Solve by factoring x^3-4x^2-3x+18=0
5 answers
You would factor by grouping
I tried it didn't work
Man, you're a better grouper than I am. The 18 made me think of 9*2, so I got
x(x-3)^2 = x^3-6x^2+9x
hmm. too many x^2, so
2(x-3)^2 = 2x^2-12x+18
aha - (x+2)(x-3)^2
x(x-3)^2 = x^3-6x^2+9x
hmm. too many x^2, so
2(x-3)^2 = 2x^2-12x+18
aha - (x+2)(x-3)^2
I see, for this problem you would have to plug in numbers with trial and error, by doing this you get x=-2 as one of the roots, then you would factor (x+2)out of the problem. Then you would get x²-6x+9. This is a perfect square, so when you factor that, you get (x-3)^2.
So you would get (x+2)(x-3)^2
So you would get (x+2)(x-3)^2
ok thanks for your help:D
1 answer