Answer :
Sure, let's factor the difference of two squares for each given expression. The difference of two squares formula is:
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]
We'll apply this formula to each expression:
39. [tex]\(x^2 - 100\)[/tex]:
- Here, [tex]\(a = x\)[/tex] and [tex]\(b = 10\)[/tex] (since [tex]\(10^2 = 100\)[/tex]).
- So, it factors to [tex]\((x - 10)(x + 10)\)[/tex].
40. [tex]\(x^2 - 144\)[/tex]:
- Here, [tex]\(a = x\)[/tex] and [tex]\(b = 12\)[/tex] (since [tex]\(12^2 = 144\)[/tex]).
- So, it factors to [tex]\((x - 12)(x + 12)\)[/tex].
41. [tex]\(36x^2 - 49\)[/tex]:
- Here, [tex]\(a = 6x\)[/tex] (since [tex]\((6x)^2 = 36x^2\)[/tex]) and [tex]\(b = 7\)[/tex] (since [tex]\(7^2 = 49\)[/tex]).
- So, it factors to [tex]\((6x - 7)(6x + 7)\)[/tex].
42. [tex]\(64x^2 - 81\)[/tex]:
- Here, [tex]\(a = 8x\)[/tex] (since [tex]\((8x)^2 = 64x^2\)[/tex]) and [tex]\(b = 9\)[/tex] (since [tex]\(9^2 = 81\)[/tex]).
- So, it factors to [tex]\((8x - 9)(8x + 9)\)[/tex].
43. [tex]\(9x^2 - 25y^2\)[/tex]:
- Here, [tex]\(a = 3x\)[/tex] (since [tex]\((3x)^2 = 9x^2\)[/tex]) and [tex]\(b = 5y\)[/tex] (since [tex]\((5y)^2 = 25y^2\)[/tex]).
- So, it factors to [tex]\((3x - 5y)(3x + 5y)\)[/tex].
44. [tex]\(36x^2 - 49y^2\)[/tex]:
- Here, [tex]\(a = 6x\)[/tex] (since [tex]\((6x)^2 = 36x^2\)[/tex]) and [tex]\(b = 7y\)[/tex] (since [tex]\((7y)^2 = 49y^2\)[/tex]).
- So, it factors to [tex]\((6x - 7y)(6x + 7y)\)[/tex].
45. [tex]\(x^4 - 16\)[/tex]:
- Here, [tex]\(a = x^2\)[/tex] and [tex]\(b = 4\)[/tex] (since [tex]\(4^2 = 16\)[/tex]).
- It first factors to [tex]\((x^2 - 4)(x^2 + 4)\)[/tex].
- Further breaking down [tex]\(x^2 - 4\)[/tex] using the difference of squares formula: [tex]\(x^2 - 4 = (x - 2)(x + 2)\)[/tex].
- So, the complete factorization is [tex]\((x - 2)(x + 2)(x^2 + 4)\)[/tex].
46. [tex]\(x^4 - 1\)[/tex]:
- Here, [tex]\(a = x^2\)[/tex] and [tex]\(b = 1\)[/tex].
- It factors to [tex]\((x^2 - 1)(x^2 + 1)\)[/tex].
- Further breaking down [tex]\(x^2 - 1\)[/tex] using the difference of squares formula: [tex]\(x^2 - 1 = (x - 1)(x + 1)\)[/tex].
- So, the complete factorization is [tex]\((x - 1)(x + 1)(x^2 + 1)\)[/tex].
47. [tex]\(16x^4 - 81\)[/tex]:
- Here, [tex]\(a = 4x^2\)[/tex] and [tex]\(b = 9\)[/tex].
- It factors to [tex]\((4x^2 - 9)(4x^2 + 9)\)[/tex].
- Further breaking down [tex]\(4x^2 - 9\)[/tex] using the difference of squares formula: [tex]\(4x^2 - 9 = (2x - 3)(2x + 3)\)[/tex].
- So, the complete factorization is [tex]\((2x - 3)(2x + 3)(4x^2 + 9)\)[/tex].
48. [tex]\(81x^4 - 1\)[/tex]:
- Here, [tex]\(a = 9x^2\)[/tex] and [tex]\(b = 1\)[/tex].
- It factors to [tex]\((9x^2 - 1)(9x^2 + 1)\)[/tex].
- Further breaking down [tex]\(9x^2 - 1\)[/tex] using the difference of squares formula: [tex]\(9x^2 - 1 = (3x - 1)(3x + 1)\)[/tex].
- So, the complete factorization is [tex]\((3x - 1)(3x + 1)(9x^2 + 1)\)[/tex].
