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Can someone show and give the answers to these math problems using Decartes' rule of signs?

P(x)=2x^4+x^3-14x^2-19x-6 AND P(x)=x^4-x^3+7x^2-9x-18 Using Decartes' rule of signs, determine the number of possible positive,negative, or imaginary zeros. List all real zeros. Find all the zeros.

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1. +2x^4+x^3-14x^2-19x-6 The leading term is positive. One instance of sign changing ( a positive number next to a negative number WHEN IN ORDER) so there is One positive root. Then we replace x with -x and simplify. (All even exponents with x will stay the same, but all odd exponents will have their value switched from negative to positive or positive to negative. All numbers without x will stay the same) +2x^4-x^3-14x^2+19x-6 There are 3 instances of sign changes, so we have 3 possible negative roots. Since Descartes' rule of signs says to subtract two from our possible number until we reach 0 or -1 to find every other possibility, we also get 1 negative root as possible (3-2). So the choices are 1 positive, 3 negative, 0 imaginary (since the exponent is 4, there can only be 4 roots) OR 1 positive, 1 negative, and 2 imaginary. To find the list of possible real zeros, you take the last number in the original equation (a 6) and divide its factors by the factors of the first coefficient in the equation (a 2). So we have 1, 2, 3, and 6 being divided by 1 and 2. The possible real zeros are then 1, 2, 3, 6, ½, 2/2 (or simply 1 again), 3/2, and 6/2 (or simply 3 again) THE OPPOSITE/NEGATIVE VALUES ARE ALSO POSSIBILITIES! using synthetic division (explained http://www.purplemath.com/modules/synthdiv.htm ) we must find enough roots to expand the equation to become all quadratic and linear terms. After using synthetic division to test the real possible zeros (the correct possibilities will give you a remainder of 0) we'll end up with (x+2)(x+1)(x-3)(x+ ½), or zeros of -2, -1, -1/2, and 3. The second equation is positive with 3sign changes. There are either 3 or 1 (3-2) positive roots. Replacing x with -x, we get x^4+x^3+7x^2+9x-18 so one sign change. There will be 1 negative root. We have either 3 (2+1) roots or 1 root (0+1) accounted for, but we need 4, so we are left with 3 positive, 1 negative, 0 imaginary or 1 positive, 1 negative, 2 imaginary. The possible zeros are all the factors of 18, divided by all the factors of 1, so 1, 2, 3, 6, 9, 18 and their opposites. Use synthetic division to reduce the equation to (x+1)(x-2)(x^2+9) So use the quadratic formula of (-b + (b^2-4ac)^1/2)/(2a)= x and (-b – (b^2-4ac)^1/2)/(2a) = x to find the other two zeros. In this case a = 1, b = 0, and c = 9. we get (0+(0^2-4(1)(9)^1/2) / (2(1)) and (0-(0^2-4(1)(9)^1/2) / (2(1)) simplifies to (+(0-36)^1/2) / 2 and (-(0-36)^1/2) / 2 simplifies to ((-36)^1/2) / 2 and (-(-36)^1/2) / 2 simplifies to 6i/2 and -6i/2 simplifies to 3i and -3i. The zeros are -3i, -1, 2, and 3i.