Question
Answer
$$(x^2+6x-3)(3x^2-4x+2)=3x^4+14x^3-31x^2+24x-6$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x^2+6x-3)(3x^2-4x+2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3x^4+14x^3-31x^2+24x-6\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^2+6x-3}\right) $ by each term in $ \left( 3x^2-4x+2\right) $. $$ \left( \color{blue}{x^2+6x-3}\right) \cdot \left( 3x^2-4x+2\right) = 3x^4-4x^3+2x^2+18x^3-24x^2+12x-9x^2+12x-6 $$ |
| ② | Combine like terms: $$ 3x^4 \color{blue}{-4x^3} + \color{red}{2x^2} + \color{blue}{18x^3} \color{green}{-24x^2} + \color{orange}{12x} \color{green}{-9x^2} + \color{orange}{12x} -6 = \\ = 3x^4+ \color{blue}{14x^3} \color{green}{-31x^2} + \color{orange}{24x} -6 $$ |
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