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**Hint**: Let there be any cubic equation as \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] which has three roots as \[\alpha ,\beta ,\gamma \]

So we can write this cubic equation as \[a(x-\alpha )(x-\beta )(x-\gamma )=0\], Now we can also say that \[(x-\alpha ),(x-\beta )and(x-\gamma )\] are the factors of cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] . We will obtain an equation in terms of a and b substituting factor which is x+1=0 or x=-1 in the given equation. We have already been given another equation in a and b, so now using elimination method, we can compute values of a and b.

**:**

__Complete step-by-step answer__In the question we are given a cubic equation as \[2{{x}^{3}}+a{{x}^{2}}+2bx+1=0\] which has a factor \[x+1\]

Assuming the roots are \[\alpha ,\beta ,\gamma \] so we can write our cubic equation as

\[2(x-\alpha )(x-\beta )(x-\gamma )=0\]

And factors of this equation will be \[(x-\alpha ),(x-\beta )and(x-\gamma )\]

But one factor given as \[x+1\]

So, replacing \[(x-\alpha )\] with \[x+1\]

Now our equation can be written as \[2(x+1)(x-\beta )(x-\gamma )=0\]

Looking this carefully if we make \[(x+1)=0\] then this cubic equation will become 0

It means \[(x+1)=0\] gives \[x=-1\] is a solution of this cubic equation

putting \[x=-1\] in equation \[2{{x}^{3}}+a{{x}^{2}}+2bx+1\] will make this equation 0

\[2{{(-1)}^{3}}+a{{(-1)}^{2}}+2b(-1)+1=0\]

\[\to (-2+a-2b+1)=0\]

On solving gives

\[a-2b=1.......(1)\]

Now \[a\]and \[b\]are unknown

But in the question, we are given an equation \[2a-3b=4.......(2)\]

So, we have 2 equation and 2 variables \[a\] and \[b\]

We can calculate them but elimination method

So first we eliminate \[a\]

Multiplying the equation (1) with 2

Now equation (1) will look like

\[2a-4b=2.........(3)\]

On subtracting equation (3) from equation (2), we get

\[2a-3b-(2a-4b)=4-2\]

\[2a-3b-2a+4b=2\]

\[b=2\]

Now we get \[b=2\] so we can put this value in equation (1) to get the value of \[a\]

Equation (1) \[a-2b=1\]

\[a-2(2)=1\]

\[a=5\]

Hence, we get the value \[a=5\] and \[b=2\]

**Note**: In this question students can make mistakes, by taking its root as 1 and not -1. This question is very basic; you just need to solve equations carefully. You can also calculate the values of \[a\] and \[b\] by substitution method.

Like we are given these two equations so we find value of \[a\] from equation (1) and put it in equation (2)

\[a-2b=1.......(1)\] \[2a-3b=4.......(2)\]

\[a=1+2b\] from equation (1)

Putting value of \[a\] into equation (2)

We get

\[2(1+2b)-3b=4\]

\[\to 2+4b-3b=4\]

\[\to 2+4b-3b=4\]

\[\to b=2\]

And we have \[a=1+2b\] so putting value \[b=2\]

\[\to a=5\]

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