Answer :
Alright class, let's tackle the task step by step. We are asked to factorize the given polynomials. Let's break them down one by one:
### Problem 1: Factor [tex]\( c^2 - 25 \)[/tex]
This is a classic difference of squares problem. The general form for the difference of squares is:
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]
In this case, we have:
[tex]\[ c^2 - 25 = c^2 - 5^2 \][/tex]
Using the difference of squares formula:
[tex]\[ c^2 - 25 = (c - 5)(c + 5) \][/tex]
So, the factorization of [tex]\( c^2 - 25 \)[/tex] is [tex]\( (c - 5)(c + 5) \)[/tex].
### Problem 2: Factor [tex]\( w^4 - 1 \)[/tex]
Again, this is a difference of squares. Recognize that [tex]\( 1 \)[/tex] is [tex]\( 1^2 \)[/tex], so we can write:
[tex]\[ w^4 - 1 = w^4 - 1^2 \][/tex]
Using the difference of squares formula:
[tex]\[ w^4 - 1 = (w^2 - 1)(w^2 + 1) \][/tex]
Notice that [tex]\( w^2 - 1 \)[/tex] is also a difference of squares:
[tex]\[ w^2 - 1 = (w - 1)(w + 1) \][/tex]
So, combining these, we get the full factorization:
[tex]\[ w^4 - 1 = (w - 1)(w + 1)(w^2 + 1) \][/tex]
### Problem 3: Factor [tex]\( 36 - h^6 \)[/tex]
We can again use the difference of squares. Notice [tex]\( 36 \)[/tex] is [tex]\( 6^2 \)[/tex] and [tex]\( h^6 \)[/tex] is [tex]\( (h^3)^2 \)[/tex], so:
[tex]\[ 36 - h^6 = 6^2 - (h^3)^2 \][/tex]
Applying the difference of squares formula:
[tex]\[ 36 - h^6 = (6 - h^3)(6 + h^3) \][/tex]
So, the factorization of [tex]\( 36 - h^6 \)[/tex] is [tex]\( (6 - h^3)(6 + h^3) \)[/tex].
### Problem 4: Factor [tex]\( 1 - r^{10} \)[/tex]
Finally, we can treat this as another difference of squares problem. Recognize [tex]\( 1 \)[/tex] is [tex]\( 1^2 \)[/tex] and [tex]\( r^{10} \)[/tex] is [tex]\( (r^5)^2 \)[/tex], so:
[tex]\[ 1 - r^{10} = 1^2 - (r^5)^2 \][/tex]
Applying the difference of squares formula:
[tex]\[ 1 - r^{10} = (1 - r^5)(1 + r^5) \][/tex]
Next, notice that [tex]\( 1 - r^5 \)[/tex] can be further factorized as a difference of squares:
[tex]\[ 1 - r^5 = (r - 1)(1 + r + r^2 + r^3 + r^4) \][/tex]
[tex]\[ 1 + r^5 = (r + 1)(1 - r + r^2 - r^3 + r^4) \][/tex]
Combining these results gives:
[tex]\[ 1 - r^{10} = (r - 1)(r + 1)(r^4 - r^3 + r^2 - r + 1)(r^4 + r^3 + r^2 + r + 1) \][/tex]
### Summary of Factors:
1. [tex]\( c^2 - 25 \)[/tex] factors to [tex]\( (c - 5)(c + 5) \)[/tex]
2. [tex]\( w^4 - 1 \)[/tex] factors to [tex]\( (w - 1)(w + 1)(w^2 + 1) \)[/tex]
3. [tex]\( 36 - h^6 \)[/tex] factors to [tex]\( (6 - h^3)(6 + h^3) \)[/tex]
4. [tex]\( 1 - r^{10} \)[/tex] factors to [tex]\( (r - 1)(r + 1)(r^4 - r^3 + r^2 - r + 1)(r^4 + r^3 + r^2 + r + 1) \)[/tex]
Putting it all together, the factors for each polynomial are:
[tex]\[ \boxed{(c - 5)(c + 5), (w - 1)(w + 1)(w^2 + 1), (6 - h^3)(6 + h^3), (r - 1)(r + 1)(r^4 - r^3 + r^2 - r + 1)(r^4 + r^3 + r^2 + r + 1)} \][/tex]
### Problem 1: Factor [tex]\( c^2 - 25 \)[/tex]
This is a classic difference of squares problem. The general form for the difference of squares is:
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]
In this case, we have:
[tex]\[ c^2 - 25 = c^2 - 5^2 \][/tex]
Using the difference of squares formula:
[tex]\[ c^2 - 25 = (c - 5)(c + 5) \][/tex]
So, the factorization of [tex]\( c^2 - 25 \)[/tex] is [tex]\( (c - 5)(c + 5) \)[/tex].
