Question

find all the zeroes of 2x^4-3x^3-3x^2+6x-2 if you know that two of its zeroes are √2 and -√2.​

Answer 1

Question :

Find all the zeroes of 2x⁴ - 3x³ - 3x² + 6x - 2. If you know that two of its zeroes are √2 & -√2

Answer

Given : -

Multinomial 2x⁴ - 3x³ - 3x² + 6x - 2

Zeroes are √2 and -√2

Required to find : -

• All zeroes ?

Solution : -

Let the given multinomial be;

p(x) = 2x⁴ - 3x³ - 3x² + 6x - 2

Here,

√2 and -√2 are zeroes

So,

x = √2 or -√2

This implies;

x = √2 (or) x = -√2

x - √2 = 0 (or) x + √2 = 0

Now,

(x - √2) (x + √2)

• (a + b)(a - b) = a² - b²

(x)² - (√2)²

x² - 2

Let this be ; g(x) = x² - 2

Now,

Let's divide p(x) with g(x)

We have;

Division algorithm !

p(x) = g(x) x q(x) + r(x)

{ r(x) = 0 }

p(x) = ( x² - 2 ) q(x)

2x⁴ - 3x³ - 3x² + 6x - 2 ÷ (x² - 2) = q(x)

2x² - 3x + 1 = q(x)

Let's factorize q(x)

2x² - 3x + 1

2x² - 2x - 1x + 1

2x(x - 1) + 1(x - 1)

(x - 1) (2x + 1)

Now,

(x - 1) = 0 & 2x + 1 = 0

x = 1 & x = (-1)/(2)

Therefore,

All zeroes of p(x) are √2 , -√2 , 1 , (-1)/(2)