# if (x+1) and (x+2) are both factors of x^3+ax^2+bx-10 , find the values of a and bhelp !

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As we have (x+1) (x+2) we can use substitution to solve and find the third factor. Remember that these factors are true when y or f(x) = 0

x^3+ax^2+bx-10 Substitute the two values you have (x=-1; x= -2)

`f(-1)= (-1)^3 + a(-1)^2 + b(-1) - 10`

`therefore 0= -1 +a - b - 10` multiply this line by 2. You'll see why later (b)

`therefore 0= -2 + 2a - 2b -20` Next use the other factor to create your second equation

`f(-2) = (-2)^3 + a (-2)^2 + b(-2) - 10 `

`therefore 0 =-8 + 4a - 2b -10`

Now solve simultaneously:``

`(0=-2 +2a - 2b -20)` - `(0= -8 +4a -2b -10` )

The bs will cancel out (which is why we multiplied earlier)and you will be left with

`0= 6-2a -10`

`therefore 2a= -4`

`therefore a = -2`

`therefore` `y= x^3 -2x^2 + bx - 10` As these points are either turning points or intercepts y=0 and x = (-1) or you could use (-2)

`therefore 0=(-1)^3 -2(-1)^2 + b(-1) - 10`

0= -1 -2 -b -10

b= -13

`therefore` **y= `x^3-2x^2 -13x-10` **

**a=-2 b= -13**