Answer :
Sure! Let's factor each of these expressions step by step.
1) Factor [tex]\(28x^3 + 49x^2 - 16x - 28\)[/tex]:
First, we can group and factor by grouping:
[tex]\[
(28x^3 + 49x^2) + (-16x - 28)
\][/tex]
Factor out the common factor from each group:
[tex]\[
7x^2(4x + 7) - 4(4x + 7)
\][/tex]
Now, notice that [tex]\(4x + 7\)[/tex] is a common factor:
[tex]\[
(7x^2 - 4)(4x + 7)
\][/tex]
This is the completely factored form.
2) Factor [tex]\(16g^4 - 625\)[/tex]:
This is a difference of squares:
[tex]\[
(4g^2)^2 - 25^2
\][/tex]
Apply the difference of squares formula: [tex]\(a^2 - b^2 = (a - b)(a + b)\)[/tex]:
[tex]\[
(4g^2 - 25)(4g^2 + 25)
\][/tex]
The first factor, [tex]\(4g^2 - 25\)[/tex], is also a difference of squares:
[tex]\[
(2g - 5)(2g + 5)
\][/tex]
Thus, the complete factorization is:
[tex]\[
(2g - 5)(2g + 5)(4g^2 + 25)
\][/tex]
3) Factor [tex]\(64x^3 + 27\)[/tex]:
This is a sum of cubes:
[tex]\[
(4x)^3 + 3^3
\][/tex]
Apply the sum of cubes formula: [tex]\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)[/tex]:
[tex]\[
(4x + 3)((4x)^2 - (4x)(3) + 3^2)
\][/tex]
Simplify:
[tex]\[
(4x + 3)(16x^2 - 12x + 9)
\][/tex]
4) Factor [tex]\(2x^4 - 12x^3 + 18x^2\)[/tex]:
First, factor out the greatest common factor:
[tex]\[
2x^2(x^2 - 6x + 9)
\][/tex]
Next, notice that [tex]\(x^2 - 6x + 9\)[/tex] is a perfect square trinomial:
[tex]\[
(x - 3)^2
\][/tex]
Therefore, the complete factorization is:
[tex]\[
2x^2(x - 3)^2
\][/tex]
That completes the factorization of all given expressions! Let me know if you need further clarification or help.
1) Factor [tex]\(28x^3 + 49x^2 - 16x - 28\)[/tex]:
First, we can group and factor by grouping:
[tex]\[
(28x^3 + 49x^2) + (-16x - 28)
\][/tex]
Factor out the common factor from each group:
[tex]\[
7x^2(4x + 7) - 4(4x + 7)
\][/tex]
Now, notice that [tex]\(4x + 7\)[/tex] is a common factor:
[tex]\[
(7x^2 - 4)(4x + 7)
\][/tex]
This is the completely factored form.
2) Factor [tex]\(16g^4 - 625\)[/tex]:
This is a difference of squares:
[tex]\[
(4g^2)^2 - 25^2
\][/tex]
Apply the difference of squares formula: [tex]\(a^2 - b^2 = (a - b)(a + b)\)[/tex]:
[tex]\[
(4g^2 - 25)(4g^2 + 25)
\][/tex]
The first factor, [tex]\(4g^2 - 25\)[/tex], is also a difference of squares:
[tex]\[
(2g - 5)(2g + 5)
\][/tex]
Thus, the complete factorization is:
[tex]\[
(2g - 5)(2g + 5)(4g^2 + 25)
\][/tex]
3) Factor [tex]\(64x^3 + 27\)[/tex]:
This is a sum of cubes:
[tex]\[
(4x)^3 + 3^3
\][/tex]
Apply the sum of cubes formula: [tex]\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)[/tex]:
[tex]\[
(4x + 3)((4x)^2 - (4x)(3) + 3^2)
\][/tex]
Simplify:
[tex]\[
(4x + 3)(16x^2 - 12x + 9)
\][/tex]
4) Factor [tex]\(2x^4 - 12x^3 + 18x^2\)[/tex]:
First, factor out the greatest common factor:
[tex]\[
2x^2(x^2 - 6x + 9)
\][/tex]
Next, notice that [tex]\(x^2 - 6x + 9\)[/tex] is a perfect square trinomial:
[tex]\[
(x - 3)^2
\][/tex]
Therefore, the complete factorization is:
[tex]\[
2x^2(x - 3)^2
\][/tex]
That completes the factorization of all given expressions! Let me know if you need further clarification or help.
