Question

N+ n- 2n=0Can you help me out with this please!!

Answer 1

we have the equation

`n^3+n^2-2n=0`

step 1

Factor n

so

`n\lbrack n^2+n-2\rbrack=0`

Solve the quadratic equation, using the formula

we have

a=1

b=1

c=-2

substitute

`n=\frac{-1\pm\sqrt[\square]{(1^2)-4(1)(-2)}}{2(1)}`
`\begin{gathered} n=\frac{-1\pm\sqrt[\square]{9}}{2} \\ \\ n=(-1\pm3)/(2) \end{gathered}`

the values of n are

n=1 and n=-2

therefore

`n^3+n^2-2n=n(n+2)(n-1)`

the values of n are

n=0

n=-2

n=1