can someone do this in long division

(x^2-x-6)/(x-3)  factor the numerator

[ (x + 2) (x - 3) ]  / (x - 3)      cancel the(x - 3)'s

x + 2

Step 1: Divide the first term under the long division line by the first term outside it, \(\frac{x^2}{x}\) in this case

Step 2: Multiply the newest term ontop of the division line with each term on the outside -> \(x*x + x*-3\)

Step 3: Subtract the 2nd line from the 1st -> \((x^2-x) - (x^2-3x) = 2x\)

Step 4: Repeat steps 1 to 3 eg: \(\frac{2x}{x}=2\) ...

Whats left up top is the factorised version.