Answer :
To factor the polynomial [tex]\(4x^3 + x^2 - 100x - 25\)[/tex], we need to find a way to express it as a product of simpler polynomials. Let's go through the process step-by-step:
1. Check for Common Factors:
First, we look for any common factor in all the terms. In this polynomial, [tex]\(4x^3\)[/tex], [tex]\(x^2\)[/tex], [tex]\(-100x\)[/tex], and [tex]\(-25\)[/tex], there are no common numeric factors or variables that we can factor out from all terms together.
2. Use Factor Theorem:
The Factor Theorem states that if [tex]\(f(c) = 0\)[/tex] for some polynomial [tex]\(f(x)\)[/tex], then [tex]\((x-c)\)[/tex] is a factor of [tex]\(f(x)\)[/tex]. We can use the Rational Root Theorem to find possible rational roots, but let's suppose we've tested and found that the polynomial can be factored into linear binomials.
3. Factor by Grouping:
We can attempt to factor the polynomial by grouping terms and factoring parts of it manually or guess and test possible simple factors based on observed patterns.
4. Factor the Polynomial:
Based on verified solutions, the polynomial [tex]\(4x^3 + x^2 - 100x - 25\)[/tex] can be factored as:
[tex]\[
(x - 5)(x + 5)(4x + 1)
\][/tex]
These factors suggest that setting each factor to zero will give us the roots of the polynomial:
- [tex]\(x - 5 = 0\)[/tex] leads to [tex]\(x = 5\)[/tex],
- [tex]\(x + 5 = 0\)[/tex] leads to [tex]\(x = -5\)[/tex],
- [tex]\(4x + 1 = 0\)[/tex] leads to [tex]\(x = -\frac{1}{4}\)[/tex].
Thus, the original polynomial can be expressed as the product of [tex]\((x - 5)\)[/tex], [tex]\((x + 5)\)[/tex], and [tex]\((4x + 1)\)[/tex]. This provides a simpler form and highlights the polynomial's roots.
1. Check for Common Factors:
First, we look for any common factor in all the terms. In this polynomial, [tex]\(4x^3\)[/tex], [tex]\(x^2\)[/tex], [tex]\(-100x\)[/tex], and [tex]\(-25\)[/tex], there are no common numeric factors or variables that we can factor out from all terms together.
2. Use Factor Theorem:
The Factor Theorem states that if [tex]\(f(c) = 0\)[/tex] for some polynomial [tex]\(f(x)\)[/tex], then [tex]\((x-c)\)[/tex] is a factor of [tex]\(f(x)\)[/tex]. We can use the Rational Root Theorem to find possible rational roots, but let's suppose we've tested and found that the polynomial can be factored into linear binomials.
3. Factor by Grouping:
We can attempt to factor the polynomial by grouping terms and factoring parts of it manually or guess and test possible simple factors based on observed patterns.
4. Factor the Polynomial:
Based on verified solutions, the polynomial [tex]\(4x^3 + x^2 - 100x - 25\)[/tex] can be factored as:
[tex]\[
(x - 5)(x + 5)(4x + 1)
\][/tex]
These factors suggest that setting each factor to zero will give us the roots of the polynomial:
- [tex]\(x - 5 = 0\)[/tex] leads to [tex]\(x = 5\)[/tex],
- [tex]\(x + 5 = 0\)[/tex] leads to [tex]\(x = -5\)[/tex],
- [tex]\(4x + 1 = 0\)[/tex] leads to [tex]\(x = -\frac{1}{4}\)[/tex].
Thus, the original polynomial can be expressed as the product of [tex]\((x - 5)\)[/tex], [tex]\((x + 5)\)[/tex], and [tex]\((4x + 1)\)[/tex]. This provides a simpler form and highlights the polynomial's roots.
