Question

find the zeroes.
it will help if u will do it in copy, click &send...​

Answer 1

Question :-

• Polynomial = 2x⁴ - 3x³ - 3x² + 6x - 2.
• Two Roots are = √2 & (-√2).

Concept used :-

• if a & b are the zeros of Polynomial than (x - a)(x - b) is Completely Divide The given polynomial , or Remainder will be Zero.

Solution :-

Given zeros = √2 and -√2

so,

→ (x - √2) (x + √2)

→ (x² - 2) will Divide the Polynomial .

now,

Divide Process :-

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

2x⁴ -4x²

-3x³ +x² + 6x

-3x³ + 6x

x² - 2

x². - 2

__0__

we get the quotient :- 2x² - 3x + 1

Putting Quotient Equal to Zero , & splitting the middle term, we get,

→ 2x² - 3x + 1 = 0

→ 2x² - 2x - x + 1 = 0

→ 2x(x - 1) -1(x - 1) = 0

→ (2x - 1)(x - 1) = 0

Putting Both Equal to Zero Now,

→ 2x - 1 = 0

→ 2x = 1

→ x = 1/2

and,

→ x - 1 = 0

→ x = 1

Hence, all the zeroes of the given Polynomial are {√2 , -√2 , ½ , 1} .

Answer 2

Given

Polynomial = 2x⁴ - 3x³ - 3x² + 6x - 2

Two zeros = √2 & -√2

To find

All the zeros of polynomial.

Solution

Here, two zeros are √2 & -√2 .

According to remainder theorem,(x - a)(x - b) will be the required divisor of the given polynomial.

➛ (x - a)(x - b) = Divisor

➛ (x - √2)(x - (-√2)) = Divisor

➛ (x - √2)(x + √2) = Divisor

➛ x² - (√2)² = Divisor

➛ x² - 2 = Divisor

Now dividing given polynomial by (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

So, quotient = 2x² - 3x + 1

⇒ 2x² - 3x + 1 = 0

⇒ 2x² - 2x - x + 1 = 0

⇒ 2x(x - 1) - 1(x - 1) = 0

⇒ (2x - 1)(x - 1) = 0

⇒ 2x = 1 or, x = 1

⇒ x = ½ or, x = 1

So, other two zeros = ½ & 1

Therefore,

All zeros of polynomial are: √2 , -√2, ½ & 1