Question

Find the zeros of 2x^4-3x^3-3x^2+6x-2x^2-10x-5 if two zeros are root2 and-root2

Answer 1

Correct 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 :-

Zeros of equation 2x⁴ - 3x³ - 3x² + 6x - 2x² - 10x - 5 are √2 and -√2

To Find :-

The remaining zeros of equation .

Solution :-

Firstly what's the correct polynomial is →

→ 2x⁴ - 3x³ - 3x² + 6x - 2

Now we are given with 2 zeros which are

→ x = √2

→ x -√2 = 0 ...... eq.1st

→ x = - √2

→ x +√2 = 0 .......eq2nd

Now the product of eq 1st and eq. 2nd will divide completely the given quadratic equation .

So there product will be :-

→ ( x+√2 ) × ( x-√2)

→ x² - 2

Now let's divide 2x⁴ - 3x³ - 3x² +6x + 2 by x² - 2

Which is attached in the attachment above .

From division we concluded that quotient is

2x² - 3x + 1 .

Now splitting the middle term of quotient .

→ 2x² - 2x - x + 1 = 0

→ 2x ( x - 1 ) - 1 ( x - 1 ) = 0

→ ( x-1 ) . ( 2x - 1 ) = 0

→ x = 1

→ x = 1/2

So x = 1 or x = 1/2 .

So by this method we got our two remaining zeros.

Zeros are 1 , 1/2 , √2 , -√2 .