When \(10x^3+mx^2- x+10\) is divided by\(5x-3\), the quotient is \(2x^2+nx-2\)and the remainder is 4.
The value of m=-21 and n=-3
Given :
When \(10x^3+mx^2- x+10\) is divided by\(5x-3\), the quotient is \(2x^2+nx-2\)and the remainder is 4.
We frame equation using quotient and remainder
Dividend = quotient (divisor)+ remainder
\(10x^3+mx^2- x+10=(2x^2+nx-2)(5x-3)+4\\10x^3+mx^2- x+10=10x^3-6x^2+5nx^2-3nx-10x+6+4\\\\10x^3+mx^2- x+10=10x^3+x^2(5n-6)+x(-3n-10)+10\\\\\)
Now we make x^2 parts are equal and solve for m and n
Also we make x parts are equal
\(-3n-10=-1\\-3n=-1+10\\-3n=9\
=-3\)
Now make x^2 terms equal
\(5n-6=m\\5(-3)-6=m\\m=-21\)
The value of m and n are
\(m=-21 and n=-3\)
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