3k^2+1k-2/4k-2 divided by k^2+3k+2/2k^2+5k-3

Question

campbell_st · Answer

2 fractions, invert the 2nd and multiply

$$(3k^2 + k -2)/(4k-2) \times(2k^2 + 5k -2)/(k^2 +3k +2)$$

anonymous · Answer

I need the answer step by step

campbell_st · Answer

oops should be -3 in the numerator of the 2nd fraction and not -2

campbell_st · Answer

factorise both

$$((3k -2)(k+1))/2(2k-1) \times ((2k -1)(k+3))/((k+1)(k+2))$$

campbell_st · Answer

looks like (k+1) and (2k -1) can be eliminated as common factors which leaves

$$((3k -2)(k+3))/(2(k+2)$$

just expand the numerator and denominator to get the answer

anonymous · Answer

so what wuld the answer  be

campbell_st · Answer

just expand then both... you'll end up with a quadratic divided by a linear ... good luck

anonymous · Answer

can you just say the answer

campbell_st · Answer

well I've done the hard bit... you can do the rest