Question

find all the zeroes the
of the polynomial x^4_3x^3+6x_4 if two of its zeroes are √2 and -√2​

Answer 1

Answer:

2&1

Step-by-step explanation:

x=root2

x= - root2

then

(x-root2) × ( x+root2)

=(x^2-2)

divide x^4-3x^3+6x-4 with (x^2-2)

you will get an equation:

x^2-3x+2

then solve it by spliting method:

x^2-2x-x+2

x(x-2) - 1(x-2)

( x-2)(x-1)

x-2 =0

x=2

x-1=0

x=1

Answer 2

Answer:

1 and 2

Step-by-step explanation:

p(X)= x⁴-3x³+6x-4

Two zeroes are  $\sqrt{2\\}$ and -$\sqrt{2\\}$

= x+$\sqrt{2\\}$ and x-$\sqrt{2\\}$ are two zeroes

Multiplying (x+$\sqrt{2\\}$)(x-$\sqrt{2\\}$)

We get x²-2

Now we have to divide p(x) by x²-2

Then, we get an equation

x²-3x-2

Now, we have to use middle term split method here

x²-2x-x-2

x(x-2)-1(x-2)

(x-1)(x-2) are two zeroes

Now putting them equal to 0

We get

x=1  and x=2

Hope this will help you.

Thanks