1-(9c²+6c+4)=1-9c²-6c-4=-9c²-6c-3=-3(3c²-2c-1)=-3(3(c+1/3)(c-1))=-3(3c+1)(c-1)
b1+b2=140, b1+b1q=140, b1(1+q)=140 (1)
b2+b3=105, b1q+b1q^2=105,b1q(1+q)=105 (2)
разделим (2) на (1) получим q=105/140=3/4
b1(1+3/4)=140, b1=140*4/7=80
b2=80*3/4=60
b3=60*3/4=45
<span>y=x³+2x²-x-2
y=0
</span><span>x³+2x²-x-2=0
</span><span> x³+2x²-x-2</span><u>| x-1</u>
- <u> x³-x²</u> x²+3x+2=(x+1)(x+2)
3x² -x
<u>- 3x² -3x</u>
2x -2
<u> - 2x -2</u>
0
Получаем:
![x_{1}=1, x_{2}=-1, x_{2}=-2](https://tex.z-dn.net/?f=x_%7B1%7D%3D1%2C+x_%7B2%7D%3D-1%2C+x_%7B2%7D%3D-2)