Question

if the zeroes of the polynomial f(x)=2x^44-3x^3-3x^2+6x-2. fond two of ots zeroes are root 2 and -root 2​

Answer 1

Answer :-

Two zeroes = √2, - √2

Sum of zeroes = $\alpha + \beta$

= √ 2 + (- √2)

= 0

Product of zeroes = ${\alpha \times \beta}$

= √2 × - √2

= -2

$\sf{[{x}^{2} + (\alpha + \beta)x + (\alpha \times \beta)]}$

$\implies{\sf{{x}^{2} - 2}}$

Divide 2x⁴ - 3x³ - 3x² + 6x - 2 with x² - 2.

_______________________

x² - 2 )2x⁴ - 3x³ - 3x² + 6x - 2( 2x²-3x+1

2x⁴ - 4x²

- +

_____________________

-3x³ + x² + 6x - 2

-3x³ + 6x

+ -

________________

x² - 2

x² - 2

- +

________________

0

Now, Factorise the quotient ;

$\sf{{2x}^{2} - 3x + 1 = 0}$

$\implies{\sf{{2x}^{2} - 2x - x + 1 = 0}}$

$\implies{\sf{2x (x - 1) - 1 (x - 1) = 0}}$

$\implies{\sf{(2x - 1) (x -1) = 0}}$

$\implies{\sf{2x - 1 = 0}}$

$\implies{\sf{2x = 1}}$

$\implies{\sf{x = \dfrac{1}{2}}}$

Also,

$\implies{\sf{x - 1 = 0}}$

$\implies{\sf{x = 1}}$

So, the four zeroes are :-

$\boxed{\sf{\sqrt{2} , - \sqrt{2} , \dfrac{1}{2}\ and\ 1}}$