Originally Posted by

**mathaddict** Alright , lets come back to this equation .

$\displaystyle x^3-7x+4=x^3-x^2a-3x^2+3xa+2x-2a-2$

DO some regrouping .

$\displaystyle x^3+0x^2-7x+4=x^3-(a+3)x^2+(3a+2)x-2a-2$

So now we compare ..

$\displaystyle x^3=x^3$ ( there is nothing to compare)

$\displaystyle 0x^2=-(a+3)x^2$ , cancel the $\displaystyle x^2$ and solve for a .

$\displaystyle -7x=(3a+2)x$ , cancel the x and solve for a

the last one , the constant , do the same comparing .

All of these comparisons give a so you will only have to compare one of them .

Hope this helps. Any more doubts ?