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In order to find the number of 6-number combinations out of 49 numbers, I need to consider that each number can be drawn only once and that the order in which they are drawn doesn’t matter.
Then, I can conclude that when I am picking 6 numbers, my first choice can include any of 49 numbers available, second choice – any of 48 numbers available (one number is already drawn, so it is not available any more), third choice – any of 47 numbers available, fourth choice – any of 46 numbers available, fifth choice – any of 45 numbers available, sixth choice – any of 44 numbers available.

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As zero remainder is received for the synthetic division by -4, x = -4 is zero and (x+4) is the factor of the given polynomial. However, the subsequent division by 2 produced the non-zero remainder. x=2 is not a zero.
a) For f(x) = 2x 3 -5x2-4x+3 from the Rational roots test we can suggest that +/-3 (multiple of the factor 3/2) can be the possible solutions. Using synthetic division we find that x=3 is a zero of the polynomial:

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