Answer :
To find the x-intercepts of the function [tex]\( f(x) = x^4 - x^3 + x^2 - x \)[/tex], we need to set the function equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ f(x) = x^4 - x^3 + x^2 - x = 0 \][/tex]
We will solve this polynomial equation step-by-step.
1. Factoring out the common term:
Notice that each term in the polynomial has an [tex]\( x \)[/tex] in it. We can factor [tex]\( x \)[/tex] out:
[tex]\[ x (x^3 - x^2 + x - 1) = 0 \][/tex]
This gives us one solution:
[tex]\[ x = 0 \][/tex]
2. Solving the remaining cubic polynomial:
Now we need to solve the cubic equation inside the parentheses:
[tex]\[ x^3 - x^2 + x - 1 = 0 \][/tex]
To solve this, we can look for roots of the cubic polynomial. Let's rewrite it for clarity:
[tex]\[ x^3 - x^2 + x - 1 = (x - 1)(x^2 + 1) \][/tex]
This factorization helps us identify the roots. Let's break it into parts:
- For the factor [tex]\( x - 1 \)[/tex]:
[tex]\[ x - 1 = 0 \implies x = 1 \][/tex]
- For the factor [tex]\( x^2 + 1 \)[/tex]:
[tex]\[ x^2 + 1 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x^2 = -1 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm i \][/tex]
where [tex]\( i \)[/tex] is the imaginary unit ([tex]\( i = \sqrt{-1} \)[/tex]).
3. Listing all solutions:
Now, we have found all the solutions to the original equation [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ x = 0, \quad x = 1, \quad x = -i, \quad x = i \][/tex]
4. Counting the x-intercepts:
Therefore, we have four solutions. The x-intercepts are the real solutions where the polynomial touches or crosses the x-axis. In this case, the real x-intercepts are:
[tex]\[ x = 0 \quad \text{and} \quad x = 1 \][/tex]
And the complex solutions (which are not x-intercepts on the real number line) are:
[tex]\[ x = -i \quad \text{and} \quad x = i \][/tex]
Hence, the function [tex]\( f(x) = x^4 - x^3 + x^2 - x \)[/tex] has:
- 2 real x-intercepts ([tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex])
- Total of 4 solutions in the complex plane ([tex]\( x = 0, x = 1, x = -i, x = i \)[/tex])
Therefore, the answer to the given question is:
[tex]\( 4 \ x\text{-intercepts} \)[/tex]
[tex]\[ f(x) = x^4 - x^3 + x^2 - x = 0 \][/tex]
We will solve this polynomial equation step-by-step.
1. Factoring out the common term:
Notice that each term in the polynomial has an [tex]\( x \)[/tex] in it. We can factor [tex]\( x \)[/tex] out:
[tex]\[ x (x^3 - x^2 + x - 1) = 0 \][/tex]
This gives us one solution:
[tex]\[ x = 0 \][/tex]
2. Solving the remaining cubic polynomial:
Now we need to solve the cubic equation inside the parentheses:
[tex]\[ x^3 - x^2 + x - 1 = 0 \][/tex]
To solve this, we can look for roots of the cubic polynomial. Let's rewrite it for clarity:
[tex]\[ x^3 - x^2 + x - 1 = (x - 1)(x^2 + 1) \][/tex]
This factorization helps us identify the roots. Let's break it into parts:
- For the factor [tex]\( x - 1 \)[/tex]:
[tex]\[ x - 1 = 0 \implies x = 1 \][/tex]
- For the factor [tex]\( x^2 + 1 \)[/tex]:
[tex]\[ x^2 + 1 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x^2 = -1 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm i \][/tex]
where [tex]\( i \)[/tex] is the imaginary unit ([tex]\( i = \sqrt{-1} \)[/tex]).
3. Listing all solutions:
Now, we have found all the solutions to the original equation [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ x = 0, \quad x = 1, \quad x = -i, \quad x = i \][/tex]
4. Counting the x-intercepts:
Therefore, we have four solutions. The x-intercepts are the real solutions where the polynomial touches or crosses the x-axis. In this case, the real x-intercepts are:
[tex]\[ x = 0 \quad \text{and} \quad x = 1 \][/tex]
And the complex solutions (which are not x-intercepts on the real number line) are:
[tex]\[ x = -i \quad \text{and} \quad x = i \][/tex]
Hence, the function [tex]\( f(x) = x^4 - x^3 + x^2 - x \)[/tex] has:
- 2 real x-intercepts ([tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex])
- Total of 4 solutions in the complex plane ([tex]\( x = 0, x = 1, x = -i, x = i \)[/tex])
Therefore, the answer to the given question is:
[tex]\( 4 \ x\text{-intercepts} \)[/tex]
