Question

if y^2-1 is a factor of y^4+my^3+2y^2-3y+n find value of m and n​

Answer 1

Question :-

If y² - 1 is the factor of y⁴ + my³ + 2y² - 3y + n .

Find the value of m and n .

Answer :-

Given :-

p ( y ) = y⁴ + my³ + 2y² - 3y + n

If y² - 1 is the factor .

Required to find :-

Values of m and n ?

Solution :-

Given :-

p (y) = y⁴ + my³ + 2y² - 3y + n

If y² - 1 is the factor ,

we know that,

y² - 1 = 0

y² = 1

y = √1

y = +1 or -1

So,

Let's substitute this value in place of y in p(y) .

p ( y ) = y⁴ + my³ + 2y² - 3y + n

p(1) =

(1)⁴ + m(1)³ + 2(1)² - 3(1) + n = 0

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

1 + m + 2 - 3 + n = 0

3 + m - 3 + n = 0

+3 & -3 get cancelled

m + n = 0

$\red{\tt{ n = - m }}{\longrightarrow{equation - 1 }}$

Similarly,

Substitute the value of - 1 in p ( y )

So,

p ( y ) = y⁴ + my³ + 2y² - 3y + n

p ( -1 ) =

( -1 )⁴ + m( -1 )³ + 2( -1 )² - 3( -1 ) + n = 0

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

1 - m + 2 + 3 + n = 0

Substitute the value of n from equation 1

1 - m + 2 + 3 - m = 0

6 - 2m = 0

- 2m = - 6

negative signs get cancelled on both sides

$\longrightarrow{\tt{ 2m = 6 }}$

$\rightarrow{\tt{ m = \dfrac{6}{2}}}$

$\implies{\red{\underline{\tt{ m = 3 }}}}$

So,

Now substitute this value of m in equation 1 .

$\longrightarrow{\tt{ n = - m }}$

$\rightarrow{\tt{n = - ( 3 ) }}$

$\implies{\red{\underline{\tt{ n = -3 }}}}$

Therefore,

$\underline{\large{\leadsto{\boxed{\mathrm{ Value \; of \; m = 3 }}}}{\bigstar}}$

$\underline{\large{\leadsto{\boxed{\mathrm{ Value \; of \; n = -3 }}}}{\bigstar}}$

Answer 2

Answer:

-3 is a right answer friend