Question
Answer
$$(3x^2-1)(2x^2-5x+3)+12=6x^4-15x^3+7x^2+5x+9$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3x^2-1)(2x^2-5x+3)+12& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}6x^4-15x^3+9x^2-2x^2+5x-3+12 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^4-15x^3+7x^2+5x+9\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3x^2-1}\right) $ by each term in $ \left( 2x^2-5x+3\right) $. $$ \left( \color{blue}{3x^2-1}\right) \cdot \left( 2x^2-5x+3\right) = 6x^4-15x^3+9x^2-2x^2+5x-3 $$ |
| ② | Combine like terms: $$ 6x^4-15x^3+ \color{blue}{9x^2} \color{blue}{-2x^2} +5x \color{red}{-3} + \color{red}{12} = 6x^4-15x^3+ \color{blue}{7x^2} +5x+ \color{red}{9} $$ |
This page was created using
Simplify polynomials calculator