Question

f(x)=2x4-3x3-3x2+6x-2;√2 and -√2​

Answer 1

SolutioN :

Let's,

• f( x ) = √2.

$\tt\dagger \: \: \: \: \: \fbox{ f(x) = 2 {x}^{4} - 3 {x}^{3} - 3 {x}^{2} + 6x- 2.}$

$\tt : \implies f( \sqrt{2} ) = 2 {x}^{4} - 3 {x}^{3} - 3 {x}^{2} + 6x- 2$

$\tt : \implies 2 { (\sqrt{2} )}^{4} - 3 { (\sqrt{2}) }^{3} - 3 { (\sqrt{2} )}^{2} + 6( \sqrt{2} )- 2$

$\tt : \implies 2 \times( 4) - 3 \times (2 \sqrt{2}) - 3 ( \sqrt{2}) + 6 \sqrt{2} - 2.$

$\tt: \implies 8 - 8.$

$\tt : \implies 0.$

When.

• f( x ) = -√2.

$\tt : \implies f( - \sqrt{2} ) = 2 {x}^{4} - 3 {x}^{3} - 3 {x}^{2} + 6x- 2$

$\tt : \implies 2 {( - \sqrt{2}) }^{4} - 3 { ( - \sqrt{2}) }^{3} - 3 {( - \sqrt{2} )}^{2} - 6 \sqrt{2} - 2$

$\tt : \implies 8 - 8.$

$\tt : \implies 0.$

★ Yes, When we putting given value f( x ) it becomes 0.

So, We can say that √2 and -√2 is the Zeros of p( x ).