### Problem 2: Factor [tex]\( w^4 - 1 \)[/tex]
Again, this is a difference of squares. Recognize that [tex]\( 1 \)[/tex] is [tex]\( 1^2 \)[/tex], so we can write:
[tex]\[ w^4 - 1 = w^4 - 1^2 \][/tex]
Using the difference of squares formula:
[tex]\[ w^4 - 1 = (w^2 - 1)(w^2 + 1) \][/tex]
Notice that [tex]\( w^2 - 1 \)[/tex] is also a difference of squares:
[tex]\[ w^2 - 1 = (w - 1)(w + 1) \][/tex]
So, combining these, we get the full factorization:
[tex]\[ w^4 - 1 = (w - 1)(w + 1)(w^2 + 1) \][/tex]
### Problem 3: Factor [tex]\( 36 - h^6 \)[/tex]
We can again use the difference of squares. Notice [tex]\( 36 \)[/tex] is [tex]\( 6^2 \)[/tex] and [tex]\( h^6 \)[/tex] is [tex]\( (h^3)^2 \)[/tex], so:
[tex]\[ 36 - h^6 = 6^2 - (h^3)^2 \][/tex]
Applying the difference of squares formula:
[tex]\[ 36 - h^6 = (6 - h^3)(6 + h^3) \][/tex]
So, the factorization of [tex]\( 36 - h^6 \)[/tex] is [tex]\( (6 - h^3)(6 + h^3) \)[/tex].
### Problem 4: Factor [tex]\( 1 - r^{10} \)[/tex]
Finally, we can treat this as another difference of squares problem. Recognize [tex]\( 1 \)[/tex] is [tex]\( 1^2 \)[/tex] and [tex]\( r^{10} \)[/tex] is [tex]\( (r^5)^2 \)[/tex], so:
[tex]\[ 1 - r^{10} = 1^2 - (r^5)^2 \][/tex]
Applying the difference of squares formula:
[tex]\[ 1 - r^{10} = (1 - r^5)(1 + r^5) \][/tex]
Next, notice that [tex]\( 1 - r^5 \)[/tex] can be further factorized as a difference of squares:
[tex]\[ 1 - r^5 = (r - 1)(1 + r + r^2 + r^3 + r^4) \][/tex]
[tex]\[ 1 + r^5 = (r + 1)(1 - r + r^2 - r^3 + r^4) \][/tex]
Combining these results gives:
[tex]\[ 1 - r^{10} = (r - 1)(r + 1)(r^4 - r^3 + r^2 - r + 1)(r^4 + r^3 + r^2 + r + 1) \][/tex]
### Summary of Factors:
1. [tex]\( c^2 - 25 \)[/tex] factors to [tex]\( (c - 5)(c + 5) \)[/tex]
2. [tex]\( w^4 - 1 \)[/tex] factors to [tex]\( (w - 1)(w + 1)(w^2 + 1) \)[/tex]
3. [tex]\( 36 - h^6 \)[/tex] factors to [tex]\( (6 - h^3)(6 + h^3) \)[/tex]
4. [tex]\( 1 - r^{10} \)[/tex] factors to [tex]\( (r - 1)(r + 1)(r^4 - r^3 + r^2 - r + 1)(r^4 + r^3 + r^2 + r + 1) \)[/tex]
Putting it all together, the factors for each polynomial are:
[tex]\[ \boxed{(c - 5)(c + 5), (w - 1)(w + 1)(w^2 + 1), (6 - h^3)(6 + h^3), (r - 1)(r + 1)(r^4 - r^3 + r^2 - r + 1)(r^4 + r^3 + r^2 + r + 1)} \][/tex]
