate one of the unknowns:

Say x=2 `therefore 2-17 = A(2-2) + B(2-7)` Simplifying:

`therefore -15 = -5B`

`therefore -15/-5 = B`

`therefore B=3`

Now substituting into our other equation when x-0:

`-17 = -2A -7B` becomes `-17 = -2A -7(3)`

`therefore -17+21= -2A`

`therefore 4=-2A`

`therefore A=4/-2`

`therefore A=-2`

**Ans: `(x-17)/(x^2-9x+14) = 3/(x-2) - 2/(x-7)` **

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