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Answer :
To factor the polynomial [tex]\(16x^3 + 4x^2 - 100x - 25\)[/tex] completely, we can follow a structured approach:
1. Look for Patterns or Grouping: Start by grouping the terms to make factoring easier. Group the first two terms together and the last two terms together:
[tex]\[
(16x^3 + 4x^2) + (-100x - 25)
\][/tex]
2. Factor Out Common Factors: Within each group, factor out the greatest common factor (GCF).
- For the first group [tex]\(16x^3 + 4x^2\)[/tex], the GCF is [tex]\(4x^2\)[/tex]:
[tex]\[
4x^2(4x + 1)
\][/tex]
- For the second group [tex]\(-100x - 25\)[/tex], the GCF is [tex]\(-25\)[/tex]:
[tex]\[
-25(4x + 1)
\][/tex]
3. Factor by Grouping: Notice that the expression now has the same binomial factor [tex]\((4x + 1)\)[/tex]:
[tex]\[
4x^2(4x + 1) - 25(4x + 1)
\][/tex]
You can factor out [tex]\((4x + 1)\)[/tex]:
[tex]\[
(4x + 1)(4x^2 - 25)
\][/tex]
4. Further Factor: See if the quadratic [tex]\(4x^2 - 25\)[/tex] can be factored further. Recognize it as a difference of squares:
[tex]\[
4x^2 - 25 = (2x)^2 - (5)^2
\][/tex]
Applying the difference of squares formula [tex]\(a^2 - b^2 = (a - b)(a + b)\)[/tex], this factors to:
[tex]\[
(2x - 5)(2x + 5)
\][/tex]
5. Complete Factorization: Combine all the factors:
[tex]\[
(4x + 1)(2x - 5)(2x + 5)
\][/tex]
So, the polynomial [tex]\(16x^3 + 4x^2 - 100x - 25\)[/tex] factors completely to [tex]\((2x - 5)(2x + 5)(4x + 1)\)[/tex].
1. Look for Patterns or Grouping: Start by grouping the terms to make factoring easier. Group the first two terms together and the last two terms together:
[tex]\[
(16x^3 + 4x^2) + (-100x - 25)
\][/tex]
2. Factor Out Common Factors: Within each group, factor out the greatest common factor (GCF).
- For the first group [tex]\(16x^3 + 4x^2\)[/tex], the GCF is [tex]\(4x^2\)[/tex]:
[tex]\[
4x^2(4x + 1)
\][/tex]
- For the second group [tex]\(-100x - 25\)[/tex], the GCF is [tex]\(-25\)[/tex]:
[tex]\[
-25(4x + 1)
\][/tex]
3. Factor by Grouping: Notice that the expression now has the same binomial factor [tex]\((4x + 1)\)[/tex]:
[tex]\[
4x^2(4x + 1) - 25(4x + 1)
\][/tex]
You can factor out [tex]\((4x + 1)\)[/tex]:
[tex]\[
(4x + 1)(4x^2 - 25)
\][/tex]
4. Further Factor: See if the quadratic [tex]\(4x^2 - 25\)[/tex] can be factored further. Recognize it as a difference of squares:
[tex]\[
4x^2 - 25 = (2x)^2 - (5)^2
\][/tex]
Applying the difference of squares formula [tex]\(a^2 - b^2 = (a - b)(a + b)\)[/tex], this factors to:
[tex]\[
(2x - 5)(2x + 5)
\][/tex]
5. Complete Factorization: Combine all the factors:
[tex]\[
(4x + 1)(2x - 5)(2x + 5)
\][/tex]
So, the polynomial [tex]\(16x^3 + 4x^2 - 100x - 25\)[/tex] factors completely to [tex]\((2x - 5)(2x + 5)(4x + 1)\)[/tex].
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