Find the values of a and b so that (x + 1) and (x - 1) are the factors of

Solution:

Here, $f(x)=x^{4}+a x^{3}-3 x^{2}+2 x+b$

The factors are (x + 1) and (x - 1)

From factor theorem, if x = 1, -1 are the factors of f(x) then f(1) = 0 and f(-1) = 0

Let, us take x + 1

⟹ x + 1 = 0

⟹ x = -1

Substitute value of x in f(x)

$f(-1)=(-1)^{4}+a(-1)^{3}-3(-1)^{2}+2(-1)+b$

= 1 - a - 3 - 2 + b

= -a + b - 4 ... 1

Let, us take x - 1

⟹ x - 1 = 0

⟹ x = 1

Substitute value of x in f(x)

$f(1)=(1)^{4}+a(1)^{3}-3(1)^{2}+2(1)+b$

= 1 + a - 3 + 2 + b

= a + b .... 2

Solve equations 1 and 2

- a + b = 4

a + b = 0

2b = 4

b = 2

substitute value of b in eq 2

a + 2 = 0

a = - 2

the values are a = -2 and b = 2