How do I find $A$ $A$ and $B$ $B$ in the partial fraction decomposition $91 x + 269 = A {(x + 4)}^{2} + B (x + 4)$ $91x+269=A(x+4)^2+B(x+4)$ ?

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How do I find $A$ $A$ and $B$ $B$ in the partial fraction decomposition $91 x + 269 = A {(x + 4)}^{2} + B (x + 4)$ $91x+269=A(x+4)^2+B(x+4)$ ?

I know how to do PFDs when there are two factors that are different, but how can I do this when there are two factors that are the same? If I set $x = - 4$ $x=-4$ then I remove both terms, and then I have nothing left to solve for.

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Answer 1 · 2018-05-19T17:49:17

$2 x - 16 + \frac{91}{x + 4} - \frac{95}{{(x + 4)}^{2}}$ $2x-16+ frac{91}{x+4} - frac{95}{(x+4)^2}$

Explanation:

If you take the original question: $\frac{2 x^{3} - 5 x + 13}{{(x + 4)}^{2}}$ $\frac{2x^3−5x+13}{(x+4)^2}$
and you apply long division to it you do indeed get:

$2 x - 16 + \frac{91 x + 269}{{(x + 4)}^{2}}$ $2x-16+\frac{91x+269}{(x+4)^2$ and the equation for the partial fraction decomposition is indeed:

$\frac{91 x + 269}{{(x + 4)}^{2}} = \frac{A}{x + 4} + \frac{B}{{(x + 4)}^{2}}$ $\frac{91x+269}{(x+4)^2}= \frac{A}{x+4}+\frac{B}{(x+4)^2}$

However, once you clear the denominator by multiplying everything by ${(x + 4)}^{2}$ $(x+4)^2$ , you get:

$91 x + 269 = A (x + 4) + B$ $91x + 269 = A(x+4)+B$ , and now subbing in $x = - 4$ $x =-4$

you get:

$B = 91 (- 4) + 269 = - 364 + 269 = - 95$ $B = 91(-4) + 269 = -364 + 269 = -95$
and to solve for A:

$91 x + 269 = A x + 4 A + B$ $91x + 269 = Ax + 4A + B$
$91 x = A x$ $91x = Ax$
$A = 91$ $A = 91$

Thus, the partial fraction decomposition of $\frac{2 x^{3} - 5 x + 13}{{(x + 4)}^{2}}$ $\frac{2x^3−5x+13}{(x+4)^2}$

is $2 x - 16 + \frac{91}{x + 4} - \frac{95}{{(x + 4)}^{2}}$ $2x-16+ frac{91}{x+4} - frac{95}{(x+4)^2}$

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How do I find $A$ $A$ and $B$ $B$ in the partial fraction decomposition $91 x + 269 = A {(x + 4)}^{2} + B (x + 4)$ $91x+269=A(x+4)^2+B(x+4)$ ?

I know how to do PFDs when there are two factors that are different, but how can I do this when there are two factors that are the same? If I set $x = - 4$ $x=-4$ then I remove both terms, and then I have nothing left to solve for.

1 Answer

Explanation:

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Humanities

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How do I find AA and BB in the partial fraction decomposition 91x+269=A(x+4)2+B(x+4)91x+269=A(x+4)^2+B(x+4)?

I know how to do PFDs when there are two factors that are different, but how can I do this when there are two factors that are the same? If I set x=−4x=-4 then I remove both terms, and then I have nothing left to solve for.

1 Answer

Explanation:

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Impact of this question

How do I find $A$ $A$ and $B$ $B$ in the partial fraction decomposition $91 x + 269 = A {(x + 4)}^{2} + B (x + 4)$ $91x+269=A(x+4)^2+B(x+4)$ ?

I know how to do PFDs when there are two factors that are different, but how can I do this when there are two factors that are the same? If I set $x = - 4$ $x=-4$ then I remove both terms, and then I have nothing left to solve for.