To find the x-intercepts of the polynomial function f ( x ) = x 4 − x 3 + x 2 − x , we:
Factor the polynomial as f ( x ) = x ( x − 1 ) ( x 2 + 1 ) .
Find the roots by setting each factor to zero, resulting in x = 0 , 1 , i , − i .
Identify the real roots, which are x = 0 and x = 1 .
Conclude that there are 2 x-intercepts.
Explanation
Understanding the Problem We are given the polynomial function f ( x ) = x 4 − x 3 + x 2 − x and asked to find the number of x-intercepts. X-intercepts occur where the graph of the function intersects the x-axis, which means we need to find the values of x for which f ( x ) = 0 .
Setting up the Equation To find the x-intercepts, we set f ( x ) = 0 and solve for x : x 4 − x 3 + x 2 − x = 0
Factoring out x We can factor out an x from each term: x ( x 3 − x 2 + x − 1 ) = 0
Factoring by Grouping Now, let's factor the cubic expression x 3 − x 2 + x − 1 . We can use factoring by grouping: x 2 ( x − 1 ) + 1 ( x − 1 ) = ( x 2 + 1 ) ( x − 1 )
Complete Factorization So, the factored form of the polynomial is: f ( x ) = x ( x − 1 ) ( x 2 + 1 )
Finding the Roots Now we set each factor equal to zero to find the roots:
x = 0
x − 1 = 0 ⇒ x = 1
x 2 + 1 = 0 ⇒ x 2 = − 1 ⇒ x = ± i
The roots are x = 0 , x = 1 , x = i , and x = − i .
Identifying Real Roots Since we are looking for x-intercepts, we only consider the real roots. The real roots are x = 0 and x = 1 . The roots x = i and x = − i are imaginary, so they do not correspond to x-intercepts.
Conclusion Therefore, there are 2 x-intercepts on the graph of the polynomial function f ( x ) = x 4 − x 3 + x 2 − x .
Examples
Understanding x-intercepts is crucial in many real-world applications. For example, in physics, the x-intercepts of a projectile's trajectory can represent the points where the projectile lands. In economics, the x-intercepts of a cost function can represent the break-even points where the cost equals zero. By finding and interpreting x-intercepts, we can gain valuable insights into the behavior of functions and their practical implications.
A current of 15.0 A flowing for 30 seconds corresponds to a total charge of 450 C . This charge equals approximately 2.81 × 1 0 21 electrons. Thus, about 2.81 × 1 0 21 electrons flow through the device during the specified time.
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