Question

Multiply k+3/4k-2 times (12k^2-3)

Answer 1

Simplify

 k+34k−2⋅(12k2−3) Step-By-Step Solution:Factor 2 out of 4k−2.k+32(2k−1)⋅(12k2−3)Multiply k+32(2k−1) by 12k2−3 to get k+32(2k−1)(12k2−3).k+32(2k−1)(12k2−3)Apply the distributive property.k+32(2k−1)(12k2)+k+32(2k−1)⋅−3Cancel the common factor of 2.More Stepsk+32k−16k21+k+32(2k−1)⋅−3Simplify.Less StepsMultiply k+32k−1 and 6k21 to get (k+3)(6k2)2k−1.(k+3)(6k2)2k−1+k+32(2k−1)⋅−3Remove parentheses around 6k2.(k+3)⋅6k22k−1+k+32(2k−1)⋅−3Simplify k+32(2k−1)⋅−3.More Steps(k+3)⋅6k22k−1+(k+3)⋅−32(2k−1)Simplify each term.More Steps6(k+3)k22k−1−3(k+3)2(2k−1)To write 6(k+3)k22k−1 as a fraction with a common denominator, multiply by 22.6(k+3)k22k−122−3(k+3)2(2k−1)Write each expression with a common denominator of (2k−1)⋅2, by multiplying each by an appropriate factor of 1.More Steps6(k+3)k2⋅22(2k−1)−3(k+3)2(2k−1)Combine the numerators over the common denominator.6(k+3)k2⋅2−(3(k+3))2(2k−1)Multiply k+32k−1 and 6k21 to get (k+3)(6k2)2k−1.(k+3)(6k2)2k−1+k+32(2k−1)⋅−3k+32k−16k21+k+32(2k−1)Factor 2 out of 4k−2.k+3/2(2k−1)⋅(12k2−3)Multiply k+3/2(2k−1) by 12k2−3 to get k+3/2(2k−1)(12k2−3).k+3/2(2k−1)(12k2−3)Apply the distributive property.k+3/2(2k−1)(12k2)+k+3/2(2k−1)⋅−3

k+3/2k−16k21+k+3/2(2k−1)⋅−3
Remove parentheses around 6k2.(k+3)⋅6k2/2k−1+k+3/2(2k−1)⋅−3Simplify k+32(2k−1)⋅−3.
6(k+3)k22k−1−3(k+3)2(2k−1)To write 6(k+3)k2/2k−1 as a fraction with a common denominator, multiply by 2/2.6(k+3)k22k−122−3(k+3)2(2k−1)

k+3/4k-2*(12k^2-3)Popular Problems > Algebra > SimplifyProblem:Simplify

 k+34k−2⋅(12k2−3) Step-By-Step Solution:Factor 2 out of 4k−2.k+32(2k−1)⋅(12k2−3)Multiply k+32(2k−1) by 12k2−3 to get k+32(2k−1)(12k2−3).k+32(2k−1)(12k2−3)Apply the distributive property.k+32(2k−1)(12k2)+k+32(2k−1)⋅−3Cancel the common factor of 2.More Stepsk+32k−16k21+k+32(2k−1)⋅−3Simplify.Less StepsMultiply k+32k−1 and 6k21 to get (k+3)(6k2)2k−1.(k+3)(6k2)2k−1+k+32(2k−1)⋅−3Remove parentheses around 6k2.(k+3)⋅6k22k−1+k+32(2k−1)⋅−3Simplify k+32(2k−1)⋅−3.More Steps(k+3)⋅6k22k−1+(k+3)⋅−32(2k−1)Simplify each term.More Steps6(k+3)k22k−1−3(k+3)2(2k−1)To write 6(k+3)k22k−1 as a fraction with a common denominator, multiply by 22.6(k+3)k22k−122−3(k+3)2(2k−1)Write each expression with a common denominator of (2k−1)⋅2, by multiplying each by an appropriate factor of 1.More Steps6(k+3)k2⋅2/2(2k−1)−3(k+3)2(2k−1)Combine the numerators over the common denominator.6(k+3)k2⋅2−(3(k+3))/2(2k−1)
3(k+3)(2k+1)/2

Answer 2

$\dfrac{k+3}{4k-2}\cdot(12k^2-3)=\dfrac{k(12k^2)+k(-3)+3(12k^2)+3(-3)}{4k-2}\\\\=\dfrac{12k^3-3k+36k^2-9}{4k-2}=\dfrac{12k^3+36k^2-3k-9}{4k-2}$