answer：To determine Minji's delivery speed after 3 weeks of training, I'll substitute ( x = 3 ) into the polynomial function ( f(x) = 2x^3 - 5x^2 + 3x + 4 ).First, calculate ( 2x^3 ):( 2(3)^3 = 2(27) = 54 ).Next, calculate ( -5x^2 ):( -5(3)^2 = -5(9) = -45 ).Then, calculate ( 3x ):( 3(3) = 9 ).Finally, add the constant term 4.Adding all these together:( 54 - 45 + 9 + 4 = 22 ).So, Minji's delivery speed after 3 weeks is 22 meters per second.For the recovery period modeled by ( g(x) = x^2 - 6x + 8 ), I'll solve the quadratic equation ( x^2 - 6x + 8 = 0 ).First, factor the quadratic:( x^2 - 6x + 8 = (x - 2)(x - 4) = 0 ).Setting each factor equal to zero gives:( x - 2 = 0 ) → ( x = 2 ),( x - 4 = 0 ) → ( x = 4 ).Therefore, Minji will be fully healed and ready to return to training after 2 weeks and 4 weeks of recovery.

answer：Alright, so I have this problem about user complaints related to untested features in software. The equation given is a quadratic one: ( C = 2x^2 - 5x + 3 ), where ( C ) is the number of complaints and ( x ) is the number of untested features. First, the problem asks me to find the number of untested features that would result in zero user complaints. That means I need to solve for ( x ) when ( C = 0 ). So, I'll set up the equation:( 2x^2 - 5x + 3 = 0 )Hmm, quadratic equations. I remember they can be solved using the quadratic formula, factoring, or completing the square. Let me see if this one factors nicely. The quadratic is ( 2x^2 -5x +3 ). I need two numbers that multiply to ( 2*3 = 6 ) and add up to -5. Let me think: -2 and -3 multiply to 6 and add up to -5. So, maybe I can factor it as:( 2x^2 - 2x - 3x + 3 = 0 )Grouping the terms:( (2x^2 - 2x) + (-3x + 3) = 0 )Factor out common terms:( 2x(x - 1) -3(x - 1) = 0 )Now, factor out the common binomial:( (2x - 3)(x - 1) = 0 )So, setting each factor equal to zero:1. ( 2x - 3 = 0 ) => ( x = 3/2 ) or 1.52. ( x - 1 = 0 ) => ( x = 1 )Wait, so the solutions are ( x = 1 ) and ( x = 1.5 ). But ( x ) represents the number of untested features, which should be a whole number, right? Because you can't have half a feature. Hmm, this is interesting. So, does that mean that at 1 untested feature, there are zero complaints, and also at 1.5 features? But since 1.5 isn't a whole number, maybe only ( x = 1 ) is the valid solution in this context.But let me double-check my factoring. Maybe I made a mistake there. Let's see:Original equation: ( 2x^2 -5x +3 )Trying to factor:Looking for two numbers that multiply to 6 (since 2*3) and add to -5. Those numbers are -2 and -3.So, split the middle term:( 2x^2 -2x -3x +3 )Group:( (2x^2 -2x) + (-3x +3) )Factor:( 2x(x -1) -3(x -1) )Which gives ( (2x -3)(x -1) ). Yeah, that seems right. So, the roots are at 1 and 1.5. But since we can't have half a feature, maybe only x=1 is acceptable? Or perhaps the model allows for fractional features, but in reality, it's discrete. Hmm, the problem doesn't specify, so maybe both solutions are mathematically valid, but in practical terms, only x=1 makes sense. But the question just asks to solve the equation, so perhaps both are acceptable. I'll note that.Moving on to the second part. The advocate wants to keep complaints under 8, so we need to solve the inequality:( 2x^2 -5x +3 < 8 )First, subtract 8 from both sides to set it to less than zero:( 2x^2 -5x +3 -8 < 0 )Simplify:( 2x^2 -5x -5 < 0 )So, now we have a quadratic inequality: ( 2x^2 -5x -5 < 0 ). To solve this, I need to find the roots of the quadratic equation ( 2x^2 -5x -5 = 0 ) and then determine the intervals where the quadratic is negative.Using the quadratic formula:( x = frac{5 pm sqrt{(-5)^2 - 4*2*(-5)}}{2*2} )Calculate discriminant:( D = 25 + 40 = 65 )So,( x = frac{5 pm sqrt{65}}{4} )Compute approximate values:( sqrt{65} ) is approximately 8.0623.So,( x = frac{5 + 8.0623}{4} = frac{13.0623}{4} ≈ 3.2656 )and( x = frac{5 - 8.0623}{4} = frac{-3.0623}{4} ≈ -0.7656 )So, the roots are approximately at x ≈ -0.7656 and x ≈ 3.2656.Since the quadratic opens upwards (coefficient of x² is positive), the quadratic will be below zero between the two roots. So, the solution to the inequality ( 2x^2 -5x -5 < 0 ) is:( -0.7656 < x < 3.2656 )But since x represents the number of untested features, it can't be negative. So, the valid range is:( 0 ≤ x < 3.2656 )But since x must be an integer (assuming you can't have a