Question

Find the all the zeroes of 2x⁴
- 3x³
- 3x²
+ 6x – 2, if you know that two of its
zeroes are √2 and - √2.

Answer 1

EXPLANATION.

All the zeroes of the polynomial.

⇒ 2x⁴ - 3x³ - 3x² + 6x - 2.

If two of its zeroes is √2 and -√2.

As we know that,

Zeroes of the polynomial.

⇒ x = √2.

⇒ x - √2. - - - - - (1).

⇒ x = -√2.

⇒ x + √2. - - - - - (2).

Products of the zeroes of the quadratic polynomial.

⇒ (x - √2)(x + √2).

As we know that,

Formula of :

⇒ (x² - y²) = (x - y)(x + y).

Using this formula in the equation, we get.

⇒ (x² - 2).

Divide : 2x⁴ - 3x³ - 3x² + 6x - 2  by  x² - 2.

We get,

⇒ 2x² - 3x + 1.

Factorizes the equation into middle term splits, we get.

⇒ 2x² - 3x + 1 = 0.

⇒ 2x² - 2x - x + 1 = 0.

⇒ 2x(x - 1) - 1(x - 1) = 0.

⇒ (2x - 1)(x - 1) = 0.

⇒ x = 1/2  and  x = 1.

All zeroes are = √2, -√2, 1/2, 1.

Answer 2

Zeroes -

$\pink{ \sqrt{2} } \\ \orange{ - \sqrt{2}} \\ \blue{ 1 }\\ \red{ \frac{1}{2} }$

See the attachment for explanation.

hope it helps.