Question

verify that √2 & -√2 are zeros of given polynomial 2x⁴-3x³-3x²+6x-2 .​

Answer 1

Given :-

• Polynomial : 2x⁴-3x³-3x²+6x-2

To Find :

• √2 & -√2 are zeros of given polynomial 2x⁴-3x³-3x²+6x-2 .

Solution:

Given f(x) = $\tt \: 2x⁴ - 3x³ - 3x² + 6x - 2$

Let, x = √2

$\sf\huge\implies$ $\tt \: f(x) = 2(√2)⁴ - 3(√2)³ - 3(√2)² + 6(√2) - 2$

$\sf\huge\implies$ $\tt \: 2(4) - 6√2 - 3(2) + 6√2 - 2$

$\sf\huge\implies$ $\tt \: 8 - 8 = 0$

$\star$ When x = -√2 ,

$\sf\huge\implies$ $\tt \: f (-√2) = 2(-√2)⁴ - 3(-√2)³ - 3(-√2)² + 6(-√2) -2$

$\sf\huge\implies$ 2(4) +6√2 - 3(2) - 6√2 - 2

$\sf\huge\implies$ 8 - 8 = 0

So, we can say that √2 and -√2

are the zeroes of polynomial

2x⁴-3x³-3x²+6x-2

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Answer 2

Step-by-step explanation:

Given :-

Polynomial : 2x⁴-3x³-3x²+6x-2

To Find :

√2 & -√2 are zeros of given polynomial 2x⁴-3x³-3x²+6x-2 .

Solution:

Given f(x) = $\tt \: 2x⁴ - 3x³ - 3x² + 6x - 22x⁴−3x³−3x²+6x−2$

Let, x = √2

$\sf\\tt \: f(x) = 2(√2)⁴ - 3(√2)³ - 3(√2)² + 6(√2) -$

2f(x)=2(√2)⁴−3(√2)³−3(√2)²+6(√2)−2

$\sf\tt \: 2(4) - 6√2 - 3(2) + 6√2 - 22(4)−6√2−3(2)+6√2−2$

$\sf\tt \: 8 - 8 = 08−8=0$

$\star⋆ When x = -√2 ,$

$\sf\tt \: f (-√2) = 2(-√2)⁴ - 3(-√2)³ - 3(-√2)² + 6(-√2)$

$\sf 2(4) +6√2 - 3(2) - 6√2 - 2$

$\sf 8 - 8 = 0$

So, we can say that √2 and -√2

are the zeroes of polynomial

2x⁴-3x³-3x²+6x-2