Answer :
Certainly! Let's go through the factorization process for each expression step-by-step:
### 1. [tex]\( xy - 2y + 3x + 6 \)[/tex]
Group terms to factor by grouping:
[tex]\[ (xy - 2y) + (3x + 6) \][/tex]
Factor out common factors from each group:
[tex]\[ y(x - 2) + 3(x - 2) \][/tex]
Now, factor out the common binomial factor:
[tex]\[ (y + 3)(x - 2) \][/tex]
So, the factorized form is:
[tex]\[ (y + 3)(x - 2) \][/tex]
### 2. [tex]\( P q\left(r^2 + 1\right) - r\left(p^2 + q^2\right) \)[/tex]
This expression does not easily simplify through simple factorization methods. However, we can rewrite it for clarity:
[tex]\[ pqr^2 + pq - rp^2 - rq^2 \][/tex]
Here, there's no common continuous factor or group to factor by. It remains in its current form.
### 3. [tex]\( 2x^2 + 7x + 6 \)[/tex]
To factor this quadratic expression, we find two numbers that multiply to [tex]\( 2 \times 6 = 12 \)[/tex] and add to [tex]\( 7 \)[/tex]:
[tex]\[ 2x^2 + 7x + 6 = 2x^2 + 3x + 4x + 6 \][/tex]
Group terms for factorization:
[tex]\[ = (2x^2 + 3x) + (4x + 6) \][/tex]
Factor out the common factors:
[tex]\[ = x(2x + 3) + 2(2x + 3) \][/tex]
Factor out the common binomial factor:
[tex]\[ = (2x + 3)(x + 2) \][/tex]
So, the factorized form is:
[tex]\[ (2x + 3)(x + 2) \][/tex]
### 4. [tex]\( 12 \frac{a^2}{b^2} + \frac{a}{b} - 20 \)[/tex]
First, rewrite the expression to make it easier to manage:
[tex]\[ 12 \frac{a^2}{b^2} + \frac{a}{b} - 20 = \frac{12a^2 + ab - 20b^2}{b^2} \][/tex]
This quadratic expression in the numerator can be factorized by finding two numbers that multiply to [tex]\( 12 \times (-20) = -240 \)[/tex] and add to [tex]\( 1 \)[/tex]:
[tex]\[ \frac{(4a - 5b)(3a + 4b)}{b^2} \][/tex]
So, the factorized form is:
[tex]\[ \frac{(4a - 5b)(3a + 4b)}{b^2} \][/tex]
### 5. [tex]\( \frac{x^2}{y^2} - 2 - \frac{8y^2}{z^2} \)[/tex]
This expression involves different variables and terms. It can be simplified but not easily factored in a typical sense.
### Additional Factorization Problems:
#### i. [tex]\( 25a^2b^2 - 1 \)[/tex]
This is a difference of squares:
[tex]\[ (5ab)^2 - 1^2 = (5ab - 1)(5ab + 1) \][/tex]
So, the factorized form is:
[tex]\[ (5ab - 1)(5ab + 1) \][/tex]
#### ii. [tex]\( 4 - (m - n)^2 \)[/tex]
This is another difference of squares:
[tex]\[ 4 - (m - n)^2 = 2^2 - (m - n)^2 = (2 + (m - n))(2 - (m - n)) \][/tex]
Simplify:
[tex]\[ (2 + m - n)(2 - m + n) \][/tex]
So, the factorized form is:
[tex]\[ (2 + m - n)(2 - m + n) \][/tex]
#### iii. [tex]\( a^2 - b^2 - 45 - 4b + 4\)[/tex]
Rearrange terms to recognize another potential difference of squares or groupwise simplification:
[tex]\[ a^2 - b^2 + 4 - 45 - 4b \][/tex]
This does not straightforwardly simplify further without context.
#### iv. [tex]\( 16a^4 - 4a^2 - 4a - 1 \)[/tex]
Group terms appropriately:
[tex]\[ 16a^4 - (4a^2 + 4a + 1) = (4a^2)^2 - (2a + 1)^2 \][/tex]
Recognize the difference of squares:
[tex]\[ (4a^2 - (2a + 1))(4a^2 + (2a + 1)) \][/tex]
So, the factorized form is:
[tex]\[ (4a^2 - 2a - 1)(4a^2 + 2a + 1) \][/tex]
#### v. [tex]\( x^3 + 6x + 5 - 4y - y^2 \)[/tex]
This expression involves multiple variables and terms. Factoring directly isn't straightforward without context or further grouping.
Each factorization has specific techniques; some are easily grouped, while others require recognizing patterns or more advanced algebraic methods.
