Question

Pls help solve this ASAP

Answer 1

k = 1 , k = 2

Explanation:

if (x - a) is a factor of f(x) then f(a) = 0

here

f(x) = k²
`x^(4)` - 3kx² + 2

given (x - 1) is a factor of f(x) then f(1) = 0 , so

k²
`(1)^(4)` - 3k
`(1)^(2)` + 2 = 0 , that is

k² - 3k + 2 = 0 ← quadratic in standard form

(k - 1)(k - 2) = 0 ← in factored form

equate each factor to zero and solve for k

k - 1 = 0 ⇒ k = 1

k - 2 = 0 ⇒ k = 2

Answer 2

k = 1, 2

Explanation:

Let the given polynomial be a function.

So,

f(x) = k²x⁴ - 3kx² + 2

Given factor = x - 1

Let,

x - 1 = 0

So,

x = 1

Put x = 1 in the above function.

So,

f(1) = k²(1)⁴ - 3k(1)² + 2

According to remainder theorem, f(1) = 0.

So,

0 = k² - 3k + 2

k² - 3k + 2 = 0

Applying mid-term break formula.

k² - 2k - k + 2 = 0

Take common factors.

k(k - 2) - 1(k - 2) = 0

Take (k - 2) as the common factor.

(k - 2)(k - 1) = 0

Either,

k - 2 = 0 OR k - 1 = 0 [Zero's method]

So,

k = 2 OR k = 1

`\rule[225]{225}{2}`