Question

Apply the Fundamental Theorem of Algebra to find the number of imaginary roots for the polynomial f(x)=4x^2−14+x^8 , if you know its graph has two x-intercepts.

Answer 1

The Fundamental Theorem of Algebra states that a polynomial of degree \( n \) will have exactly \( n \) roots (including real and complex roots, counted with multiplicity).

In the given polynomial \( f(x) = 4x^2 - 14 + x^8 \), the degree is determined by the term with the highest power of \( x \), which is \( x^8 \). Hence, the degree of the polynomial is 8.

Since the polynomial is of degree 8, it has a total of 8 roots. We also know that the graph of the polynomial has two x-intercepts. This means that there are 2 real roots.

The total number of roots is given by the degree of the polynomial:
\[
\text{Total roots} = 8
\]
\[
\text{Real roots} = 2
\]

To find the number of imaginary (or complex) roots, we can use the relationship:
\[
\text{Complex roots} = \text{Total roots} - \text{Real roots}
\]

Since complex roots come in conjugate pairs (and every complex root counts as two in the root count), the remaining roots after accounting for the 2 real roots must be complex.

Thus, we calculate:
\[
\text{Complex roots} = 8 - 2 = 6
\]

Therefore, since complex roots occur in pairs, we can specifically say:
\[
\text{Imaginary roots} = \frac{6}{2} = 3 \text{ pairs of complex roots}
\]

So the total number of imaginary roots is 6.

Thus, the polynomial \( f(x) = 4x^2 - 14 + x^8 \) has **6 imaginary roots**.