### 1. [tex]\( xy - 2y + 3x + 6 \)[/tex]
Group terms to factor by grouping:
[tex]\[ (xy - 2y) + (3x + 6) \][/tex]
Factor out common factors from each group:
[tex]\[ y(x - 2) + 3(x - 2) \][/tex]
Now, factor out the common binomial factor:
[tex]\[ (y + 3)(x - 2) \][/tex]
So, the factorized form is:
[tex]\[ (y + 3)(x - 2) \][/tex]
### 2. [tex]\( P q\left(r^2 + 1\right) - r\left(p^2 + q^2\right) \)[/tex]
This expression does not easily simplify through simple factorization methods. However, we can rewrite it for clarity:
[tex]\[ pqr^2 + pq - rp^2 - rq^2 \][/tex]
Here, there's no common continuous factor or group to factor by. It remains in its current form.
### 3. [tex]\( 2x^2 + 7x + 6 \)[/tex]
To factor this quadratic expression, we find two numbers that multiply to [tex]\( 2 \times 6 = 12 \)[/tex] and add to [tex]\( 7 \)[/tex]:
[tex]\[ 2x^2 + 7x + 6 = 2x^2 + 3x + 4x + 6 \][/tex]
Group terms for factorization:
[tex]\[ = (2x^2 + 3x) + (4x + 6) \][/tex]
Factor out the common factors:
[tex]\[ = x(2x + 3) + 2(2x + 3) \][/tex]
Factor out the common binomial factor:
[tex]\[ = (2x + 3)(x + 2) \][/tex]
So, the factorized form is:
[tex]\[ (2x + 3)(x + 2) \][/tex]
### 4. [tex]\( 12 \frac{a^2}{b^2} + \frac{a}{b} - 20 \)[/tex]
First, rewrite the expression to make it easier to manage:
[tex]\[ 12 \frac{a^2}{b^2} + \frac{a}{b} - 20 = \frac{12a^2 + ab - 20b^2}{b^2} \][/tex]
This quadratic expression in the numerator can be factorized by finding two numbers that multiply to [tex]\( 12 \times (-20) = -240 \)[/tex] and add to [tex]\( 1 \)[/tex]:
[tex]\[ \frac{(4a - 5b)(3a + 4b)}{b^2} \][/tex]
So, the factorized form is:
[tex]\[ \frac{(4a - 5b)(3a + 4b)}{b^2} \][/tex]
### 5. [tex]\( \frac{x^2}{y^2} - 2 - \frac{8y^2}{z^2} \)[/tex]
This expression involves different variables and terms. It can be simplified but not easily factored in a typical sense.
### Additional Factorization Problems:
#### i. [tex]\( 25a^2b^2 - 1 \)[/tex]
This is a difference of squares:
[tex]\[ (5ab)^2 - 1^2 = (5ab - 1)(5ab + 1) \][/tex]
So, the factorized form is:
[tex]\[ (5ab - 1)(5ab + 1) \][/tex]
#### ii. [tex]\( 4 - (m - n)^2 \)[/tex]
This is another difference of squares:
[tex]\[ 4 - (m - n)^2 = 2^2 - (m - n)^2 = (2 + (m - n))(2 - (m - n)) \][/tex]
Simplify:
[tex]\[ (2 + m - n)(2 - m + n) \][/tex]
So, the factorized form is:
[tex]\[ (2 + m - n)(2 - m + n) \][/tex]
#### iii. [tex]\( a^2 - b^2 - 45 - 4b + 4\)[/tex]
Rearrange terms to recognize another potential difference of squares or groupwise simplification:
[tex]\[ a^2 - b^2 + 4 - 45 - 4b \][/tex]
This does not straightforwardly simplify further without context.
#### iv. [tex]\( 16a^4 - 4a^2 - 4a - 1 \)[/tex]
Group terms appropriately:
[tex]\[ 16a^4 - (4a^2 + 4a + 1) = (4a^2)^2 - (2a + 1)^2 \][/tex]
Recognize the difference of squares:
[tex]\[ (4a^2 - (2a + 1))(4a^2 + (2a + 1)) \][/tex]
So, the factorized form is:
[tex]\[ (4a^2 - 2a - 1)(4a^2 + 2a + 1) \][/tex]
#### v. [tex]\( x^3 + 6x + 5 - 4y - y^2 \)[/tex]
This expression involves multiple variables and terms. Factoring directly isn't straightforward without context or further grouping.
Each factorization has specific techniques; some are easily grouped, while others require recognizing patterns or more advanced algebraic methods.
