Question

if x-1 and x+2 are two factors of the polynomial 2x^3 +mx^2 - x - n then the value of m^2-n^2​

Answer 1

Answer:

m^2-n^2 = -155\25 (or) -6.2

Step-by-step explanation:

x-1 and x+2 are 2 factors of p(x) = 2x^3+mx^2 -x-n

let

x-1 = 0

x = 1

let

x+2 = 0

x = -2

p(x) = 2x^3+mx^2-x-n

let x = 1

p(1) = 2(1)^3+m(1)^2-1-n = 0

2+m-1-n =0

1+m-n = 0

m-n = -1 -----(1)

let x = -2

p(-2) = 2(-2)^3+m(-2)^2-(-2)-n = 0

-16+4m+2-n = 0

-14+4m-n = 0

4m-n = 14 ------(2)

by adding (1) & (2)

m-n = -1

4m-n = 14

__________

5m = 13

m = 13\5

substitute 'm' value in (1)

m-n = -1

(13\5)- n = -1

13-5n\5 = -1

13-5n = -1×5

13-5n = -5

-5n = -5-13

-5n = -18

n = 18\5

Then,

m^2 - n^2 = (13\5)^2-(18\5)^2

= (169\25)-(324\25)

=169-324\25

m^2-n^2 = -155\25 (or) -6.2