Find the values of m and n when the polynomial f\left( x \right)={{x}^{3}}-2{{x}^{2}}+mx+n has a factor \left( x+2 \right) and leaves a remainder 9 when divided by \left( x+1 \right).
Find the values of m and n when the polynomial f\left( x \right)={{x}^{3}}-2{{x}^{2}}+mx+n has a factor \left( x+2 \right) and leaves a remainder 9 when divided by \left( x+1 \right).

Remainder theorem is a theorem in algebra: if f(x) is a polynomial in x then the remainder on dividing f(x) by x-a is f(a).

In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is special case of the polynomial remainder theorem.

Solution:

From the question it is given that,

f\left( x \right)={{x}^{3}}-2{{x}^{2}}+mx+n

Remainder = 9

Let us assume that, x+2=0,x=-2

x+1=0,x+-1

Now, substitute the value of xin f\left( x \right) we get,

f\left( -2 \right)={{\left( -2 \right)}^{3}}-2{{\left( -2 \right)}^{2}}+m\left( -2 \right)+n=0

-8-8-2m+n=0

-16-2m+n=0

n=2m+16 … [equation (i)]

Then, f\left( -1 \right)={{\left( -1 \right)}^{3}}-2{{\left( -1 \right)}^{2}}+m\left( -1 \right)+n=9

-1-2-m+n=9

-3-m+n=9

m=-3-9+n

m=n-12 … [equation (ii)]

Now, combining both equation (i) and equation (ii) we get,

n=2\left( n-12 \right)+16

n=2n-24+16

n=2n-8

2n-n=8

n=8

Consider the equation (ii) to find out the value of m,

m=n-12

m=8-12

m=-4

Therefore, the value of n is 8 and m is -4.