Question

P(x) = 2x4 - 3x3 - 3x2 + 6x - 2
Zeroes are √2&-√2 find all the zeroes

Answer 1

Answer:

$2 {x}^{4} - 9 - 6 + 6x - 2 = 0$

$2 {x}^{4} - 9 - 8 + 6x = 0$

$2 {x}^{4} - 17 + 6x = 0$

$2 {x}^{4} + 0 {x}^{3} + 0 {x}^{2} + 6x = 0$

the above is the quadratic equation

now,

the zeroes are

Step-by-step explanation:

$\sqrt{2} \: and \: - \sqrt{2}$

$\alpha + \beta = \sqrt{2 } - \sqrt{2 } \: = 0$

$\alpha \beta = \sqrt{2 } \times - \sqrt{2} = - 2$

the general formula for polynomial is

${x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0$

${x}^{2} - (0) - 2 = 0$

${x}^{2} - 2 = 0$

divide,

$2 {x}^{4} + 0 {x}^{3} + 0 {x}^{2} + 6x - 17 \div {x}^{2} - 2$

the answer will be available