Question
Answer
$$\dfrac{2(x-1)(x-3)(x-4)(x-5)}{(2-1)\cdot(2-3)\cdot(2-4)\cdot(2-5)}=\dfrac{2x^4-26x^3+118x^2-214x+120}{-6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2(x-1)(x-3)(x-4)(x-5)}{(2-1)\cdot(2-3)\cdot(2-4)\cdot(2-5)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{(2x-2)(x-3)(x-4)(x-5)}{\frac{8-16-12+24-4+8+6-12}{1}\cdot(-3)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{(2x^2-6x-2x+6)(x-4)(x-5)}{\frac{2}{1}\cdot(-3)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{(2x^2-8x+6)(x-4)(x-5)}{2\cdot(-3)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{(2x^3-8x^2-8x^2+32x+6x-24)(x-5)}{-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} } }}}\frac{(2x^3-16x^2+38x-24)(x-5)}{-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle11}{\textcircled {11}} \htmlClass{explanationCircle explanationCircle12}{\textcircled {12}} } }}}\frac{2x^4-26x^3+118x^2-214x+120}{-6}\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{2} $ by $ \left( x-1\right) $ $$ \color{blue}{2} \cdot \left( x-1\right) = 2x-2 $$ |
| ② | $$ \color{blue}{ \left( 4-6-2 + 3\right) } \cdot \left( 2-4\right) = \color{blue}{4} \cdot2+\color{blue}{4} \cdot-4\color{blue}{-6} \cdot2\color{blue}{-6} \cdot-4\color{blue}{-2} \cdot2\color{blue}{-2} \cdot-4+\color{blue}{3} \cdot2+\color{blue}{3} \cdot-4 = \\ = 8-16-12 + 24-4 + 8 + 6-12 $$$$ \color{blue}{ 1 } \cdot 1 = 1 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{2x-2}\right) $ by each term in $ \left( x-3\right) $. $$ \left( \color{blue}{2x-2}\right) \cdot \left( x-3\right) = 2x^2-6x-2x+6 $$ |
| ④ | Simplify numerator $$ \color{blue}{8} \color{red}{-16} \color{green}{-12} + \color{orange}{24} \color{blue}{-4} + \color{red}{8} + \color{green}{6} \color{green}{-12} = \color{green}{2} $$ |
| ⑤ | Combine like terms: $$ 2x^2 \color{blue}{-6x} \color{blue}{-2x} +6 = 2x^2 \color{blue}{-8x} +6 $$ |
| ⑥ | Remove 1 from denominator. |
| ⑦ | Multiply each term of $ \left( \color{blue}{2x^2-8x+6}\right) $ by each term in $ \left( x-4\right) $. $$ \left( \color{blue}{2x^2-8x+6}\right) \cdot \left( x-4\right) = 2x^3-8x^2-8x^2+32x+6x-24 $$ |
| ⑧ | $ 2 \cdot ( -3 ) = -6 $ |
| ⑨ | Combine like terms: $$ 2x^3 \color{blue}{-8x^2} \color{blue}{-8x^2} + \color{red}{32x} + \color{red}{6x} -24 = 2x^3 \color{blue}{-16x^2} + \color{red}{38x} -24 $$ |
| ⑩ | $ 2 \cdot ( -3 ) = -6 $ |
| ⑪ | Multiply each term of $ \left( \color{blue}{2x^3-16x^2+38x-24}\right) $ by each term in $ \left( x-5\right) $. $$ \left( \color{blue}{2x^3-16x^2+38x-24}\right) \cdot \left( x-5\right) = 2x^4-10x^3-16x^3+80x^2+38x^2-190x-24x+120 $$ |
| ⑫ | Combine like terms: $$ 2x^4 \color{blue}{-10x^3} \color{blue}{-16x^3} + \color{red}{80x^2} + \color{red}{38x^2} \color{green}{-190x} \color{green}{-24x} +120 = \\ = 2x^4 \color{blue}{-26x^3} + \color{red}{118x^2} \color{green}{-214x} +120 $$ |
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