Question
Answer
$$(x+3)^4-15(x+3)^3+91(x+3)^2-249(x+3)+252=x^4-3x^3+10x^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+3)^4-15(x+3)^3+91(x+3)^2-249(x+3)+252& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^4+12x^3+54x^2+108x+81-15(x^3+9x^2+27x+27)+91(x^2+6x+9)-249(x+3)+252 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}x^4+12x^3+54x^2+108x+81-(15x^3+135x^2+405x+405)+91x^2+546x+819-(249x+747)+252 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}x^4+12x^3+54x^2+108x+81-15x^3-135x^2-405x-405+91x^2+546x+819-(249x+747)+252 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}x^4-3x^3-81x^2-297x-324+91x^2+546x+819-(249x+747)+252 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}x^4-3x^3+10x^2+249x+495-(249x+747)+252 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}x^4-3x^3+10x^2+249x+495-249x-747+252 \xlongequal{ } \\[1 em] & \xlongequal{ }x^4-3x^3+10x^2+ \cancel{249x}+495 -\cancel{249x}-747+252 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} } }}}x^4-3x^3+10x^2\end{aligned} $$ | |
| ① | $$ (x+3)^4 = (x+3)^2 \cdot (x+3)^2 $$ |
| ② | Find $ \left(x+3\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(x+3\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 3 + \color{red}{3^2} = x^2+6x+9\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{x^2+6x+9}\right) $ by each term in $ \left( x^2+6x+9\right) $. $$ \left( \color{blue}{x^2+6x+9}\right) \cdot \left( x^2+6x+9\right) = x^4+6x^3+9x^2+6x^3+36x^2+54x+9x^2+54x+81 $$ |
| ④ | Combine like terms: $$ x^4+ \color{blue}{6x^3} + \color{red}{9x^2} + \color{blue}{6x^3} + \color{green}{36x^2} + \color{orange}{54x} + \color{green}{9x^2} + \color{orange}{54x} +81 = \\ = x^4+ \color{blue}{12x^3} + \color{green}{54x^2} + \color{orange}{108x} +81 $$Find $ \left(x+3\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = x $ and $ B = 3 $. $$ \left(x+3\right)^3 = x^3+3 \cdot x^2 \cdot 3 + 3 \cdot x \cdot 3^2+3^3 = x^3+9x^2+27x+27 $$Find $ \left(x+3\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(x+3\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 3 + \color{red}{3^2} = x^2+6x+9\end{aligned} $$ |
| ⑤ | Multiply $ \color{blue}{15} $ by $ \left( x^3+9x^2+27x+27\right) $ $$ \color{blue}{15} \cdot \left( x^3+9x^2+27x+27\right) = 15x^3+135x^2+405x+405 $$Multiply $ \color{blue}{91} $ by $ \left( x^2+6x+9\right) $ $$ \color{blue}{91} \cdot \left( x^2+6x+9\right) = 91x^2+546x+819 $$Multiply $ \color{blue}{249} $ by $ \left( x+3\right) $ $$ \color{blue}{249} \cdot \left( x+3\right) = 249x+747 $$ |
| ⑥ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 15x^3+135x^2+405x+405 \right) = -15x^3-135x^2-405x-405 $$ |
| ⑦ | Combine like terms: $$ x^4+ \color{blue}{12x^3} + \color{red}{54x^2} + \color{green}{108x} + \color{orange}{81} \color{blue}{-15x^3} \color{red}{-135x^2} \color{green}{-405x} \color{orange}{-405} = \\ = x^4 \color{blue}{-3x^3} \color{red}{-81x^2} \color{green}{-297x} \color{orange}{-324} $$ |
| ⑧ | Combine like terms: $$ x^4-3x^3 \color{blue}{-81x^2} \color{red}{-297x} \color{green}{-324} + \color{blue}{91x^2} + \color{red}{546x} + \color{green}{819} = \\ = x^4-3x^3+ \color{blue}{10x^2} + \color{red}{249x} + \color{green}{495} $$ |
| ⑨ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 249x+747 \right) = -249x-747 $$ |
| ⑩ | Combine like terms: $$ x^4-3x^3+10x^2+ \, \color{blue}{ \cancel{249x}} \,+ \color{green}{495} \, \color{blue}{ -\cancel{249x}} \, \color{orange}{-747} + \color{orange}{252} = x^4-3x^3+10x^2 $$ |
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