Question

Find the values of m and n if y2-1 is a factor of y4 + my3 +2y2-3y+n.

Answer 1

Step-by-step explanation:

factorize y^2 - 1

                             = ( y^2) - ( 1^2)

identity used= a^2 - b^2 = (a+b) (a-b)

                    = (y+1) (y-1)

now we have the factors,

if we say that a term is completely divisible by say, x then it is completely divisible also by the factors of x

now p(y) = y^4+m(y^3)+2y^2-3y+n

       p(1)= 1^4+m(1^3)+2*1^2-3*1+n

        0= 1+m+2-3+n

        0=m+n-------> eq.1

now p(y)=y^4+m(y^3)+2y^2-3y+n

        p(-1)=(-1)^4+m((-1^3)+2*-1^2-3*-1+n

             0 = 1-m+2-3+n

             0=n-m-------> eq.2

adding eq.1 & eq.2

n-m=0

n+m=0

n     =0

m=0

Hope it helps :)-