Question
Answer
$$(x^2-3x)(4x^2+2x-9)=4x^4-10x^3-15x^2+27x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x^2-3x)(4x^2+2x-9)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4x^4+2x^3-9x^2-12x^3-6x^2+27x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4x^4-10x^3-15x^2+27x\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^2-3x}\right) $ by each term in $ \left( 4x^2+2x-9\right) $. $$ \left( \color{blue}{x^2-3x}\right) \cdot \left( 4x^2+2x-9\right) = 4x^4+2x^3-9x^2-12x^3-6x^2+27x $$ |
| ② | Combine like terms: $$ 4x^4+ \color{blue}{2x^3} \color{red}{-9x^2} \color{blue}{-12x^3} \color{red}{-6x^2} +27x = 4x^4 \color{blue}{-10x^3} \color{red}{-15x^2} +27x $$ |
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