Factor the ff with solution 

1) X²+12x+20

3) X²+9x+18

4) X²-16x+15

5) X²-14x+24

6) X²-2x-24

7) X²+6x+16

Let's factor each of the provided quadratic expressions step by step. I'll be using the method of factoring trinomials, and where necessary, checking if a quadratic formula is required.

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### **1) \(x^2 + 12x + 20\)**

We need to find two numbers that multiply to 20 and add to 12.

The numbers are \(10\) and \(2\) because:
\[
10 \times 2 = 20 \quad \text{and} \quad 10 + 2 = 12
\]

Thus, the factorization is:
\[
x^2 + 12x + 20 = (x + 10)(x + 2)
\]

---

### **3) \(x^2 + 9x + 18\)**

We need two numbers that multiply to 18 and add to 9.

The numbers are \(6\) and \(3\) because:
\[
6 \times 3 = 18 \quad \text{and} \quad 6 + 3 = 9
\]

Thus, the factorization is:
\[
x^2 + 9x + 18 = (x + 6)(x + 3)
\]

---

### **4) \(x^2 - 16x + 15\)**

We need two numbers that multiply to 15 and add to \(-16\).

The numbers are \(-15\) and \(-1\) because:
\[
(-15) \times (-1) = 15 \quad \text{and} \quad (-15) + (-1) = -16
\]

Thus, the factorization is:
\[
x^2 - 16x + 15 = (x - 15)(x - 1)
\]

---

### **5) \(x^2 - 14x + 24\)**

We need two numbers that multiply to 24 and add to \(-14\).

The numbers are \(-12\) and \(-2\) because:
\[
(-12) \times (-2) = 24 \quad \text{and} \quad (-12) + (-2) = -14
\]

Thus, the factorization is:
\[
x^2 - 14x + 24 = (x - 12)(x - 2)
\]

---

### **6) \(x^2 - 2x - 24\)**

We need two numbers that multiply to \(-24\) and add to \(-2\).

The numbers are \(4\) and \(-6\) because:
\[
4 \times (-6) = -24 \quad \text{and} \quad 4 + (-6) = -2
\]

Thus, the factorization is:
\[
x^2 - 2x - 24 = (x + 4)(x - 6)
\]

---

### **7) \(x^2 + 6x + 16\)**

We need two numbers that multiply to 16 and add to 6. However, no real numbers satisfy this condition (since 16 is positive and the factors will not add up to 6). Therefore, this expression cannot be factored over the real numbers.

To solve it using the quadratic formula:

The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For \(x^2 + 6x + 16\), we have \(a = 1\), \(b = 6\), and \(c = 16\).

First, calculate the discriminant:
\[
\Delta = b^2 - 4ac = 6^2 - 4(1)(16) = 36 - 64 = -28
\]
Since the discriminant is negative, there are no real solutions. The solutions are complex.

Thus, the factorization over the complex numbers is:
\[
x^2 + 6x + 16 = (x - (-3 + i\sqrt{7}))(x - (-3 - i\sqrt{7}))
\]

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Let me know if you'd like further explanation on any of these. Here are a few related questions to deepen your understanding:

1. Can you verify the discriminant for the quadratic \(x^2 + 6x + 16\)?
2. How do you identify if a quadratic equation can be factored over real numbers?
3. What are the steps to use the quadratic formula effectively?
4. How does the sign of the discriminant impact the nature of solutions?
5. Can you factor quadratic expressions by completing the square?

**Tip:** Always check the discriminant (\(b^2 - 4ac\)) when factoring seems impossible—it tells you if the solutions are real or complex!

Factor the ff with solution: 1) X²+12x+20, 3) X²+9x+18, 4) X²-16x+15, 5) X²-14x+24, 6) X²-2x-24, 7) X²+6x+16

Step-by-step guide to factoring quadratic equations like X²+12x+20, X²+9x+18, X²-16x+15, and more. Learn the process of factoring and solving quadratic equations, including handling cases with complex numbers, for Grades 8-10.

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