find all zeros of P, real and complex. Factor P completely. P(x)= x^4 +6x^2 +9

Question

akashdeepdeb · Answer

$$P(x)= x^4 +6x^2 +9$$

Now P(x) is a bi-quadratic polynomial BUT we CAN write it as a quadratic polynomial by taking x^2 = y

Now
$$ P(y) = y^2 + 6y + 9 $$

Where y = x^2

Now find out the roots of 
P(y) = -3,-3

y = x^2 = -3
x = +√3i , -√3i

y = x^2 = -3
x = +√3i , - √3i

So the roots pf P(x) are --> +√3i , -√3i, +√3i , - √3i

Understood? :)

anonymous · Answer

why do we find the roots of -3?

akashdeepdeb · Answer

Because ORIGINALLY we need to find the roots of x and NOT y.

anonymous · Answer

ooohh okay thank you so much! I don't know if clicking the best response helps you in any way but I clicked it just in case!

akashdeepdeb · Answer

Helps a lot.
You're welcome. :)