Question

please help me with this question below

Answer 1

Answer:

$m=1\\ \\n = -\dfrac{10}{13}\\ \\p=-\dfrac{3}{13}$

Step-by-step explanation:

Given

$\dfrac{x-37}{(x-1)(x+3)(x-10)}$

Rewrite it in the form

$\dfrac{m}{x-1}+\dfrac{n}{x+3}+\dfrac{p}{x-10}$

To find $m,\ n$ and $p,$ add these three fractions:

$\dfrac{m(x+3)(x-10)+n(x-1)(x-10)+p(x-1)(x+3)}{(x-1)(x+3)(x-10)}\\ \\ \\=\dfrac{m(x^2-10x+3x-30)+n(x^2-10x-x+10)+p(x^2+3x-x-3)}{(x-1)(x+3)(x-10)}\\ \\ \\=\dfrac{mx^2-7mx-30m+nx^2-11nx+10n+px^2+2px-3p}{(x-1)(x+3)(x-10)}\\ \\ \\=\dfrac{x^2(m+n+p)+x(-7m-11n+2p)+(-30m+10n-3p)}{(x-1)(x+3)(x-10)}$

This fraction and initial fraction are equal, they have the same denominators, so they have the same numerators:

$x^2(m+n+p)+x(-7m-11n+2p)+(-30m+10n-3p)=x-37$

Equate coefficients:

$at \ x^2:\ \ m+n+p=0\\ \\at \ x:\ \ -7m-11n+2p=1\\ \\at\ 1:\ \ -30m+10n-3p=-37$

from the first equation:

$m=-n-p,$

then

$\left\{\begin{array}{l}-7(-n-p)-11n+2p=1\\ \\-30(-n-p)+10n-3p=-37\end{array}\right.\\ \\  \\\left\{\begin{array}{l}7n+7p-11n+2p=1\\ \\30n+30p+10n-3p=-37\end{array}\right.\\ \\ \\\left\{\begin{array}{l}-4n+9p=1\\ \\40n+27p=-37\end{array}\right.$

Multiply the first equation by 10 and add it to the second equation:

$-40n+90p+40n+27p=10-37\\ \\117p=-27\\ \\13p=-3\\ \\p=-\dfrac{3}{13}$

Then

$-4n+9\cdor\left(-\dfrac{3}{13}\right)=1\\ \\ \\-4n=1+\dfrac{27}{13}\\ \\ \\-4n=\dfrac{40}{13}\\ \\ \\ n=-\dfrac{10}{13}$

Hence,

$m=-\left(-\dfrac{10}{13}\right)-\left(-\dfrac{3}{13}\right)\\ \\m=1$

So,

$m=1\\ \\n = -\dfrac{10}{13}\\ \\p=-\dfrac{3}{13}$