Question

Find all zeroes of 2x⁴ -3x³ - 3x² + 6x - 2 , if you know that two of its zeroes are √2 and -√2 .

Answer 1

$\huge\bigstar\underline\mathfrak\purple{Answer}$

All the zeroes of this polynomial are :

√2, -√2, 1 and 1/2

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$\huge\bigstar\underline\mathfrak\purple{Explanation}$

Given : √2 and -√2 are the zeroes of this polynomial, 2x⁴ -3x³ - 3x² + 6x - 2

To find : All the zeroes of this polynomial, 2x⁴ -3x³ - 3x² + 6x - 2

Solution : It is given that √2 and -√2 are the zeroes of this polynomial, 2x⁴ -3x³ - 3x² + 6x - 2

Therefore, (x+√2)(x-√2) i.e x² - 2 is the factor of this polynomial, 2x⁴ -3x³ - 3x² + 6x - 2.

Now, let us divide the given polynomial by x²-2 [Refer to the attachment]

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For other zeroes,

2x² - 3x + 1 = 0

By splitting middle term,

=> 2x² - 2x - x + 1 = 0

=> 2x ( x - 1 ) - 1 ( x - 1 ) = 0

=> (2x-1)(x-1) = 0

=> 2x-1 = 0 and x-1 = 0

=> x = 1/2 and 1

Therefore, all the zeroes of this polynomial, 2x⁴ -3x³ - 3x² + 6x - 2 are 1, 1/2, √2 and -√2.

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Answer 2

Answer:

Concept:

Polynomials, finding all zeros using the given equation.

Step-by-step explanation:

Given:

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

Find:

all zeros of the given equation.

Solution:

$Let \ p(x)=2 x^{4}-3 x^{3}-3 x^{2}+6 x-2 \\Since x=\sqrt{2} and \sqrt{2} are the zeroes of the given polynomial p(x)$

$(x+\sqrt{2})(x-\sqrt{2}) will be a factor or x^{2}-2 will be a factor of p(x)$

we have,

Check the image for calculation

$\begin{aligned}&2 x^{2}-3 x+1\\&=2 x^{2}-2 x-x+1 \\&=2 x(x-1)-1(x-1) \\&=(2 x-1)(x-1)\end{aligned}$

$So, \ p(x)=(x+\sqrt{2})(x-\sqrt{2})(2 x-1)(x-1) \\Thus, the other zeroes of p(x) are \frac{1}{2} and 1 .$

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