fraction of a feature), the possible integer values are x = 0, 1, 2, 3.Wait, but let me check if x=3 is included. Let's plug x=3 into the original inequality:( 2*(3)^2 -5*(3) +3 = 18 -15 +3 = 6 < 8 ). Yes, 6 is less than 8.x=4: ( 2*16 -20 +3 = 32 -20 +3 =15, which is greater than 8. So, x=4 is not allowed.So, the range is x from 0 to 3 inclusive.But wait, the quadratic inequality solution was x < 3.2656, so x can be up to just below 3.2656, which includes x=3 as an integer.So, summarizing:1. The number of untested features that result in zero complaints are x=1 and x=1.5, but since x must be an integer, only x=1 is valid.2. To keep complaints under 8, x must be between 0 and 3 inclusive.But wait, in the first part, if x=1.5 is a solution, but x must be an integer, so maybe x=1 is the only solution. However, if the model allows x to be a real number, then both 1 and 1.5 are solutions. The problem doesn't specify, so perhaps both are acceptable.Similarly, for the inequality, since x is a count, it's an integer, so the range is x=0,1,2,3.But let me confirm the first part again. If x=1.5 is a solution, but in reality, you can't have half a feature, so maybe only x=1 is the valid solution. Alternatively, maybe the model is considering x as a real variable, so both are acceptable. The problem doesn't specify, so perhaps both are correct.So, final answers:1. x=1 and x=1.52. x is between 0 and 3.2656, but since x is an integer, x=0,1,2,3.But let me write the exact forms for the roots in the first part.For the first equation:( 2x^2 -5x +3 =0 )Solutions are x=1 and x=3/2.So, exact solutions are x=1 and x=3/2.For the inequality, the exact roots are (5 ±√65)/4, so the solution is between (5 -√65)/4 and (5 +√65)/4.But since x must be non-negative, the interval is from 0 to (5 +√65)/4.But as an integer, x can be 0,1,2,3.So, to present the answers:1. The number of untested features resulting in zero complaints are x=1 and x=3/2.2. The range of x to keep complaints under 8 is 0 ≤ x < (5 +√65)/4, which is approximately 0 ≤ x < 3.2656. Since x must be an integer, x can be 0,1,2,3.But the problem might expect the exact form for the inequality solution, so perhaps writing it in terms of the roots.Alternatively, maybe the problem expects the answer in interval notation without considering x as integer. Let me check the problem statement again.The problem says "the number of untested features x", which is a count, so it's discrete. So, x must be a non-negative integer.Therefore, for part 1, x=1 is the only valid solution because x=1.5 isn't an integer.Wait, but the quadratic equation can have non-integer solutions, but in the context, x must be an integer. So, perhaps only x=1 is the solution where C=0.Similarly, for the inequality, x must be integers from 0 to 3 inclusive.So, perhaps in part 1, the answer is x=1, and in part 2, x=0,1,2,3.But let me verify by plugging x=1 into the original equation:C=2*(1)^2 -5*(1)+3=2-5+3=0. Correct.x=1.5: C=2*(2.25) -5*(1.5)+3=4.5-7.5+3=0. So, mathematically correct, but in practice, x=1.5 isn't possible.So, perhaps the answer expects both solutions, but in the context, only x=1 is valid.But the problem doesn't specify whether x must be an integer, so maybe both are acceptable.I think I'll present both solutions for part 1, and for part 2, present the exact interval and then note the integer solutions.So, to sum up:1. Solve ( 2x^2 -5x +3 =0 ). The solutions are x=1 and x=3/2.2. Solve ( 2x^2 -5x +3 <8 ). Subtract 8: ( 2x^2 -5x -5 <0 ). The roots are (5 ±√65)/4, approximately -0.7656 and 3.2656. Since x≥0, the solution is 0 ≤x < (5 +√65)/4 ≈3.2656. If x must be integer, then x=0,1,2,3.But the problem doesn't specify whether x must be integer, so perhaps the answer is the interval ( (5 -√65)/4, (5 +√65)/4 ), but since x can't be negative, it's [0, (5 +√65)/4 ).But let me write the exact forms.So, final answers:1. x=1 and x=3/22. 0 ≤x < (5 +√65)/4But if x must be integer, then x=0,1,2,3.But the problem doesn't specify, so perhaps both parts should be answered with exact solutions, regardless of x being integer.So, for part 1, x=1 and x=3/2.For part 2, the range is 0 ≤x < (5 +√65)/4.But let me write the exact forms.Alternatively, maybe the problem expects the answers in a specific way, like for part 1, two solutions, and for part 2, the interval.I think that's the way to go.

