Question
Answer
$$-2(x+1)^4(x-2)^3(x+3)=-2x^8-2x^7+24x^6+20x^5-82x^4-90x^3+76x^2+136x+48$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-2(x+1)^4(x-2)^3(x+3)& \xlongequal{ }-2(x^4+4x^3+6x^2+4x+1)(x^3-6x^2+12x-8)(x+3) \xlongequal{ } \\[1 em] & \xlongequal{ }-(2x^4+8x^3+12x^2+8x+2)(x^3-6x^2+12x-8)(x+3) \xlongequal{ } \\[1 em] & \xlongequal{ }-(2x^7-4x^6-12x^5+16x^4+34x^3-12x^2-40x-16)(x+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-(2x^8+2x^7-24x^6-20x^5+82x^4+90x^3-76x^2-136x-48) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-2x^8-2x^7+24x^6+20x^5-82x^4-90x^3+76x^2+136x+48\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2x^7-4x^6-12x^5+16x^4+34x^3-12x^2-40x-16}\right) $ by each term in $ \left( x+3\right) $. $$ \left( \color{blue}{2x^7-4x^6-12x^5+16x^4+34x^3-12x^2-40x-16}\right) \cdot \left( x+3\right) = \\ = 2x^8+6x^7-4x^7-12x^6-12x^6-36x^5+16x^5+48x^4+34x^4+102x^3-12x^3-36x^2-40x^2-120x-16x-48 $$ |
| ② | Combine like terms: $$ 2x^8+ \color{blue}{6x^7} \color{blue}{-4x^7} \color{red}{-12x^6} \color{red}{-12x^6} \color{green}{-36x^5} + \color{green}{16x^5} + \color{orange}{48x^4} + \color{orange}{34x^4} + \color{blue}{102x^3} \color{blue}{-12x^3} \color{red}{-36x^2} \color{red}{-40x^2} \color{green}{-120x} \color{green}{-16x} -48 = \\ = 2x^8+ \color{blue}{2x^7} \color{red}{-24x^6} \color{green}{-20x^5} + \color{orange}{82x^4} + \color{blue}{90x^3} \color{red}{-76x^2} \color{green}{-136x} -48 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left(2x^8+2x^7-24x^6-20x^5+82x^4+90x^3-76x^2-136x-48 \right) = -2x^8-2x^7+24x^6+20x^5-82x^4-90x^3+76x^2+136x+48 $$ |
This page was created using
Simplify polynomials calculator