Answer:
![9c^3 - 12c^2 - 18c - 24= [3c^2 - 6][3c - 4]](https://tex.z-dn.net/?f=9c%5E3%20-%2012c%5E2%20-%2018c%20-%2024%3D%20%5B3c%5E2%20-%206%5D%5B3c%20-%204%5D)
Step-by-step explanation:
Given

Required
Factor
Group into 2
![[9c^3 - 12c^2] - [18c + 24]](https://tex.z-dn.net/?f=%5B9c%5E3%20-%2012c%5E2%5D%20-%20%5B18c%20%2B%2024%5D)
Factorize each group
![3c^2[3c - 4] - 6[3c - 4]](https://tex.z-dn.net/?f=3c%5E2%5B3c%20-%204%5D%20-%206%5B3c%20-%204%5D)
Factor out 3c - 4
![[3c^2 - 6][3c - 4]](https://tex.z-dn.net/?f=%5B3c%5E2%20-%206%5D%5B3c%20-%204%5D)
Hence:
![9c^3 - 12c^2 - 18c - 24= [3c^2 - 6][3c - 4]](https://tex.z-dn.net/?f=9c%5E3%20-%2012c%5E2%20-%2018c%20-%2024%3D%20%5B3c%5E2%20-%206%5D%5B3c%20-%204%5D)
Answer:
b
Step-by-step explanation:
Answer:
Step-by-step explanation:
<u>Given </u>
<u>In the first quadrant both sine and cosine are positive, so as tangent.</u>
- tanθ = sinθ / cosθ =
θ / cosθ =
/ 4/5 =
/ 4/5 = - 3/5*5/4 =
- 3/4
Answer:
x= 2
x= -3
Step-by-step explanation:
<u>Solving as normal quadratic equation:</u>
- 2x^2+2x=12
- x^2 +x = 6
- x^2 + x - 6= 0
- x = (-1 ± √(1+4*1*6)/2
- x = ( - 1 ± √25)/2
- x = (-1 ± 5)/2
- x= 2
- x= -3
<u>Roots are</u> 2 and -3