answer：Okay, so I'm trying to help optimize a production line. The production rate is given by this polynomial: P(x) = -2x³ + 3x² + 12x. My task is to find the critical points by taking the derivative and setting it equal to zero, and then figure out the optimal number of hours to maximize production.First, I remember that critical points occur where the derivative is zero or undefined. Since this is a polynomial, the derivative will be defined everywhere, so I just need to find where the derivative equals zero.Let me start by finding the derivative of P(x). The derivative of a polynomial term by term is straightforward. The derivative of -2x³ is -6x², right? Because you bring down the exponent as a coefficient and reduce the exponent by one. Then, the derivative of 3x² is 6x, and the derivative of 12x is 12. So putting it all together, P'(x) should be -6x² + 6x + 12.Wait, let me double-check that. Yes, that seems correct. So P'(x) = -6x² + 6x + 12.Now, I need to set this derivative equal to zero and solve for x. So the equation is:-6x² + 6x + 12 = 0.Hmm, this is a quadratic equation. I can solve it using the quadratic formula, factoring, or completing the square. Let me see if it factors first. Let's factor out a common factor. All coefficients are divisible by -6, but that might complicate things. Alternatively, I can factor out a -6, but maybe it's easier to just use the quadratic formula.The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a). In this equation, a = -6, b = 6, and c = 12.Plugging these into the formula:x = [-6 ± sqrt((6)² - 4*(-6)*12)] / (2*(-6)).Calculating the discriminant first: 6² is 36, and 4*a*c is 4*(-6)*12, which is -288. So the discriminant is 36 - (-288) = 36 + 288 = 324. That's a perfect square, which is good because it means the roots will be rational.So sqrt(324) is 18. Then, plugging back in:x = [-6 ± 18] / (-12).Let's compute both possibilities.First, with the plus sign:x = (-6 + 18) / (-12) = (12)/(-12) = -1.Second, with the minus sign:x = (-6 - 18)/(-12) = (-24)/(-12) = 2.So the critical points are at x = -1 and x = 2.But wait, x represents the number of hours the production line operates each day. It doesn't make sense for x to be negative, so x = -1 is not a feasible solution in this context. Therefore, the only critical point we need to consider is x = 2.Now, to determine whether this critical point is a maximum or a minimum, I can use the second derivative test. Let me find the second derivative P''(x).The first derivative was P'(x) = -6x² + 6x + 12. Taking the derivative of that gives P''(x) = -12x + 6.Now, evaluate P''(x) at x = 2:P''(2) = -12*(2) + 6 = -24 + 6 = -18.Since P''(2) is negative (-18), this means the function is concave down at x = 2, which indicates a local maximum. Therefore, x = 2 is the point where the production rate is maximized.So, the optimal number of hours to operate the production line each day to maximize the production rate is 2 hours.Wait, hold on a second. Let me think about this again. If the production line is operated for 2 hours, is that really the maximum? Let me check the behavior of the function.The original function is a cubic polynomial with a negative leading coefficient, which means as x approaches infinity, P(x) approaches negative infinity, and as x approaches negative infinity, P(x) approaches positive infinity. But since x can't be negative, we're only concerned with x ≥ 0.So, the function increases to a certain point and then decreases after that. The critical point at x = 2 is where it changes from increasing to decreasing, so that should indeed be the maximum.Just to be thorough, let me calculate the production rate at x = 2 and maybe at a point before and after to see the trend.At x = 0: P(0) = -2*(0)^3 + 3*(0)^2 + 12*(0) = 0.At x = 1: P(1) = -2*(1)^3 + 3*(1)^2 + 12*(1) = -2 + 3 + 12 = 13.At x = 2: P(2) = -2*(8) + 3*(4) + 12*(2) = -16 + 12 + 24 = 20.At x = 3: P(3) = -2*(27) + 3*(9) + 12*(3) = -54 + 27 + 36 = 9.So, production increases from 0 to 13 at x=1, then to 20 at x=2, and then drops to 9 at x=3. That confirms that x=2 is indeed the maximum.Therefore, the optimal number of hours is 2.

