Question
Answer
$$\dfrac{(k+1)(2k+1)(2k+3)}{3}=\dfrac{4k^3+12k^2+11k+3}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(k+1)(2k+1)(2k+3)}{3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{(2k^2+k+2k+1)(2k+3)}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{(2k^2+3k+1)(2k+3)}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4k^3+6k^2+6k^2+9k+2k+3}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{4k^3+12k^2+11k+3}{3}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{k+1}\right) $ by each term in $ \left( 2k+1\right) $. $$ \left( \color{blue}{k+1}\right) \cdot \left( 2k+1\right) = 2k^2+k+2k+1 $$ |
| ② | Combine like terms: $$ 2k^2+ \color{blue}{k} + \color{blue}{2k} +1 = 2k^2+ \color{blue}{3k} +1 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{2k^2+3k+1}\right) $ by each term in $ \left( 2k+3\right) $. $$ \left( \color{blue}{2k^2+3k+1}\right) \cdot \left( 2k+3\right) = 4k^3+6k^2+6k^2+9k+2k+3 $$ |
| ④ | Simplify numerator $$ 4k^3+ \color{blue}{6k^2} + \color{blue}{6k^2} + \color{red}{9k} + \color{red}{2k} +3 = 4k^3+ \color{blue}{12k^2} + \color{red}{11k} +3 $$ |
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