Question

Find all the zeros of polynomial:-​

Answer 1

Answer:

1 & 1/2

Solution:

To find: All zeros of f(x) = 2x⁴ - 3x³ - 3x² + 6x - 2 = 0

Given: Two zeros √2 and -√2.

Procedure:

⇒ As √2 and -√2 are two zeros, (x - √2) and (x + √2) are two factors, so is (x - √2)(x + √2) = (x² - 2).

⇒ Divide f(x) by (x² - 2) and factorise the quotient formed.

⇒ The factors of the quotient thus formed are the other factors of f(x).

Method:

$2x^4-3x^3-3x^2+6x-2 \\ \\ 2x^4-3x^3-4x^2+x^2+6x-2 \\ \\ 2x^4-4x^2-3x^3+6x+x^2-2 \\ \\ 2x^2(x^2-2)-3x(x^2-2) +(x^2-2) \\ \\ (x^2-2)(2x^2-3x+1) \\ \\ (x^2-2)(2x^2-2x-x+1) \\ \\ (x^2-2)(2x(x-1)-(x-1)) \\ \\ (x^2-2)(x-1)(2x-1)$

Inference: The other two zeros are 1 and 1/2.