How do you write a polynomial in standard form given the zeros x=1, -1, √3i, -√3i?

Question

Theodore Amidon · Accepted Answer

When we know the zeros of a polynomial #z_i# , we can obtain the polynomial, by multiplying a constant different of zero, #a#, by the product of all #(x-z_i)#. If zeros are # 1, -1, sqrt(3)i, -sqrt(3)i# The polynomial will be: #a(x-1)(x+1)(x-sqrt(3)i)(x+sqrt(3)i)# #a(x^2-1)(x^2+3)# #a(x^4+3x^2-x^2-3)# #a(x^4+2x^2-3)#, where a is any real number except zero.

Sadie Ackerman · Answer

undefined To write a polynomial in standard form given the zeros $ x = 1, -1, \sqrt{3}i, -\sqrt{3}i $, you can use the fact that complex roots occur in conjugate pairs.

Start by writing the factors of the polynomial using the given zeros. For real roots $ x = 1 $ and $ x = -1 $, the factors are $ (x - 1) $ and $ (x + 1) $, respectively.

For the complex conjugate pair $ x = \sqrt{3}i $ and $ x = -\sqrt{3}i $, the factors are $ (x - \sqrt{3}i) $ and $ (x + \sqrt{3}i) $, respectively.

Now, multiply all these factors together to obtain the polynomial expression.

$$ (x - 1)(x + 1)(x - \sqrt{3}i)(x + \sqrt{3}i) $$

Expand this expression and simplify to get the polynomial in standard form.

$$ (x - 1)(x + 1)(x^2 + 3) $$

Now, distribute and multiply the factors:

$$ (x - 1)(x + 1)(x^2 + 3) = (x^2 - 1)(x^2 + 3) $$

$$ = x^4 + 3x^2 - x^2 - 3 $$

$$ = x^4 + 2x^2 - 3 $$

Thus, the polynomial in standard form is $ x^4 + 2x^2 - 3 $.