Question
Answer
$$(ix-i-3)(ix+2i-5)\cdot0=0$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(ix-i-3)(ix+2i-5)\cdot0& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(1i^2x^2+i^2x-2i^2-8ix-i+15)\cdot0 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}0i^2x^2+0i^2x+0i^2+0ix+0i+0 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}0\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{ix-i-3}\right) $ by each term in $ \left( ix+2i-5\right) $. $$ \left( \color{blue}{ix-i-3}\right) \cdot \left( ix+2i-5\right) = i^2x^2+2i^2x-5ix-i^2x-2i^2+5i-3ix-6i+15 $$ |
| ② | Combine like terms: $$ i^2x^2+ \color{blue}{2i^2x} \color{red}{-5ix} \color{blue}{-i^2x} -2i^2+ \color{green}{5i} \color{red}{-3ix} \color{green}{-6i} +15 = \\ = i^2x^2+ \color{blue}{i^2x} -2i^2 \color{red}{-8ix} \color{green}{-i} +15 $$ |
| ③ | $$ \left( \color{blue}{i^2x^2+i^2x-2i^2-8ix-i+15}\right) \cdot 0 = 0i^2x^20i^2x0i^20ix0i0 $$ |
| ④ | Combine like terms: $$ 0 = 0 $$ |
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