Answer :
- Factor the polynomial by grouping: $(4x^3 - x^2) + (-100x + 25)$.
- Factor out common factors: $x^2(4x - 1) - 25(4x - 1)$.
- Factor out the common binomial: $(4x - 1)(x^2 - 25)$.
- Factor the difference of squares: $(4x - 1)(x - 5)(x + 5)$.
- Solve for the roots: $\boxed{x = -5, \frac{1}{4}, 5}$.
### Explanation
1. Problem Analysis
We are given the polynomial $4x^3 - x^2 - 100x + 25$ and we want to find its roots.
2. Grouping Terms
We can factor this polynomial by grouping. First, group the terms as $(4x^3 - x^2) + (-100x + 25)$.
3. Factoring Common Factors
Next, factor out the common factors from each group: $x^2(4x - 1) - 25(4x - 1)$.
4. Factoring Common Binomial
Now, factor out the common binomial factor $(4x - 1)$: $(4x - 1)(x^2 - 25)$.
5. Factoring Difference of Squares
We can further factor the difference of squares: $(4x - 1)(x - 5)(x + 5)$.
6. Setting Factors to Zero
To find the roots, we set each factor to zero and solve for $x$: $4x - 1 = 0$, $x - 5 = 0$, and $x + 5 = 0$.
7. Solving for Roots
Solving each equation, we get:
$4x - 1 = 0 \Rightarrow x = \frac{1}{4}$
$x - 5 = 0 \Rightarrow x = 5$
$x + 5 = 0 \Rightarrow x = -5$
Thus, the roots are $x = \frac{1}{4}$, $x = 5$, and $x = -5$.
8. Final Answer
The roots of the polynomial $4x^3 - x^2 - 100x + 25$ are $\boxed{-\frac{1}{4}, 5, -5}$.
### Examples
Polynomials and their roots are fundamental in many areas of science and engineering. For instance, in structural engineering, finding the roots of a polynomial can help determine the stability of a bridge or building. In economics, roots can represent equilibrium points in market models. Understanding how to factor and find roots of polynomials is thus a crucial skill for solving real-world problems.
- Factor out common factors: $x^2(4x - 1) - 25(4x - 1)$.
- Factor out the common binomial: $(4x - 1)(x^2 - 25)$.
- Factor the difference of squares: $(4x - 1)(x - 5)(x + 5)$.
- Solve for the roots: $\boxed{x = -5, \frac{1}{4}, 5}$.
### Explanation
1. Problem Analysis
We are given the polynomial $4x^3 - x^2 - 100x + 25$ and we want to find its roots.
2. Grouping Terms
We can factor this polynomial by grouping. First, group the terms as $(4x^3 - x^2) + (-100x + 25)$.
3. Factoring Common Factors
Next, factor out the common factors from each group: $x^2(4x - 1) - 25(4x - 1)$.
4. Factoring Common Binomial
Now, factor out the common binomial factor $(4x - 1)$: $(4x - 1)(x^2 - 25)$.
5. Factoring Difference of Squares
We can further factor the difference of squares: $(4x - 1)(x - 5)(x + 5)$.
6. Setting Factors to Zero
To find the roots, we set each factor to zero and solve for $x$: $4x - 1 = 0$, $x - 5 = 0$, and $x + 5 = 0$.
7. Solving for Roots
Solving each equation, we get:
$4x - 1 = 0 \Rightarrow x = \frac{1}{4}$
$x - 5 = 0 \Rightarrow x = 5$
$x + 5 = 0 \Rightarrow x = -5$
Thus, the roots are $x = \frac{1}{4}$, $x = 5$, and $x = -5$.
8. Final Answer
The roots of the polynomial $4x^3 - x^2 - 100x + 25$ are $\boxed{-\frac{1}{4}, 5, -5}$.
### Examples
Polynomials and their roots are fundamental in many areas of science and engineering. For instance, in structural engineering, finding the roots of a polynomial can help determine the stability of a bridge or building. In economics, roots can represent equilibrium points in market models. Understanding how to factor and find roots of polynomials is thus a crucial skill for solving real-world problems.
