Question

please help. its urgent​

Answer 1

As (x-1) and (x+1) are the factors of f(x)=

${x}^{4} + a {x}^{3} - 3 {x}^{2} + 2x + b$

Then, f(1) and f(-1) should be equal to 0.

Hence, required equation

$1 + a - 3 + 2 + b = 1 - a- 3 - 2 + b \\ = > 2a = 0 \\ = > a = 0$

Similarly,

$1 - 3 + 2 + b = 0 \\ = > b = 0$

HOPE THIS COULD HELP!!!

Answer 2

Answer:

Given:

x+1 and x-1 are factors of the polynomial.

$x^{4}+ax^3-3x^2+2x+b$

To Find:

• The values of 'a' and 'b'

Solving Question:

We know the factor theorem that if x-a is a factor of polynomial p(x) then  p(a) = 0

We will use this to find the answer.

Solution:

x + 1 = 0

or, x = -1

substitute the values ,

$x^4+ax^3-3x^2+2x+b\\\\x=-1\\\\\implies (-1)^4+a(-1)^3-3(-1)^2+2(-1)+b=0\\\\or,1-a-3-2+b=0\\\\or,-4-a+b=0\\\\or,b-a=4......equ(1)$

x -1 = 0

or, x = 1

$x^4+ax^3-3x^2+2x+b=0\\\\x=1\\\\\implies1^4+a*1^3-3*1^2+2*1+b=0\\\\or,1+a-3+2+b=0\\\\or,a+b=0......equ(2)$

Take equ(1) and (2)

b - a = 4

b + a = 0

.....................

we get, 2b = 4

or, b = 2

substitute in equ{2)

a +b =0

or, a +2 =0

or, a = -2

∴ The value of 'a' is -2 and that of 'b' is 2