Question

Please help me with this question below

Answer 1

`m=1\\ \\n = -(10)/(13)\\ \\p=-(3)/(13)`

Explanation:

Given

`(x-37)/((x-1)(x+3)(x-10))`

Rewrite it in the form

`(m)/(x-1)+(n)/(x+3)+(p)/(x-10)`

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

`(m(x+3)(x-10)+n(x-1)(x-10)+p(x-1)(x+3))/((x-1)(x+3)(x-10))\\ \\ \\=(m(x^2-10x+3x-30)+n(x^2-10x-x+10)+p(x^2+3x-x-3))/((x-1)(x+3)(x-10))\\ \\ \\=(mx^2-7mx-30m+nx^2-11nx+10n+px^2+2px-3p)/((x-1)(x+3)(x-10))\\ \\ \\=(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=-(3)/(13)`

Then

`-4n+9\cdor\left(-(3)/(13)\right)=1\\ \\ \\-4n=1+(27)/(13)\\ \\ \\-4n=(40)/(13)\\ \\ \\ n=-(10)/(13)`

Hence,

`m=-\left(-(10)/(13)\right)-\left(-(3)/(13)\right)\\ \\m=1`

So,

`m=1\\ \\n = -(10)/(13)\\ \\p=-(3)/(13)`