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`        solve for x:2x3+3x2+x-30=0zezxdcvjjmctkjtuxyjrjkcbyuy,iluvytiyrtytrtxrecnyrvbjnkyuyjctryretestxrhtfjgch`
10 months ago

## Answers : (2)

```							step 1: by hit and trial, put some integers like 1,2,-1and-2 in the given equation RESULT: if we put 2 in the given equation, it will give us zero in RHS side, so we can say that 2 is the root of the cubic equation                    or (x-2) is the factor of equation step 2 :  now divide the cubic equation by its factor (x-2) , we will get quotient as 2x2 + 7x + 15 and remainder will be 0 . step 3 : (x-2)(2x2 + 7x + 15 ) = 2x3 + 3x2 + x -30 step 4 : now solve the quadratic equation and you will get three factors or three vaue of x                 NOTE: roots may be imaginary or real
```
10 months ago
```							Just use hit and trial method to obtain at least one zero of the equation. Here we get 2 as one of its solutions. Then x-2 is a factor of the  given equation. You wiil get 2x^2+7x+15 as its quotient. Factorise the quotient you will get another roots.
```
10 months ago
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