Answer :
Certainly! Let's factorize each of the given expressions step by step using appropriate algebraic identities.
### a. [tex]\(64a^2 - 25b^2\)[/tex]
This can be recognized as a difference of squares, which can be factored using the identity [tex]\(A^2 - B^2 = (A - B)(A + B)\)[/tex] where [tex]\(A = 8a\)[/tex] and [tex]\(B = 5b\)[/tex].
[tex]\[64a^2 - 25b^2 = (8a)^2 - (5b)^2 = (8a - 5b)(8a + 5b)\][/tex]
### b. [tex]\(x^2 + 12x + 36\)[/tex]
This is a perfect square trinomial. We can factor it using the identity [tex]\((x + b)^2 = x^2 + 2bx + b^2\)[/tex]. Here, [tex]\(b = 6\)[/tex] because [tex]\(2 \cdot 6 = 12\)[/tex] and [tex]\(6^2 = 36\)[/tex].
[tex]\[x^2 + 12x + 36 = (x + 6)^2\][/tex]
### d. [tex]\(x^2 - \frac{y^2}{121}\)[/tex]
Again, we have a difference of squares. Here, [tex]\(A = x\)[/tex] and [tex]\(B = \frac{y}{11}\)[/tex] because [tex]\(\left( \frac{y}{11} \right)^2 = \frac{y^2}{121}\)[/tex].
[tex]\[x^2 - \frac{y^2}{121} = x^2 - \left( \frac{y}{11} \right)^2 = \left( x - \frac{y}{11} \right)\left( x + \frac{y}{11} \right)\][/tex]
### e. [tex]\(a^2 - 3a - 40\)[/tex]
This is a quadratic trinomial. We need to find two numbers that multiply to [tex]\(-40\)[/tex] (the constant term) and add up to [tex]\(-3\)[/tex] (the coefficient of [tex]\(a\)[/tex]). These numbers are [tex]\(-8\)[/tex] and [tex]\(5\)[/tex].
[tex]\[a^2 - 3a - 40 = (a - 8)(a + 5)\][/tex]
### g. [tex]\(8a^2 - 32\)[/tex]
Factor out the greatest common factor (GCF), which is [tex]\(8\)[/tex]:
[tex]\[8a^2 - 32 = 8(a^2 - 4)\][/tex]
Then factor [tex]\(a^2 - 4\)[/tex] as a difference of squares:
[tex]\[a^2 - 4 = (a - 2)(a + 2)\][/tex]
Thus,
[tex]\[8a^2 - 32 = 8(a - 2)(a + 2)\][/tex]
### h. [tex]\(25m^2n^2 - 49\)[/tex]
This is another difference of squares where [tex]\(A = 5mn\)[/tex] and [tex]\(B = 7\)[/tex]:
[tex]\[25m^2n^2 - 49 = (5mn)^2 - 7^2 = (5mn - 7)(5mn + 7)\][/tex]
### j. [tex]\((x + y)^2 - (x - y)^2\)[/tex]
This is also a difference of squares with [tex]\(A = x + y\)[/tex] and [tex]\(B = x - y\)[/tex]:
[tex]\[((x + y)^2 - (x - y)^2 = (x + y - (x - y))(x + y + (x - y)) = (x + y - x + y)(x + y + x - y)\][/tex]
[tex]\[= (2y)(2x) = 4xy\][/tex]
### k. [tex]\(81x^3 - 225x^5\)[/tex]
First, factor out the greatest common factor (GCF), which is [tex]\(9x^3\)[/tex]:
[tex]\[81x^3 - 225x^5 = 9x^3(9 - 25x^2)\][/tex]
Then recognize [tex]\(9 - 25x^2\)[/tex] as a difference of squares:
[tex]\[9 - 25x^2 = (3 - 5x)(3 + 5x)\][/tex]
