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kifflom [539]
3 years ago
5

a six sided die is rolled once what is the probability of rolling the die and landing on number 5 first and then rolling it agai

n and landing on 3? pls help ​

Mathematics
1 answer:
Natalka [10]3 years ago
5 0

Answer:

i got into B or C

Step-by-step explanation:

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P(x)= 3x^3-5x^2-14x-4
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\displaystyle\\&#10;P(x)=3x^3-5x^2-14x-4\\\\&#10;D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\&#10;\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\&#10;3x^3-5x^2-14x-4=0\\\\&#10;

\displaystyle\\&#10;\text{Verification}\\\\&#10;3x^3-5x^2-14x-4=\\\\&#10;=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\&#10;=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\&#10; =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\&#10;\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\&#10;\Longrightarrow~~~P(x)~\vdots~(3x+1)


\displaystyle\\&#10;3x^3-5x^2-14x-4=0\\&#10;~~~~~-5x^2 = x^2 - 6x^2\\&#10;~~~~~-14x =-2x-12x \\&#10;3x^3+x^2 - 6x^2-2x-12x-4=0\\&#10;x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\&#10;(3x+1)(x^2-2x -4)=0\\\\&#10;\text{Solve: } x^2-2x -4=0\\\\&#10;x_{12}= \frac{-b\pm  \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm  \sqrt{4+16}}{2}=\frac{2\pm  \sqrt{20}}{2}=\frac{2\pm  2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\&#10;x_1 =1+\sqrt{5}\\&#10;x_2 =1-\sqrt{5}\\&#10;\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}



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