Pregunta
O3/02/25 Grade 10 Factorise \( \begin{array}{ll}\text { a } x^{2}-7 x+12 & \text { () } 40 x^{2}+6 x y-18 y^{2} \\ \text { (b) } 2 x^{2}+12 x+18 & \text { (c) } 24 x^{2}-2 x y-y^{2} \\ \text { (c) } x^{2}-9 x+20 & \text { (j) } 3 x^{2}+19 x+20 \\ \text { d) } 3 x^{2}-9 x+24 & \\ \text { (e) } 2 x^{2}+5 x+3 & \\ \text { (t) } 3 x^{2}+4 x-4 & \\ \text { (g) } 5 x^{3} y+25 x^{2} y-30 x y\end{array} \)
Ask by Bob Barker. in South Africa
Feb 03,2025
Solución de tutoría real
Respuesta verificada por el tutor
Responder
Here are the factorized forms of the given expressions:
a) \(x^{2} - 7x + 12 = (x - 3)(x - 4)\)
b) \(40x^{2} + 6xy - 18y^{2} = 2(5x - 3y)(4x + 3y)\)
c) \(2x^{2} + 12x + 18 = 2(x + 3)^{2}\)
d) \(24x^{2} - 2xy - y^{2} = (6x + y)(4x - y)\)
e) \(x^{2} - 9x + 20 = (x - 4)(x - 5)\)
f) \(3x^{2} + 19x + 20 = (3x + 4)(x + 5)\)
g) \(3x^{2} - 9x + 24 = 3(x^{2} - 3x + 8)\)
h) \(2x^{2} + 5x + 3 = (2x + 3)(x + 1)\)
i) \(3x^{2} + 4x - 4 = (3x - 2)(x + 2)\)
j) \(5x^{3}y + 25x^{2}y - 30xy = 5xy(x + 6)(x - 1)\)
Solución
Sure, let's factorize each of the given expressions step by step.
---
### **a) \(x^{2} - 7x + 12\)**
**Factorization:**
\[
x^{2} - 7x + 12 = (x - 3)(x - 4)
\]
**Explanation:**
We look for two numbers that multiply to **12** and add up to **-7**. These numbers are **-3** and **-4**.
---
### **b) \(40x^{2} + 6xy - 18y^{2}\)**
**Factorization:**
\[
40x^{2} + 6xy - 18y^{2} = 2(5x - 3y)(4x + 3y)
\]
**Explanation:**
1. **Factor out the Greatest Common Factor (GCF):**
\(40x^{2} + 6xy - 18y^{2} = 2(20x^{2} + 3xy - 9y^{2})\)
2. **Factor the quadratic:**
\(20x^{2} + 3xy - 9y^{2} = (5x - 3y)(4x + 3y)\)
---
### **c) \(2x^{2} + 12x + 18\)**
**Factorization:**
\[
2x^{2} + 12x + 18 = 2(x + 3)^{2}
\]
**Explanation:**
1. **Factor out the GCF:**
\(2x^{2} + 12x + 18 = 2(x^{2} + 6x + 9)\)
2. **Recognize a perfect square trinomial:**
\(x^{2} + 6x + 9 = (x + 3)^{2}\)
---
### **d) \(24x^{2} - 2xy - y^{2}\)**
**Factorization:**
\[
24x^{2} - 2xy - y^{2} = (6x + y)(4x - y)
\]
**Explanation:**
We look for two binomials that multiply to give the original quadratic:
\[
(6x + y)(4x - y) = 24x^{2} - 2xy - y^{2}
\]
---
### **e) \(x^{2} - 9x + 20\)**
**Factorization:**
\[
x^{2} - 9x + 20 = (x - 4)(x - 5)
\]
**Explanation:**
We look for two numbers that multiply to **20** and add up to **-9**. These numbers are **-4** and **-5**.
---
### **f) \(3x^{2} + 19x + 20\)**
**Factorization:**
\[
3x^{2} + 19x + 20 = (3x + 4)(x + 5)
\]
**Explanation:**
We find two numbers that multiply to \(3 \times 20 = 60\) and add up to **19**. These numbers are **15** and **4**.
\[
3x^{2} + 19x + 20 = 3x^{2} + 15x + 4x + 20 = (3x + 4)(x + 5)
\]
---
### **g) \(3x^{2} - 9x + 24\)**
**Factorization:**
\[
3x^{2} - 9x + 24 = 3(x^{2} - 3x + 8)
\]
**Explanation:**
1. **Factor out the GCF:**
\(3x^{2} - 9x + 24 = 3(x^{2} - 3x + 8)\)
2. **Note:**
The quadratic \(x^{2} - 3x + 8\) does not factor neatly with integers, so we leave it as is.
---
### **h) \(2x^{2} + 5x + 3\)**
**Factorization:**
\[
2x^{2} + 5x + 3 = (2x + 3)(x + 1)
\]
**Explanation:**
We look for two numbers that multiply to \(2 \times 3 = 6\) and add up to **5**. These numbers are **3** and **2**.
\[
2x^{2} + 5x + 3 = (2x + 3)(x + 1)
\]
---
### **i) \(3x^{2} + 4x - 4\)**
**Factorization:**
\[
3x^{2} + 4x - 4 = (3x - 2)(x + 2)
\]
**Explanation:**
We look for two numbers that multiply to \(3 \times (-4) = -12\) and add up to **4**. These numbers are **6** and **-2**.
\[
3x^{2} + 4x - 4 = 3x^{2} + 6x - 2x - 4 = (3x - 2)(x + 2)
\]
---
### **j) \(5x^{3}y + 25x^{2}y - 30xy\)**
**Factorization:**
\[
5x^{3}y + 25x^{2}y - 30xy = 5xy(x + 6)(x - 1)
\]
**Explanation:**
1. **Factor out the GCF:**
\(5xy\)
2. **Factor the quadratic:**
\(x^{2} + 5x - 6 = (x + 6)(x - 1)\)
3. **Combine:**
\(5xy(x + 6)(x - 1)\)
---
These are the fully factorized forms of the given expressions. If you have any further questions or need additional explanations, feel free to ask!
Revisado y aprobado por el equipo de tutoría de UpStudy
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Bonus Knowledge
Let's break these down with a spark of fun! For the first one, \(ax^2 - 7x + 12\): To factor this quadratic, think of two numbers that add up to -7 and multiply to 12. Those numbers are -3 and -4! So it factors to \(a(x - 3)(x - 4)\). Now, let's tackle \(5x^3y + 25x^2y - 30xy\): Here, the common factor is \(5xy\). Pull that out, and you get \(5xy(x^2 + 5x - 6)\). Next, factor the binomial as \((x + 6)(x - 1)\). So altogether, it's \(5xy(x + 6)(x - 1)\). Factor fun never ends, right?
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