Answer
$$(x-(3-4i))(x-5)(x+1)=4ix^2+x^3-16ix-7x^2-20i+7x+15$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-(3-4i))(x-5)(x+1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x-3+4i)(x-5)(x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x^2-5x-3x+15+4ix-20i)(x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(4ix+x^2-20i-8x+15)(x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}4ix^2+x^3-16ix-7x^2-20i+7x+15\end{aligned} $$ | |
| ① | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3-4i \right) = -3+4i $$ |
| ② | Multiply each term of $ \left( \color{blue}{x-3+4i}\right) $ by each term in $ \left( x-5\right) $. $$ \left( \color{blue}{x-3+4i}\right) \cdot \left( x-5\right) = x^2-5x-3x+15+4ix-20i $$ |
| ③ | Combine like terms: $$ x^2 \color{blue}{-5x} \color{blue}{-3x} +15+4ix-20i = 4ix+x^2-20i \color{blue}{-8x} +15 $$ |
| ④ | Multiply each term of $ \left( \color{blue}{4ix+x^2-20i-8x+15}\right) $ by each term in $ \left( x+1\right) $. $$ \left( \color{blue}{4ix+x^2-20i-8x+15}\right) \cdot \left( x+1\right) = 4ix^2+4ix+x^3+x^2-20ix-20i-8x^2-8x+15x+15 $$ |
| ⑤ | Combine like terms: $$ 4ix^2+ \color{blue}{4ix} +x^3+ \color{red}{x^2} \color{blue}{-20ix} -20i \color{red}{-8x^2} \color{green}{-8x} + \color{green}{15x} +15 = \\ = 4ix^2+x^3 \color{blue}{-16ix} \color{red}{-7x^2} -20i+ \color{green}{7x} +15 $$ |
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