Question

Решите уравнение: X^3+5x^2-25x-125=0

Answer 1

x^3 + 5*x^2 - 25*x - 125 =0; x^2*(x+5)-25*(x+5)=x+5)*(x-5)*(x+5)=0; x1=5; x2=-5;

Answer 2

Способ группировки 
(x3-125) -(25x-5x2)=0
раскрываем куб разности и выносим за скобку общий множитель
(х-5)(х2+5х+25)-5х(5-х)=0
меняем знак 
(х-5)(х2+5х+25)+5х(х-5)=0
выносим общий множитель
(х-5)(х2+5х+25+5х)=0
1) х-5=0
х=5
2) х2+10х+25=0
(х+5)2=0
х=-5
Ответ: -5; +5.

Answer 3

To solve the equation $x^3 + 5x^2 - 25x - 125 = 0$, we can use either numerical methods or factorization. One of the roots can be found by inspection: $x = 5$ is a solution. We can then use polynomial division or synthetic division to find the other roots. Let's proceed with polynomial division:

Step 1: Divide the polynomial by $x - 5$ to obtain a quadratic polynomial.

$$\begin{array}{rrrrr} x^3 & + 5x^2 & - 25x & - 125 & \vert \quad x - 5 \\ \end{array}$$

Step 2: Bring down the coefficient of $x^3$.

$$\begin{array}{rrrrr} x^3 & + 5x^2 & - 25x & - 125 & \vert \quad x - 5 \\ & - (x^3 & - 5x^2) & & \\ \end{array}$$

Step 3: Subtract $x^3 - 5x^2$ from $x^3 + 5x^2$.

$$\begin{array}{rrrrr} x^3 & + 5x^2 & - 25x & - 125 & \vert \quad x - 5 \\ & - (x^3 & - 5x^2) & & \\ \hline & & 10x^2 & - 25x & \\ \end{array}$$

Step 4: Bring down the coefficient of $10x^2$.

$$\begin{array}{rrrrr} x^3 & + 5x^2 & - 25x & - 125 & \vert \quad x - 5 \\ & - (x^3 & - 5x^2) & & \\ \hline & & 10x^2 & - 25x & \\ & & - (10x^2 & - 50x) & \\ \end{array}$$

Step 5: Subtract $-(10x^2 - 50x)$ from $10x^2 - 25x$.

$$\begin{array}{rrrrr} x^3 & + 5x^2 & - 25x & - 125 & \vert \quad x - 5 \\ & - (x^3 & - 5x^2) & & \\ \hline & & 10x^2 & - 25x & \\ & & - (10x^2 & - 50x) & \\ \hline & & & 25x & - 125 \\ \end{array}$$

Step 6: Bring down the coefficient of $25x$.

$$\begin{array}{rrrrr} x^3 & + 5x^2 & - 25x & - 125 & \vert \quad x - 5 \\ & - (x^3 & - 5x^2) & & \\ \hline & & 10x^2 & - 25x & \\ & & - (10x^2 & - 50x) & \\ \hline & & & 25x & - 125 \\ & & & - (25x & - 125) & \\ \end{array}$$