If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root,then $a+b+c$ can be equal to

Question

If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root, then $a+b+c$ can be equal to(where $a,b,c\in R,a\ne0$)

$(1)\;5a$
$(2)\;3b$
$(3)\;2c$
$(4)\;0$

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As $x=1$ is the root of the equation $2x^3+x^2+2x-5=0$ and its other two roots are complex in nature. So the common root is $x=1$

Put $x=1$ in $ax^3+(a+b)x^2+(b+c)x+c=0$ we get $2a+2b+2c=0$ so $(4)$ option is correct. But $(1)$ and $(3)$ options are also correct. I don't know how they are correct.

Answer 1

Note that $$ax^3+(a+b)x^2+(b+c)x+c=(x+1)(ax^2+bx+c)$$ and $$2x^3+x^2+2x-5=(x-1)(2x^2+3x+5)$$ So we can also have the case when the roots in common are the two complex roots: $a=2k$, $b=3k$ and $c=5k$, that is $a+b+c=5a=2c$.

Answer 2

The equation $2x^3+x^2+2x−5=0$ has 3 solutions, $$1, \quad \frac14 \left(-3-i\sqrt{31} \right), \quad \mbox{and} \quad \frac14 \left(-3+i\sqrt{31} \right).$$ If you put $x=1$ (as you did) you obtained $a+b+c=0$. In the same way if you put $x=\frac14 \left(-3-i\sqrt{31} \right)$ (to check your computation maybe use wolframaplha) you get $2b=3a$ and $2c=5a$, which is equivalent to $$a+b+c=5a.$$ I let you do on your own the last case.