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valentina_108 [34]
3 years ago
10

Use the Rational Root Theorem to list all possible rational roots of the polynomial equation x3 – x2 – x – 3 = 0. Do not find th

e actual roots. (1 point)
–3, –1, 1, 3
1, 3
–33
no roots
Mathematics
2 answers:
almond37 [142]3 years ago
7 0
Hello,
<span>List all possible rational roots of the polynomial equation x3 – x2 – x – 3 = 0

Answer -1,1,-3,3
but none is a root.
</span>
Sauron [17]3 years ago
3 0

The Cubic Polynomial  given here is

x³ - x² - x -3 =0

Rational root theorem :

Consider any polynomial

→ Ax^n +A_{1}x^{n-1}+A_{2}x^{n-2}+.................+A_{n}=0

So, the factors of this polynomial is those number which divides \frac{A_{n}}{A}.Which are \pm1, \pm\frac{A_{n}}{A}

Applying the same method the number which divides (-3) are,-3,+3,-1,+1. So By rational root theorem all the factors of this cubic polynomial is \pm1,\pm3.

Option 1. -3,-1,1,3 is correct option.



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Which expression is equivalent to *picture attached*
DiKsa [7]

Answer:

The correct option is;

4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3  \left (\dfrac{50(51) }{2} \right )

Step-by-step explanation:

The given expression is presented as follows;

\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3  \right )

Which can be expanded into the following form;

\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3  \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left  n^2 + 3  \times\sum\limits _{n = 1}^{50}  n

From which we have;

\sum\limits _{k = 1}^{n} \left  k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}

\sum\limits _{k = 1}^{n} \left  k = \dfrac{n \times (n+1) }{2}

Therefore, substituting the value of n = 50 we have;

\sum\limits _{n = 1}^{50} \left  k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}

\sum\limits _{k = 1}^{50} \left  k = \dfrac{50 \times (50+1) }{2}

Which gives;

4 \times \sum\limits _{n = 1}^{50} \left  n^2 =  4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}

3  \times\sum\limits _{n = 1}^{50}  n = 3  \times \dfrac{n \times (n+1) }{2} = 3  \times \dfrac{50 \times (51) }{2}

\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3  \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3  \times \dfrac{50 \times (51) }{2}

Therefore, we have;

4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3  \left (\dfrac{50(51) }{2} \right ).

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