Thus,
[tex]\[81x^3 - 225x^5 = 9x^3(3 - 5x)(3 + 5x)\][/tex]
### m. [tex]\(36p^2q^2 - 100\)[/tex]
Factor out the greatest common factor (GCF), which is [tex]\(4\)[/tex]:
[tex]\[36p^2q^2 - 100 = 4(9p^2q^2 - 25)\][/tex]
Then recognize [tex]\(9p^2q^2 - 25\)[/tex] as a difference of squares:
[tex]\[9p^2q^2 - 25 = (3pq)^2 - 5^2 = (3pq - 5)(3pq + 5)\][/tex]
Thus,
[tex]\[36p^2q^2 - 100 = 4(3pq - 5)(3pq + 5)\][/tex]
### n. [tex]\(x^4 - y^4\)[/tex]
This is another difference of squares, where [tex]\(A = x^2\)[/tex] and [tex]\(B = y^2\)[/tex]:
[tex]\[x^4 - y^4 = (x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2)\][/tex]
Now, factor [tex]\(x^2 - y^2\)[/tex] again as a difference of squares:
[tex]\[x^2 - y^2 = (x - y)(x + y)\][/tex]
Thus,
[tex]\[x^4 - y^4 = (x - y)(x + y)(x^2 + y^2)\][/tex]
### Summary
To summarize, the factorizations are:
1. [tex]\((8a - 5b)(8a + 5b)\)[/tex]
2. [tex]\((x + 6)^2\)[/tex]
3. [tex]\((x - \frac{y}{11})(x + \frac{y}{11})\)[/tex]
4. [tex]\((a - 8)(a + 5)\)[/tex]
5. [tex]\(8(a - 2)(a + 2)\)[/tex]
6. [tex]\((5mn - 7)(5mn + 7)\)[/tex]
7. [tex]\(4xy\)[/tex]
8. [tex]\(-9x^3(5x - 3)(5x + 3)\)[/tex]
9. [tex]\(4(3pq - 5)(3pq + 5)\)[/tex]
10. [tex]\((x - y)(x + y)(x^2 + y^2)\)[/tex]
### a. [tex]\(64a^2 - 25b^2\)[/tex]
This can be recognized as a difference of squares, which can be factored using the identity [tex]\(A^2 - B^2 = (A - B)(A + B)\)[/tex] where [tex]\(A = 8a\)[/tex] and [tex]\(B = 5b\)[/tex].
[tex]\[64a^2 - 25b^2 = (8a)^2 - (5b)^2 = (8a - 5b)(8a + 5b)\][/tex]
### b. [tex]\(x^2 + 12x + 36\)[/tex]
This is a perfect square trinomial. We can factor it using the identity [tex]\((x + b)^2 = x^2 + 2bx + b^2\)[/tex]. Here, [tex]\(b = 6\)[/tex] because [tex]\(2 \cdot 6 = 12\)[/tex] and [tex]\(6^2 = 36\)[/tex].
[tex]\[x^2 + 12x + 36 = (x + 6)^2\][/tex]
### d. [tex]\(x^2 - \frac{y^2}{121}\)[/tex]
Again, we have a difference of squares. Here, [tex]\(A = x\)[/tex] and [tex]\(B = \frac{y}{11}\)[/tex] because [tex]\(\left( \frac{y}{11} \right)^2 = \frac{y^2}{121}\)[/tex].
[tex]\[x^2 - \frac{y^2}{121} = x^2 - \left( \frac{y}{11} \right)^2 = \left( x - \frac{y}{11} \right)\left( x + \frac{y}{11} \right)\][/tex]
### e. [tex]\(a^2 - 3a - 40\)[/tex]
This is a quadratic trinomial. We need to find two numbers that multiply to [tex]\(-40\)[/tex] (the constant term) and add up to [tex]\(-3\)[/tex] (the coefficient of [tex]\(a\)[/tex]). These numbers are [tex]\(-8\)[/tex] and [tex]\(5\)[/tex].
