Answer :
Sure! Let's factor each of the given polynomials completely.
1. Factor the polynomial: [tex]\(12x^3 + 96x^2 + 192x\)[/tex]
- First, factor out the greatest common factor (GCF), which is [tex]\(12x\)[/tex].
- [tex]\(12x^3 + 96x^2 + 192x = 12x(x^2 + 8x + 16)\)[/tex].
- Now, factor the quadratic [tex]\(x^2 + 8x + 16\)[/tex].
- This can be written as [tex]\( (x + 4)(x + 4) \)[/tex] or [tex]\((x + 4)^2\)[/tex].
- So, the complete factorization is: [tex]\(12x(x + 4)^2\)[/tex].
2. Factor the polynomial: [tex]\(108x^6 + 32x^3\)[/tex]
- First, factor out the GCF, which is [tex]\(4x^3\)[/tex].
- [tex]\(108x^6 + 32x^3 = 4x^3(27x^3 + 8)\)[/tex].
- Recognize [tex]\(27x^3 + 8\)[/tex] as a sum of cubes. It can be factored as [tex]\((3x + 2)(9x^2 - 6x + 4)\)[/tex].
- Thus, the complete factorization is: [tex]\(4x^3(3x + 2)(9x^2 - 6x + 4)\)[/tex].
3. Factor the polynomial: [tex]\(x^3 + x^2 - 16x - 16\)[/tex]
- Group terms to factor by grouping: [tex]\((x^3 + x^2) + (-16x - 16)\)[/tex].
- Factor out [tex]\(x^2\)[/tex] from the first group and [tex]\(-16\)[/tex] from the second group: [tex]\(x^2(x + 1) - 16(x + 1)\)[/tex].
- Notice the common factor [tex]\((x + 1)\)[/tex]: [tex]\((x + 1)(x^2 - 16)\)[/tex].
- Recognize [tex]\(x^2 - 16\)[/tex] as a difference of squares: [tex]\((x - 4)(x + 4)\)[/tex].
- Therefore, the complete factorization is: [tex]\((x - 4)(x + 1)(x + 4)\)[/tex].
4. Factor the polynomial: [tex]\(7x^{12} + 49x^9 + 70x^6\)[/tex]
- Factor out the GCF, which is [tex]\(7x^6\)[/tex].
- [tex]\(7x^{12} + 49x^9 + 70x^6 = 7x^6(x^6 + 7x^3 + 10)\)[/tex].
- The expression inside the parentheses, [tex]\(x^6 + 7x^3 + 10\)[/tex], cannot be factored further using simple techniques over the integers.
- Thus, the complete factorization is: [tex]\(7x^6(x^6 + 7x^3 + 10)\)[/tex].
I hope this explanation helps you understand how to factor these polynomials completely! If you have any questions, feel free to ask.
1. Factor the polynomial: [tex]\(12x^3 + 96x^2 + 192x\)[/tex]
- First, factor out the greatest common factor (GCF), which is [tex]\(12x\)[/tex].
- [tex]\(12x^3 + 96x^2 + 192x = 12x(x^2 + 8x + 16)\)[/tex].
- Now, factor the quadratic [tex]\(x^2 + 8x + 16\)[/tex].
- This can be written as [tex]\( (x + 4)(x + 4) \)[/tex] or [tex]\((x + 4)^2\)[/tex].
- So, the complete factorization is: [tex]\(12x(x + 4)^2\)[/tex].
2. Factor the polynomial: [tex]\(108x^6 + 32x^3\)[/tex]
- First, factor out the GCF, which is [tex]\(4x^3\)[/tex].
- [tex]\(108x^6 + 32x^3 = 4x^3(27x^3 + 8)\)[/tex].
- Recognize [tex]\(27x^3 + 8\)[/tex] as a sum of cubes. It can be factored as [tex]\((3x + 2)(9x^2 - 6x + 4)\)[/tex].
- Thus, the complete factorization is: [tex]\(4x^3(3x + 2)(9x^2 - 6x + 4)\)[/tex].
3. Factor the polynomial: [tex]\(x^3 + x^2 - 16x - 16\)[/tex]
- Group terms to factor by grouping: [tex]\((x^3 + x^2) + (-16x - 16)\)[/tex].
- Factor out [tex]\(x^2\)[/tex] from the first group and [tex]\(-16\)[/tex] from the second group: [tex]\(x^2(x + 1) - 16(x + 1)\)[/tex].
- Notice the common factor [tex]\((x + 1)\)[/tex]: [tex]\((x + 1)(x^2 - 16)\)[/tex].
- Recognize [tex]\(x^2 - 16\)[/tex] as a difference of squares: [tex]\((x - 4)(x + 4)\)[/tex].
- Therefore, the complete factorization is: [tex]\((x - 4)(x + 1)(x + 4)\)[/tex].
4. Factor the polynomial: [tex]\(7x^{12} + 49x^9 + 70x^6\)[/tex]
- Factor out the GCF, which is [tex]\(7x^6\)[/tex].
- [tex]\(7x^{12} + 49x^9 + 70x^6 = 7x^6(x^6 + 7x^3 + 10)\)[/tex].
- The expression inside the parentheses, [tex]\(x^6 + 7x^3 + 10\)[/tex], cannot be factored further using simple techniques over the integers.
- Thus, the complete factorization is: [tex]\(7x^6(x^6 + 7x^3 + 10)\)[/tex].
I hope this explanation helps you understand how to factor these polynomials completely! If you have any questions, feel free to ask.
