Question
Answer
$$(r-5)(4r^2+4r-7)=4r^3-16r^2-27r+35$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(r-5)(4r^2+4r-7)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4r^3+4r^2-7r-20r^2-20r+35 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4r^3-16r^2-27r+35\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{r-5}\right) $ by each term in $ \left( 4r^2+4r-7\right) $. $$ \left( \color{blue}{r-5}\right) \cdot \left( 4r^2+4r-7\right) = 4r^3+4r^2-7r-20r^2-20r+35 $$ |
| ② | Combine like terms: $$ 4r^3+ \color{blue}{4r^2} \color{red}{-7r} \color{blue}{-20r^2} \color{red}{-20r} +35 = 4r^3 \color{blue}{-16r^2} \color{red}{-27r} +35 $$ |
This page was created using
Simplify polynomials calculator