Question

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

Answer 1

Answer: n = 0

m = 0

Step-by-step explanation: first 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

Answer 2

First of all factorize:-

y² -1

=( y²) - ( 1²)

By using identinty:-

a² - b² = (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) = y⁴+m(y³)+2y²-3y+n

p(1)= 1⁴+m(1³)+2×1²-3×1+n

0= 1+m+2-3+n

0=m+n -(1)

Now, p(y)=y⁴+m(y³)+2y²-3y+n

p(-1)=(-1)⁴+m(-1³)+2×(-1)²-3×(-1)+n

0 = 1-m+2-3+n

0=n-m -(2)

Adding (1) and (2)

n-m=0

n+m=0

n=0

m=0

Hope it helps you,

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