Question

(x+2) is factor of mx^3+nx^2+x-6.it leaves the remainder 4 when divided by (x-2).find m and n

Answer 1

Mx3+nx2+x-6 has (x+2) as a factor and it leaves a remained of 4 divide by x-2. 3(11+3m+n)=3(1) 11+3m+n=1, 3m+n=-10.

Name it as eq.2, On substracting eq.1 and 2, we get:-m=3, m=-3. Put this value of m in any eq.1 or 2, to get n=-1. So we get n=-1 as the answer by dividing with 4.

Answer 2

Answer:

m=1/2, n=3

Step-by-step explanation:

It is a factor of (x+2)

-2 is a factor

mx^3+nx^2+x-6=0

m(-8)+4n+(-2)-6=0

-8m+4n-8=0

4n-8m=8

n-2m=2----> 1)

mx^3+nx^2+x-6/(x-2)=4

m(8)+4n+2-6=4

8m+4n=8

2m+n=4----> 2)

From 1) and 2) we get

m=1/2 and n=3

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