solve the equation?

Question

solve the equation x^3 + sx^2 + x + 2 = 0, if -2 is a root


No answers like "i dont know" please

sorry, i wrote the equation wrong its actualy, x^3 + 2x^2 + x + 2 = 0 if -2 is a root.

Answer 1

x^3 + sx^2 + x + 2 = 0

if x = -2 is root, then it must satisfy the above equation

=> (-2)^3 + s (-2)^2 + (-2) + 2 = 0
=> s = 2

now x^3 + 2 x^2 + x + 2 = (x^2 + 1)*(x+2) = 0
so the roots are x = -2, x = ±i
where i = √-1

Answer 2

Hey there!

We know that -2 is the root of x^3+sx^2+x+2. Our main goal is to find the value of s. We could do that, by substituting -2 for x.

x^3+sx^2+x+2=0 -->
(-2)^3+s(-2)^2+(-2)+2=0 -->
-8+4s-2+2=0 -->
-8+4s=0 -->
4s-8=0 -->
4s=8 -->
s=2 -->
x^3+sx^2+x+2=0 -->
x^3+2x^2+x+2=0 -->
x^2(x+2)+1(x+2)=0 -->
(x^2+1)(x+2)=0 -->
x^2+1=0 or x+2=0 -->
x^2=-1 or x=-2 -->
x=±i or x=-2

Since the -2 is the root of the new equation and matches up with the root of the old equation, the value of s must have satisfied the equation. So the value of s is 2.

Hope it helps!

Answer 3

x^3 + sx^2 + x + 2 = 0 ...........(1)
As x = -2 is a root, x + 2 is a factor of the LHS.
The equation can therefore be written:
(x + 2)(x^2 + ax + b ) = 0
Expanding this:
x^3 + (2 + a)x^2 + (2a + b)x + 2b = 0 ........(2)
Comparing coefficients with those of (1) gives:
for x^2: 2 + a = s
for x: 2a + b = 1
for absolute term: 2b = 2

These three equations are easily solved as:
b = 1
2a + 1 = 1
a = 0
s = 2 + a = 2
From (2), the eauation is therefore:
x^3 + 2x^2 + x + 2 = 0.

As you know s = 2 (from your correction), you can make the solution simpler.

Answer 4

If -2 is a root, then plugging in -2 will give you zero.
Thus (-2)^3 + s * (-2)^2 + (-2) + 2 = 0
-8 + 4s - 2 + 2 = 0
4s = 8
s = 2

Answer 5

x - (-2) is a factor so using synthetic division:-
-2 |1*****2*****1******2
***|***** -2*****0**** - 2
----------------------------
****1*****0*****1*****0

(x + 2).(x² + 1) = 0
x = - 2 , x² = - 1
x = - 2 , x = i ²
x = - 2, x = ± i