Question

Find all the zeroes of the polynomial p(x)=x^4-3x^3-3x^2+6x-2, if two of its zeroes are root2 and -root2. ​

Answer 1

Step-by-step explanation:

==> In The example we have,

p(x) = 2x⁴ - 3x³ - 3x² + 6x - 2

We have also given that √2 and -√2 are the two zeroes of the given polynomial.

∴ ( x + √2) ( x - √2)

=> x² - 2

Now, divide the given polynomial by x² -2 we will use long division method.

Refer from the attachment which I have provided .

Now, you can see in the photo that we got the value of Quotient as 2x² -3x +1

So, now we can easily find the other zeroes of this polynomial by using splitting the middle term .

=> 2x² - 3x + 1

=> 2x² - 2x - x + 1

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

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

∴ ( x - 1 )= 0

∴ x = 1

and

∴ (2x -1) = 0

∴ 2x = 1

∴ x = 1/2

Therefore, the zeroes of the given polynomial are 1 , 1/2 , √2 and -√2 .

hope it helps you