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can you please show me how to do synthetic division? @Hero
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Factor by inspection.
For example, say we had x^3 + 6x^2 + 11x +6. You could spot that (x+1) is a factor (substitute x=-1 to see this) and then write the semi-factorised verision by inspection:
And continue reducing and taking factors out until you get to (x+1)(x+2)(x+3).
It's a bit odd to describe the method, but I'll explain the first step:
I took out (x+1). We have x^3 so I know the first term in the second set of bracekets will have to be x^2. When this is multiplied by x+1 we get the x^3, but only get x^2 when we need 6x^2. So the next term in the second set of brackets will be 5x (since it will multiply with the x from the first set of brackets to give a total of 6x^2).
So up till now we have (x+1)(x^2+5x+...). The 5x completed the x^2 terms, but only gives us 5x, when we need 11x. So our next term will be +6, to give the extra 6x and the +6. So at the end of this step we have (x+1)(x^2+5x+6).
You can then factorise x^2+5x+6 easily.
It seems long winded in the explanation but it really isn't, it's basically just shorthand long division. Please let me know if I can make something more clear.