These are the complete factorizations for each of the given expressions.
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]
We'll apply this formula to each expression:
39. [tex]\(x^2 - 100\)[/tex]:
- Here, [tex]\(a = x\)[/tex] and [tex]\(b = 10\)[/tex] (since [tex]\(10^2 = 100\)[/tex]).
- So, it factors to [tex]\((x - 10)(x + 10)\)[/tex].
40. [tex]\(x^2 - 144\)[/tex]:
- Here, [tex]\(a = x\)[/tex] and [tex]\(b = 12\)[/tex] (since [tex]\(12^2 = 144\)[/tex]).
- So, it factors to [tex]\((x - 12)(x + 12)\)[/tex].
41. [tex]\(36x^2 - 49\)[/tex]:
- Here, [tex]\(a = 6x\)[/tex] (since [tex]\((6x)^2 = 36x^2\)[/tex]) and [tex]\(b = 7\)[/tex] (since [tex]\(7^2 = 49\)[/tex]).
- So, it factors to [tex]\((6x - 7)(6x + 7)\)[/tex].
42. [tex]\(64x^2 - 81\)[/tex]:
- Here, [tex]\(a = 8x\)[/tex] (since [tex]\((8x)^2 = 64x^2\)[/tex]) and [tex]\(b = 9\)[/tex] (since [tex]\(9^2 = 81\)[/tex]).
- So, it factors to [tex]\((8x - 9)(8x + 9)\)[/tex].
43. [tex]\(9x^2 - 25y^2\)[/tex]:
- Here, [tex]\(a = 3x\)[/tex] (since [tex]\((3x)^2 = 9x^2\)[/tex]) and [tex]\(b = 5y\)[/tex] (since [tex]\((5y)^2 = 25y^2\)[/tex]).
- So, it factors to [tex]\((3x - 5y)(3x + 5y)\)[/tex].
44. [tex]\(36x^2 - 49y^2\)[/tex]:
- Here, [tex]\(a = 6x\)[/tex] (since [tex]\((6x)^2 = 36x^2\)[/tex]) and [tex]\(b = 7y\)[/tex] (since [tex]\((7y)^2 = 49y^2\)[/tex]).
- So, it factors to [tex]\((6x - 7y)(6x + 7y)\)[/tex].
45. [tex]\(x^4 - 16\)[/tex]:
- Here, [tex]\(a = x^2\)[/tex] and [tex]\(b = 4\)[/tex] (since [tex]\(4^2 = 16\)[/tex]).
- It first factors to [tex]\((x^2 - 4)(x^2 + 4)\)[/tex].
- Further breaking down [tex]\(x^2 - 4\)[/tex] using the difference of squares formula: [tex]\(x^2 - 4 = (x - 2)(x + 2)\)[/tex].
- So, the complete factorization is [tex]\((x - 2)(x + 2)(x^2 + 4)\)[/tex].
46. [tex]\(x^4 - 1\)[/tex]:
- Here, [tex]\(a = x^2\)[/tex] and [tex]\(b = 1\)[/tex].
- It factors to [tex]\((x^2 - 1)(x^2 + 1)\)[/tex].
- Further breaking down [tex]\(x^2 - 1\)[/tex] using the difference of squares formula: [tex]\(x^2 - 1 = (x - 1)(x + 1)\)[/tex].
- So, the complete factorization is [tex]\((x - 1)(x + 1)(x^2 + 1)\)[/tex].
47. [tex]\(16x^4 - 81\)[/tex]:
- Here, [tex]\(a = 4x^2\)[/tex] and [tex]\(b = 9\)[/tex].
- It factors to [tex]\((4x^2 - 9)(4x^2 + 9)\)[/tex].
- Further breaking down [tex]\(4x^2 - 9\)[/tex] using the difference of squares formula: [tex]\(4x^2 - 9 = (2x - 3)(2x + 3)\)[/tex].
- So, the complete factorization is [tex]\((2x - 3)(2x + 3)(4x^2 + 9)\)[/tex].
48. [tex]\(81x^4 - 1\)[/tex]:
- Here, [tex]\(a = 9x^2\)[/tex] and [tex]\(b = 1\)[/tex].
- It factors to [tex]\((9x^2 - 1)(9x^2 + 1)\)[/tex].
- Further breaking down [tex]\(9x^2 - 1\)[/tex] using the difference of squares formula: [tex]\(9x^2 - 1 = (3x - 1)(3x + 1)\)[/tex].
- So, the complete factorization is [tex]\((3x - 1)(3x + 1)(9x^2 + 1)\)[/tex].
These are the complete factorizations for each of the given expressions.
