answer:1. **Rational Root Theorem**: - The possible rational roots are the factors of the constant term (pm 1) over the factors of the leading coefficient (pm 1), giving potential roots of pm1. 2. **Descartes' Rule of Signs**: - For positive roots, evaluate the sign changes of the terms from highest to lowest degree: - x^4 (positive) to -6x^3 (negative) to 11x^2 (positive) to -6x (negative) to 1 (positive). - Sign changes: 4 (positive to negative to positive to negative to positive). - By Descartes' Rule of Signs, there could be 4 or 2 positive roots. - For negative roots, substitute x = -x: - (-x)^4 (positive) to 6(-x)^3 (negative) to 11(-x)^2 (positive) to 6(-x) (negative) to 1 (positive). - Sign changes: 3 (positive to negative to positive to negative to positive). - By Descartes' Rule of Signs, there could be 3 or 1 negative roots. 3. **Testing Rational Roots**: - Testing x = 1: (1)^4 - 6(1)^3 + 11(1)^2 - 6(1) + 1 = 1 - 6 + 11 - 6 + 1 = 1 (not a root). - Testing x = -1: (-1)^4 - 6(-1)^3 + 11(-1)^2 - 6(-1) + 1 = 1 + 6 + 11 + 6 + 1 = 25 (not a root). 4. **Conclusion**: - Based on Descartes' Rule, the polynomial should have 2 or 4 positive roots and 1 or 3 negative roots. However, rational roots testing shows no solutions at pm 1. Further testing or numerical methods may be required for irrational roots. - text{2 or 4 positive and 1 or 3 negative roots} The final answer is boxed{D) 1 or 3 negative roots, with irrational positive roots.}

answer:Let's denote the lengths of the segments between the points as follows: AB = x (the length we want to find) BC = y CD = z DE = 8 (given) Since BC = 3CD, we can write y = 3z. We are also given that AC = 11, which means AB + BC = x + y = 11. The length of AE is given as 21, which means AB + BC + CD + DE = x + y + z + 8 = 21. Now we have two equations: 1) x + y = 11 2) x + y + z + 8 = 21 From equation 2, we can subtract 8 from both sides to get: x + y + z = 13 Now we have two equations with three variables (x, y, z), but we know that y = 3z, so we can substitute y with 3z in both equations: 1) x + 3z = 11 2) x + 3z + z = 13 From equation 2, we can simplify to get: x + 4z = 13 Now we have two equations with two variables: 1) x + 3z = 11 2) x + 4z = 13 Subtract equation 1 from equation 2 to find the value of z: (x + 4z) - (x + 3z) = 13 - 11 x + 4z - x - 3z = 2 z = 2 Now that we have z, we can find y since y = 3z: y = 3 * 2 = 6 Now we can find x using equation 1: x + 3z = 11 x + 3 * 2 = 11 x + 6 = 11 x = 11 - 6 x = 5 Therefore, the length of AB is boxed{5} units.

answer:If Mina memorized 24 digits of pi and that is six times as many as Carlos memorized, we can find out how many digits Carlos memorized by dividing the number of digits Mina memorized by 6. Carlos memorized: 24 / 6 = 4 digits of pi. Since Sam memorized six more digits than Carlos, we can find out how many digits Sam memorized by adding 6 to the number of digits Carlos memorized. Sam memorized: 4 + 6 = 10 digits of pi. Nina memorized four times as many digits as Carlos, so we can find out how many digits Nina memorized by multiplying the number of digits Carlos memorized by 4. Nina memorized: 4 * 4 = 16 digits of pi. Now, to find the total sum of the digits memorized by Sam, Carlos, Mina, and Nina, we add the number of digits each person memorized: Total sum = Sam + Carlos + Mina + Nina Total sum = 10 + 4 + 24 + 16 Total sum = 54 Therefore, the total sum of the digits memorized by Sam, Carlos, Mina, and Nina is boxed{54} .

answer:Given the quadratic equation x^{2}-2x+m=1 and knowing that one of its roots is 1, we can substitute x=1 into the equation to find the value of m. Starting with the equation: x^{2}-2x+m=1 Substituting x=1: 1^{2}-2times 1+m=1 Simplifying the equation: 1-2+m=1 Further simplification leads to: -1+m=1 Solving for m: m=1+1 m=2 Therefore, the value of m that satisfies the condition is boxed{D}.