Page:Calculus Made Easy.pdf/144

But there is another way out of this difficulty. The equation must be true for all values of $x$; therefore it must be true for such values of $x$ as will cause $x-1$ and $x+1$ to become zero, that is for $x=1$ and for $x=-1$ respectively. If we make $x=1$, we get $4=(A\times 0)+(B\times 2)$, so that $B=2$; and if we make $x=-1$, we get $-2=(A\times -2)+(B\times 0)$, so that $A=1$. Replacing the $A$ and $B$ of the partial fractions by these new values, we find them to become ${\dfrac {1}{x+1}}$ and ${\dfrac {2}{x-1}}$; and the thing is done.

As a farther example, let us take the fraction ${\dfrac {4x^{2}+2x-14}{x^{3}+3x^{2}-x-3}}$. The denominator becomes zero when $x$ is given the value $1$; hence $x-1$ is a factor of it, and obviously then the other factor will be $x^{2}+4x+3$; and this can again be decomposed into $(x+1)(x+3)$. So we may write the fraction thus:

${\frac {4x^{2}+2x-14}{x^{3}+3x^{2}-x-3}}={\frac {A}{x+1}}+{\frac {B}{x-1}}+{\frac {C}{x+3}}$,

making three partial factors.

Proceeding as before, we find

$4x^{2}+2x-14=A(x-1)(x+3)+B(x+1)(x+3)+C(x+1)(x-1).$

Now, if we make $x=1$, we get:

$-8=(A\times {0})+B(2\times {4})+(C\times {0}$; that is, $B=-1$.

If $x=-1$, we get:

$-12=A(-2\times {2})+(B\times {0})+(C\times {0})$; whence $A=3$.