How to factorise by grouping for 7a, d and h
2 answers:
Step-by-step explanation:
7a
x2 + 4x + ax + 4a
x2 + ax + 4x + 4a
x (x + a) + 4 (x + a)
:(x + 4)(x+a)
7d
x2+2x-ax-2a
x2-ax+2x-2a
x (x-a)+2 (x-a)
:(x+2)(x-a)
7h
x2-2bx-5x+10b
x2-5x-2bx+10b
x (x-5)-2b(x-5)
:(x-2b)(x-5)
Answer:
You have to take out the common terms :
Question A,
![{x}^{2} + 4x + ax + 4a](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%204x%20%2B%20ax%20%2B%204a)
![= x(x + 4) + a(x + 4)](https://tex.z-dn.net/?f=%20%3D%20x%28x%20%2B%204%29%20%2B%20a%28x%20%2B%204%29)
![= (x + a)(x + 4)](https://tex.z-dn.net/?f=%20%3D%20%28x%20%2B%20a%29%28x%20%2B%204%29)
Question D,
![{x}^{2} + 2x - ax - 2a](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%202x%20-%20ax%20-%202a)
![= x(x + 2) - a(x + 2)](https://tex.z-dn.net/?f=%20%3D%20x%28x%20%20%2B%20%202%29%20-%20a%28x%20%2B%202%29)
![= (x - a)(x + 2)](https://tex.z-dn.net/?f=%20%3D%20%28x%20-%20a%29%28x%20%2B%202%29)
Question H,
![{x}^{2} - 2bx - 5x + 10b](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20-%202bx%20-%205x%20%2B%2010b)
![= x(x - 2b) - 5(x - 2b)](https://tex.z-dn.net/?f=%20%3D%20x%28x%20-%202b%29%20-%205%28x%20-%202b%29)
![= (x - 5)(x - 2b)](https://tex.z-dn.net/?f=%20%3D%20%28x%20-%205%29%28x%20-%202b%29)
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Step-by-step explanation:
Don't worry division always goes before subtraction so you don't need any parenthesis.
Tell me if I'm wrong :)