Question

i will mark as brainliest . pls help.
(x-1),(x+2) are factors of 2x^3(2x cube) +mx^2(mx square)-x-n . value of (m^2(m square)-n^2(n square)) .

Answer 1

19

Hey there, use factor theorem here.

The given expression is $2x^3+mx^2-x-n^2$

By applying division algorithm A=BQ+R

$2x^3+mx^2-x-n^2=(x-1)Q(x)$

$2x^3+mx^2-x-n^2=(x+2)Q(x)$

We should approach in a way : x=1 and x=-2 makes it zero.

x=1

$2+m-1-n^2=m-n^2+1=0$

x=-2

$-16+4m+2-n^2=4m-n^2-14=0$

The solutions are $m=5$ and $n^2=6$

Therefore, m²-n² is 25-6 or 19