Question

determine the value of m and n for mx^3 +20x^2+ nx-36 when divided by (x+1) gives remainder 0 and when divided by (x-2) gives remainder of 45.​

Answer 1

Given:

The given polynomial is

$mx^3+20x^2+nx-36$

When it divided by (x+1) gives remainder 0 and when divided by (x-2) gives remainder of 45.

To find:

The values of m and n.

Solution:

According to the remainder theorem, if a polynomial P(x) is divided by (x-c), then the remainder is P(c).

Let the given polynomial be P(x).

$P(x)=mx^3+20x^2+nx-36$

When it divided by (x+1) gives remainder 0. So, P(-1)=0.

$P(-1)=m(-1)^3+20(-1)^2+n(-1)-36$

$0=-m+20-n-36$

$0=-m-n-16$

$m+n=-16$                   ...(i)

When P(x) is divided by (x-2) gives remainder of 45. So, P(2)=0

$m(2)^3+20(2)^2+n(2)-36=45$

$8m+80+2n-36=45$

$8m+2n+44=45$

$8m+2n=45-44$

$8m+2n=1$                  ...(ii)

Multiply 2 on both sides of (i).

$2m+2n=-32$                   ...(iii)

Subtract (iii) from (ii).

$6m=33$

$m=\dfrac{33}{6}$

$m=5.5$

Putting m=5.5 in (i), we get

$5.5+n=-16$

$n=-16-5.5$

$n=-21.5$

Therefore, the values of m and n are 5.5 and -21.5 respectively.