Question

Given that the expression 2x^3 + mx^2 + nx + c leaves the same remainder when divided by x -2 or by x+1 I prove that m+n =-6

Answer 1

Given:

The expression is:

$2x^3+mx^2+nx+c$

It leaves the same remainder when divided by x -2 or by x+1.

To prove:

$m+n=-6$

Solution:

Remainder theorem: If a polynomial P(x) is divided by (x-c), thent he remainder is P(c).

Let the given polynomial is:

$P(x)=2x^3+mx^2+nx+c$

It leaves the same remainder when divided by x -2 or by x+1. By using remainder theorem, we can say that

$P(2)=P(-1)$              ...(i)

Substituting $x=-1$ in the given polynomial.

$P(-1)=2(-1)^3+m(-1)^2+n(-1)+c$

$P(-1)=-2+m-n+c$

Substituting $x=2$ in the given polynomial.

$P(2)=2(2)^3+m(2)^2+n(2)+c$

$P(2)=2(8)+m(4)+2n+c$

$P(2)=16+4m+2n+c$

Now, substitute the values of P(2) and P(-1) in (i), we get

$16+4m+2n+c=-2+m-n+c$

$16+4m+2n+c+2-m+n-c=0$

$18+3m+3n=0$

$3m+3n=-18$

Divide both sides by 3.

$\dfrac{3m+3n}{3}=\dfrac{-18}{3}$

$m+n=-6$

Hence proved.