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