Question

How many and of which kind of roots does the equation

Answer 1

Answer:

D. 1 real; 4 complex  

Step-by-step explanation:

1. Calculate the total number of roots

The fundamental theorem of algebra states that any polynomial of degree n has n roots.

Your polynomial is fifth degree, so it has five roots.

2. Calculate the number of real roots

We can use Descartes' rule of signs to determine the number of real roots:

The number of positive real roots is the same as the number of changes in the sign of the coefficients of ƒ(x) or less than by an even number.

The number of negative real roots is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this by an even number.

(a) Number of positive real roots

The coefficients of ƒ(x) are

+1 +3 +8 +14 +16 + 8  

There are no changes of sign, so there are no positive real roots.

(b) Number of negative real roots

(i) Find f(-x)

f(-x) = -x⁵+ 3x⁴ - 8x³ + 14x² - 16x + 8

(ii) Count the changes of sign

The coefficients of f(-x) are

-1 ∥ +3 ∥ -8 ∥+14 ∥ -16 ∥ +8

There are five changes of sign.

Thus, the number of negative roots is five, three, or one.

So far, we know that we have five roots. None is positive, and there could be one, three, or five negative real roots.

3. Confirm by sketching the graph

The real roots are the points at which the graph crosses or touches the x-axis.

The graph (see below) crosses the x-axis at only one point.

Thus. we have

• one real root (negative)
• four complex roots