Question
Answer
$$(2x^2-x+4)(3x^2+x+4)=6x^4-x^3+19x^2+16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x^2-x+4)(3x^2+x+4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^4-x^3+19x^2+16\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2x^2-x+4}\right) $ by each term in $ \left( 3x^2+x+4\right) $. $$ \left( \color{blue}{2x^2-x+4}\right) \cdot \left( 3x^2+x+4\right) = \\ = 6x^4+2x^3+8x^2-3x^3-x^2 -\cancel{4x}+12x^2+ \cancel{4x}+16 $$ |
| ② | Combine like terms: $$ 6x^4+ \color{blue}{2x^3} + \color{red}{8x^2} \color{blue}{-3x^3} \color{green}{-x^2} \, \color{orange}{ -\cancel{4x}} \,+ \color{green}{12x^2} + \, \color{orange}{ \cancel{4x}} \,+16 = 6x^4 \color{blue}{-x^3} + \color{green}{19x^2} +16 $$ |
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