Answer :
Let's factor the given polynomials step-by-step:
1. Factor [tex]\(12x^3 + 96x^2 + 192x\)[/tex]:
- First, identify the greatest common factor (GCF) in all the terms. The GCF is [tex]\(12x\)[/tex].
- Factoring out [tex]\(12x\)[/tex] gives us:
[tex]\[
12x(x^2 + 8x + 16)
\][/tex]
- The quadratic [tex]\(x^2 + 8x + 16\)[/tex] is a perfect square, so it can be factored further:
[tex]\[
(x + 4)^2
\][/tex]
- So, the fully factored form is:
[tex]\[
12x(x + 4)^2
\][/tex]
2. Factor [tex]\(108x^6 + 32x^3\)[/tex]:
- First, find the GCF, which is [tex]\(4x^3\)[/tex].
- Factoring out [tex]\(4x^3\)[/tex] gives:
[tex]\[
4x^3(27x^3 + 8)
\][/tex]
- Recognizing the remaining expression as a sum of cubes [tex]\(a^3 + b^3\)[/tex] where [tex]\(a = 3x\)[/tex] and [tex]\(b = 2\)[/tex], we use the formula: [tex]\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)[/tex].
- Applying the formula results in:
[tex]\[
(3x + 2)(9x^2 - 6x + 4)
\][/tex]
- Thus, the fully factored form is:
[tex]\[
4x^3(3x + 2)(9x^2 - 6x + 4)
\][/tex]
3. Factor [tex]\(x^3 + x^2 - 16x - 16\)[/tex]:
- Group terms to help in factoring:
[tex]\[
(x^3 + x^2) + (-16x - 16)
\][/tex]
- Factor by grouping:
[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 factored form is:
[tex]\[
(x - 4)(x + 1)(x + 4)
\][/tex]
4. Factor [tex]\(7x^{12} + 49x^9 + 70x^6\)[/tex]:
- The GCF is [tex]\(7x^6\)[/tex].
- Factoring out [tex]\(7x^6\)[/tex] gives:
[tex]\[
7x^6(x^6 + 7x^3 + 10)
\][/tex]
- Notice that the expression inside the parentheses can't be factored easily any further over the integers.
- Thus, the final factored expression is:
[tex]\[
7x^6(x^3 + 2)(x^3 + 5)
\][/tex]
These are the full factored forms of each polynomial.
1. Factor [tex]\(12x^3 + 96x^2 + 192x\)[/tex]:
- First, identify the greatest common factor (GCF) in all the terms. The GCF is [tex]\(12x\)[/tex].
- Factoring out [tex]\(12x\)[/tex] gives us:
[tex]\[
12x(x^2 + 8x + 16)
\][/tex]
- The quadratic [tex]\(x^2 + 8x + 16\)[/tex] is a perfect square, so it can be factored further:
[tex]\[
(x + 4)^2
\][/tex]
- So, the fully factored form is:
[tex]\[
12x(x + 4)^2
\][/tex]
2. Factor [tex]\(108x^6 + 32x^3\)[/tex]:
- First, find the GCF, which is [tex]\(4x^3\)[/tex].
- Factoring out [tex]\(4x^3\)[/tex] gives:
[tex]\[
4x^3(27x^3 + 8)
\][/tex]
- Recognizing the remaining expression as a sum of cubes [tex]\(a^3 + b^3\)[/tex] where [tex]\(a = 3x\)[/tex] and [tex]\(b = 2\)[/tex], we use the formula: [tex]\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)[/tex].
- Applying the formula results in:
[tex]\[
(3x + 2)(9x^2 - 6x + 4)
\][/tex]
- Thus, the fully factored form is:
[tex]\[
4x^3(3x + 2)(9x^2 - 6x + 4)
\][/tex]
3. Factor [tex]\(x^3 + x^2 - 16x - 16\)[/tex]:
- Group terms to help in factoring:
[tex]\[
(x^3 + x^2) + (-16x - 16)
\][/tex]
- Factor by grouping:
[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 factored form is:
[tex]\[
(x - 4)(x + 1)(x + 4)
\][/tex]
4. Factor [tex]\(7x^{12} + 49x^9 + 70x^6\)[/tex]:
- The GCF is [tex]\(7x^6\)[/tex].
- Factoring out [tex]\(7x^6\)[/tex] gives:
[tex]\[
7x^6(x^6 + 7x^3 + 10)
\][/tex]
- Notice that the expression inside the parentheses can't be factored easily any further over the integers.
- Thus, the final factored expression is:
[tex]\[
7x^6(x^3 + 2)(x^3 + 5)
\][/tex]
These are the full factored forms of each polynomial.
