Answer :
Sure! Let's factor each of the given trinomials step by step.
### 24. [tex]\(x^2 + 9x + 14\)[/tex]
To factor this trinomial, we need to find two numbers that multiply to [tex]\(14\)[/tex] (the constant term) and add up to [tex]\(9\)[/tex] (the coefficient of the middle term [tex]\(x\)[/tex]).
The numbers that work are [tex]\(7\)[/tex] and [tex]\(2\)[/tex]:
- [tex]\(7 \times 2 = 14\)[/tex]
- [tex]\(7 + 2 = 9\)[/tex]
So, we can write the trinomial as:
[tex]\[x^2 + 9x + 14 = (x + 7)(x + 2)\][/tex]
### 27. [tex]\(x^2 - 8x + 12\)[/tex]
For this trinomial, we need to find two numbers that multiply to [tex]\(12\)[/tex] (the constant term) and add up to [tex]\(-8\)[/tex] (the coefficient of the middle term [tex]\(x\)[/tex]).
The numbers that work are [tex]\(-6\)[/tex] and [tex]\(-2\)[/tex]:
- [tex]\(-6 \times -2 = 12\)[/tex]
- [tex]\(-6 + -2 = -8\)[/tex]
So, we can write the trinomial as:
[tex]\[x^2 - 8x + 12 = (x - 6)(x - 2)\][/tex]
### 30. [tex]\(b^2 + 6b - 27\)[/tex]
For this trinomial, we need to find two numbers that multiply to [tex]\(-27\)[/tex] (the constant term) and add up to [tex]\(6\)[/tex] (the coefficient of the middle term [tex]\(b\)[/tex]).
The numbers that work are [tex]\(9\)[/tex] and [tex]\(-3\)[/tex]:
- [tex]\(9 \times -3 = -27\)[/tex]
- [tex]\(9 + -3 = 6\)[/tex]
So, we can write the trinomial as:
[tex]\[b^2 + 6b - 27 = (b + 9)(b - 3)\][/tex]
### 33. [tex]\(q^2 - 7q - 10\)[/tex]
For this trinomial, we need to find two numbers that multiply to [tex]\(-10\)[/tex] (the constant term) and add up to [tex]\(-7\)[/tex] (the coefficient of the middle term [tex]\(q\)[/tex]).
The numbers that work are [tex]\(-10\)[/tex] and [tex]\(1\)[/tex]:
- [tex]\(-10 \times 1 = -10\)[/tex]
- [tex]\(-10 + 1 = -7\)[/tex]
So, we can write the trinomial as:
[tex]\[q^2 - 7q - 10 = (q - 10)(q + 1)\][/tex]
### Summary of Factorizations
1. [tex]\(x^2 + 9x + 14 = (x + 7)(x + 2)\)[/tex]
2. [tex]\(x^2 - 8x + 12 = (x - 6)(x - 2)\)[/tex]
3. [tex]\(b^2 + 6b - 27 = (b + 9)(b - 3)\)[/tex]
4. [tex]\(q^2 - 7q - 10 = (q - 10)(q + 1)\)[/tex]
I hope this helps! If you have any further questions, feel free to ask.
### 24. [tex]\(x^2 + 9x + 14\)[/tex]
To factor this trinomial, we need to find two numbers that multiply to [tex]\(14\)[/tex] (the constant term) and add up to [tex]\(9\)[/tex] (the coefficient of the middle term [tex]\(x\)[/tex]).
The numbers that work are [tex]\(7\)[/tex] and [tex]\(2\)[/tex]:
- [tex]\(7 \times 2 = 14\)[/tex]
- [tex]\(7 + 2 = 9\)[/tex]
So, we can write the trinomial as:
[tex]\[x^2 + 9x + 14 = (x + 7)(x + 2)\][/tex]
### 27. [tex]\(x^2 - 8x + 12\)[/tex]
For this trinomial, we need to find two numbers that multiply to [tex]\(12\)[/tex] (the constant term) and add up to [tex]\(-8\)[/tex] (the coefficient of the middle term [tex]\(x\)[/tex]).
The numbers that work are [tex]\(-6\)[/tex] and [tex]\(-2\)[/tex]:
- [tex]\(-6 \times -2 = 12\)[/tex]
- [tex]\(-6 + -2 = -8\)[/tex]
So, we can write the trinomial as:
[tex]\[x^2 - 8x + 12 = (x - 6)(x - 2)\][/tex]
### 30. [tex]\(b^2 + 6b - 27\)[/tex]
For this trinomial, we need to find two numbers that multiply to [tex]\(-27\)[/tex] (the constant term) and add up to [tex]\(6\)[/tex] (the coefficient of the middle term [tex]\(b\)[/tex]).
The numbers that work are [tex]\(9\)[/tex] and [tex]\(-3\)[/tex]:
- [tex]\(9 \times -3 = -27\)[/tex]
- [tex]\(9 + -3 = 6\)[/tex]
So, we can write the trinomial as:
[tex]\[b^2 + 6b - 27 = (b + 9)(b - 3)\][/tex]
### 33. [tex]\(q^2 - 7q - 10\)[/tex]
For this trinomial, we need to find two numbers that multiply to [tex]\(-10\)[/tex] (the constant term) and add up to [tex]\(-7\)[/tex] (the coefficient of the middle term [tex]\(q\)[/tex]).
The numbers that work are [tex]\(-10\)[/tex] and [tex]\(1\)[/tex]:
- [tex]\(-10 \times 1 = -10\)[/tex]
- [tex]\(-10 + 1 = -7\)[/tex]
So, we can write the trinomial as:
[tex]\[q^2 - 7q - 10 = (q - 10)(q + 1)\][/tex]
### Summary of Factorizations
1. [tex]\(x^2 + 9x + 14 = (x + 7)(x + 2)\)[/tex]
2. [tex]\(x^2 - 8x + 12 = (x - 6)(x - 2)\)[/tex]
3. [tex]\(b^2 + 6b - 27 = (b + 9)(b - 3)\)[/tex]
4. [tex]\(q^2 - 7q - 10 = (q - 10)(q + 1)\)[/tex]
I hope this helps! If you have any further questions, feel free to ask.