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multiply k+3 / 4k-2 * (12k^2+2k-4)
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\[\frac{k+3}{4k-2}\times12k^2+2k-4\]
yea
Typically this type of problem is designed so that the denominator will cancel. Therefore, if the denominator is going to cancel, it must be a factor of the numerator. So that give you a hint. Factor a 2 out of the 12k^2+2k-4 and then see if one of the factors is possible 4k-2
3k^2 + 11k + 6. factor out (4k-2) from (12k^2+2k-4) then it cancels out with the denominator or the first term, then you get (k+3)*(3k+2) which gives 3k^2 + 11k + 6.
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