Factor each polynomial.
1.q^6 - 169k⁴
2.128h^12 - 1250j^8
3.216 - m^12
4.24a^9 - 81b³
1.q^6 - 169k⁴ --- difference of squares
= (q^3 - 13k^2)(q^3 + 13k^2)
2.128h^12 - 1250j^8 --- common factor of 2, then difference of squares
= 2(64h^12 - 625j^8)
3.216 - m^12 ------ difference of cubes
= (6^3 - (m^4)^3 )
= (6 - m^4)(36 + 6m^4 + m^8)
4.24a^9 - 81b³
= 3(8a^9 - 27b^3) ---- I now see a difference of cubes
= ......
ware is the Sulotion
To factor each polynomial, let's start with the first one:
1. q^6 - 169k⁴
We can recognize that this is a difference of squares because q^6 = (q³)² and 169k⁴ = (13k²)². So the expression can be rewritten as:
(q³)² - (13k²)²
Using the formula for difference of squares: a² - b² = (a + b)(a - b), we have:
(q³ + 13k²)(q³ - 13k²)
Therefore, the factored form of the polynomial q^6 - 169k⁴ is (q³ + 13k²)(q³ - 13k²).
Now, let's move on to the second polynomial:
2. 128h^12 - 1250j^8
This polynomial does not appear to have any common factors. However, we can recognize that both terms are perfect powers. We can factor out the greatest common factor:
2(h^12) - 2(625)(j^8)
The greatest common factor is 2, so we can factor that out:
2(h^12 - 625j^8)
Now, we recognize that h^12 - 625j^8 is a difference of squares. This can be written as:
(h^6)^2 - (25j^4)^2
Using the formula for difference of squares: a² - b² = (a + b)(a - b), we have:
(h^6 + 25j^4)(h^6 - 25j^4)
Therefore, the factored form of the polynomial 128h^12 - 1250j^8 is 2(h^6 + 25j^4)(h^6 - 25j^4).
Moving on to the third polynomial:
3. 216 - m^12
This is a difference of cubes because 216 = 6³ and m^12 = (m^4)³. So the expression can be rewritten as:
(6)³ - (m^4)³
Using the formula for difference of cubes: a³ - b³ = (a - b)(a² + ab + b²), we have:
(6 - m^4)(36 + 6m^4 + m^8)
Therefore, the factored form of the polynomial 216 - m^12 is (6 - m^4)(36 + 6m^4 + m^8).
Finally, let's factor the fourth polynomial:
4. 24a^9 - 81b³
This polynomial does not have any common factors. However, we can recognize that both terms are perfect powers. We can factor out the greatest common factor:
3(8a^9) - 3(27)(b³)
The greatest common factor is 3, so we can factor that out:
3(8a^9 - 27b³)
Now, we recognize that 8a^9 - 27b³ is a difference of cubes. This can be written as:
(2a^3)^3 - (3b)^3
Using the formula for difference of cubes: a³ - b³ = (a - b)(a² + ab + b²), we have:
(2a^3 - 3b)(4a^6 + 6a^3b + 9b²)
Therefore, the factored form of the polynomial 24a^9 - 81b³ is 3(2a^3 - 3b)(4a^6 + 6a^3b + 9b²).
To factor each polynomial, you can use various methods such as factoring the difference of squares, factoring by grouping, or recognizing a common factor. Let's go through each polynomial step by step:
1. q^6 - 169k⁴
This expression represents the difference of squares. We can rewrite 169k⁴ as (13k²)². So, now we have: q^6 - (13k²)². This can be factored as a difference of squares: (q^3 - 13k²)(q^3 + 13k²).
2. 128h^12 - 1250j^8
To factor this polynomial, first, let's consider if there are any common factors. Here, we notice that both terms are divisible by 2. So, we can factor out the greatest common factor, which is 2:
2(64h^12 - 625j^8).
Now let's focus on the expression within the parentheses. We can recognize that it is a difference of squares:
2[(8h^6)^2 - (25j^4)^2].
Now we can apply the difference of squares formula: a^2 - b^2 = (a + b)(a - b)
2[(8h^6 + 25j^4)(8h^6 - 25j^4)].
3. 216 - m^12
This polynomial represents the difference of perfect cubes. We know that 6³ = 216, so we can factor it accordingly:
(6 - m^4)(36 + 6m^4 + m^8).
4. 24a^9 - 81b³
Both terms in this polynomial are divisible by 3. So, we can factor out the greatest common factor, which is 3:
3(8a^9 - 27b³).
Now let's consider the expression within the parentheses. We can recognize that it represents the difference of cubes:
3[(2a^3)^3 - (3b)^3].
Applying the difference of cubes formula: a^3 - b^3 = (a - b)(a^2 + ab + b^2)
3[(2a^3 - 3b)(4a^6 + 6a^3b + 9b^2)].
And there you have it – the factored forms of each polynomial!