[tex]\[a^2 - 3a - 40 = (a - 8)(a + 5)\][/tex]
### g. [tex]\(8a^2 - 32\)[/tex]
Factor out the greatest common factor (GCF), which is [tex]\(8\)[/tex]:
[tex]\[8a^2 - 32 = 8(a^2 - 4)\][/tex]
Then factor [tex]\(a^2 - 4\)[/tex] as a difference of squares:
[tex]\[a^2 - 4 = (a - 2)(a + 2)\][/tex]
Thus,
[tex]\[8a^2 - 32 = 8(a - 2)(a + 2)\][/tex]
### h. [tex]\(25m^2n^2 - 49\)[/tex]
This is another difference of squares where [tex]\(A = 5mn\)[/tex] and [tex]\(B = 7\)[/tex]:
[tex]\[25m^2n^2 - 49 = (5mn)^2 - 7^2 = (5mn - 7)(5mn + 7)\][/tex]
### j. [tex]\((x + y)^2 - (x - y)^2\)[/tex]
This is also a difference of squares with [tex]\(A = x + y\)[/tex] and [tex]\(B = x - y\)[/tex]:
[tex]\[((x + y)^2 - (x - y)^2 = (x + y - (x - y))(x + y + (x - y)) = (x + y - x + y)(x + y + x - y)\][/tex]
[tex]\[= (2y)(2x) = 4xy\][/tex]
### k. [tex]\(81x^3 - 225x^5\)[/tex]
First, factor out the greatest common factor (GCF), which is [tex]\(9x^3\)[/tex]:
[tex]\[81x^3 - 225x^5 = 9x^3(9 - 25x^2)\][/tex]
Then recognize [tex]\(9 - 25x^2\)[/tex] as a difference of squares:
[tex]\[9 - 25x^2 = (3 - 5x)(3 + 5x)\][/tex]
Thus,
[tex]\[81x^3 - 225x^5 = 9x^3(3 - 5x)(3 + 5x)\][/tex]
### m. [tex]\(36p^2q^2 - 100\)[/tex]
Factor out the greatest common factor (GCF), which is [tex]\(4\)[/tex]:
[tex]\[36p^2q^2 - 100 = 4(9p^2q^2 - 25)\][/tex]
Then recognize [tex]\(9p^2q^2 - 25\)[/tex] as a difference of squares:
[tex]\[9p^2q^2 - 25 = (3pq)^2 - 5^2 = (3pq - 5)(3pq + 5)\][/tex]
Thus,
[tex]\[36p^2q^2 - 100 = 4(3pq - 5)(3pq + 5)\][/tex]
### n. [tex]\(x^4 - y^4\)[/tex]
This is another difference of squares, where [tex]\(A = x^2\)[/tex] and [tex]\(B = y^2\)[/tex]:
[tex]\[x^4 - y^4 = (x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2)\][/tex]
Now, factor [tex]\(x^2 - y^2\)[/tex] again as a difference of squares:
[tex]\[x^2 - y^2 = (x - y)(x + y)\][/tex]
Thus,
[tex]\[x^4 - y^4 = (x - y)(x + y)(x^2 + y^2)\][/tex]
### Summary
To summarize, the factorizations are:
1. [tex]\((8a - 5b)(8a + 5b)\)[/tex]
2. [tex]\((x + 6)^2\)[/tex]
3. [tex]\((x - \frac{y}{11})(x + \frac{y}{11})\)[/tex]
4. [tex]\((a - 8)(a + 5)\)[/tex]
5. [tex]\(8(a - 2)(a + 2)\)[/tex]
6. [tex]\((5mn - 7)(5mn + 7)\)[/tex]
7. [tex]\(4xy\)[/tex]
8. [tex]\(-9x^3(5x - 3)(5x + 3)\)[/tex]
9. [tex]\(4(3pq - 5)(3pq + 5)\)[/tex]
10. [tex]\((x - y)(x + y)(x^2 + y^2)\)[/tex]