answer：Okay, so I have this problem about a cryptographer designing an encryption algorithm using a polynomial. The polynomial is given as ( P(x) = 2x^3 - 5x^2 + ax + 7 ), and they mention that one of the roots is ( x = 2 ). I need to find the secret constant ( a ) and then simplify the polynomial by dividing it by ( (x - 2) ) to get a quadratic.Starting with part 1: finding ( a ). Since ( x = 2 ) is a root, that means when I plug in 2 into the polynomial, it should equal zero. So, I can substitute ( x = 2 ) into ( P(x) ) and solve for ( a ).Let me write that out:( P(2) = 2(2)^3 - 5(2)^2 + a(2) + 7 = 0 )Calculating each term step by step:First, ( 2(2)^3 ). ( 2^3 ) is 8, so 2 times 8 is 16.Next, ( -5(2)^2 ). ( 2^2 ) is 4, so -5 times 4 is -20.Then, ( a(2) ) is just 2a.And the constant term is +7.Putting it all together:16 - 20 + 2a + 7 = 0Simplify the constants:16 - 20 is -4, and -4 + 7 is 3.So, 3 + 2a = 0Subtract 3 from both sides:2a = -3Divide both sides by 2:a = -3/2Wait, that seems straightforward. Let me double-check my calculations to make sure I didn't make a mistake.So, ( 2(2)^3 = 16 ), correct.( -5(2)^2 = -20 ), correct.Then, 16 - 20 is indeed -4, plus 7 is 3. So, 3 + 2a = 0, so a = -3/2. Yeah, that seems right.So, the value of ( a ) is -3/2.Moving on to part 2: simplifying ( P(x) ) divided by ( (x - 2) ) to get a quadratic polynomial.Since ( x = 2 ) is a root, we know that ( (x - 2) ) is a factor of ( P(x) ). So, dividing ( P(x) ) by ( (x - 2) ) should give us a quadratic polynomial without any remainder.I can use polynomial long division or synthetic division for this. Since the divisor is linear, synthetic division might be quicker.Let me set up synthetic division with root 2.The coefficients of ( P(x) ) are: 2 (for ( x^3 )), -5 (for ( x^2 )), a (which we found to be -3/2) for ( x ), and 7 as the constant term.So, writing them out: 2, -5, -3/2, 7.Bring down the 2.Multiply 2 by 2 (the root), which is 4. Add to the next coefficient: -5 + 4 = -1.Multiply -1 by 2 = -2. Add to the next coefficient: -3/2 + (-2) = -3/2 - 4/2 = -7/2.Multiply -7/2 by 2 = -7. Add to the last coefficient: 7 + (-7) = 0. Perfect, no remainder.So, the coefficients of the quotient polynomial are 2, -1, -7/2.Therefore, the quadratic polynomial is ( 2x^2 - x - frac{7}{2} ).Alternatively, to make it look nicer, I can write it as ( 2x^2 - x - frac{7}{2} ) or multiply through by 2 to eliminate the fraction, but since the question just asks to express it as a quadratic, either form is acceptable. But probably better to leave it as is.Wait, let me check if I did the synthetic division correctly.Starting with coefficients: 2, -5, -1.5, 7.Bring down the 2.Multiply 2 by 2: 4. Add to -5: -1. Correct.Multiply -1 by 2: -2. Add to -1.5: -3.5, which is -7/2. Correct.Multiply -7/2 by 2: -7. Add to 7: 0. Correct.Yes, that seems right.Alternatively, I could perform polynomial long division to verify.Dividing ( 2x^3 -5x^2 - frac{3}{2}x +7 ) by ( x - 2 ).First term: ( 2x^3 ) divided by ( x ) is ( 2x^2 ). Multiply ( 2x^2 ) by ( x - 2 ): ( 2x^3 -4x^2 ).Subtract that from the original polynomial:( (2x^3 -5x^2 - frac{3}{2}x +7) - (2x^3 -4x^2) = (-5x^2 +4x^2) - frac{3}{2}x +7 = (-x^2) - frac{3}{2}x +7 ).Next term: ( -x^2 ) divided by ( x ) is ( -x ). Multiply ( -x ) by ( x - 2 ): ( -x^2 + 2x ).Subtract that:( (-x^2 - frac{3}{2}x +7) - (-x^2 + 2x) = (- frac{3}{2}x - 2x) +7 = (- frac{7}{2}x) +7 ).Next term: ( -frac{7}{2}x ) divided by ( x ) is ( -frac{7}{2} ). Multiply ( -frac{7}{2} ) by ( x - 2 ): ( -frac{7}{2}x +7 ).Subtract that:( (-frac{7}{2}x +7) - (-frac{7}{2}x +7) = 0 ). Perfect, no remainder.So, the quotient is ( 2x^2 -x - frac{7}{2} ), same as before.Therefore, the quadratic polynomial is ( 2x^2 - x - frac{7}{2} ).Alternatively, if I wanted to write it without fractions, I could multiply each term by 2:( 4x^2 - 2x -7 ). But since the original polynomial had a coefficient of 2 for ( x^3 ), and we divided by ( x - 2 ), the leading coefficient of the quadratic is 2, so it's fine as is.So, summarizing:1. The value of ( a ) is ( -frac{3}{2} ).2. The polynomial divided by ( (x - 2) ) simplifies to ( 2x^2 - x - frac{7}{2} ).I think that's all. Let me just make sure I didn't make any arithmetic errors.For part 1:( P(2) = 2*(8) -5*(4) + a*(2) +7 = 16 -20 +2a +7 = (16 -20 +7) +2a = 3 +2a =0 implies a= -3/2 ). Correct.For part 2, both synthetic division and polynomial long division gave the same result, so that seems solid.Yeah, I think that's it.**Final Answer**1. The value of ( a ) is boxed{-dfrac{3}{2}}.2. The simplified quadratic polynomial is boxed{2x^2 - x - dfrac{7}{2}}.

answer：Okay, so I have this problem about cocoa farming yields, and I need to figure out two things. Let me take it step by step.First, the problem says that Farm A's yield is modeled by the polynomial ( Y_A(x) = x^3 - 6x^2 + 11x - 6 ). Farm B's yield is consistently 2 tons higher than Farm A's every year. I need to express ( Y_B(x) ) as a polynomial. Hmm, okay, so if Farm B is always 2 tons higher, that should be straightforward. I think I just need to add 2 to the entire polynomial of Farm A. So, ( Y_B(x) = Y_A(x) + 2 ). Let me write that out:( Y_B(x) = (x^3 - 6x^2 + 11x - 6) + 2 )Simplifying that, I combine the constants:( -6 + 2 = -4 ), so the polynomial becomes:( Y_B(x) = x^3 - 6x^2 + 11x - 4 )Wait, is that all? It seems too simple, but I think that's correct because adding a constant to a polynomial just shifts it vertically by that constant. So, yeah, adding 2 to every term's constant part. Okay, that seems right.Now, moving on to the second part. After promoting fair trade practices, both farms increased their yield by 20%. I need to express the new yield equation ( Y'_B(x) ) after this increase. Hmm, okay, so a 20% increase on Farm B's current yield. So, if the original yield is ( Y_B(x) ), then the new yield would be ( Y_B(x) ) plus 20% of ( Y_B(x) ). Alternatively, it's 1.2 times ( Y_B(x) ). Let me write that:( Y'_B(x) = 1.2 times Y_B(x) )Since we already have ( Y_B(x) = x^3 - 6x^2 + 11x - 4 ), I can substitute that in:( Y'_B(x) = 1.2 times (x^3 - 6x^2 + 11x - 4) )Now, I need to distribute the 1.2 across each term. Let me do that term by term.First term: ( 1.2 times x^3 = 1.2x^3 )Second term: ( 1.2 times (-6x^2) = -7.2x^2 )Third term: ( 1.2 times 11x = 13.2x )Fourth term: ( 1.2 times (-4) = -4.8 )So, putting it all together:( Y'_B(x) = 1.2x^3 - 7.2x^2 + 13.2x - 4.8 )Hmm, that looks good. But I wonder if I should write it with fractions instead of decimals to make it more precise. Let me see:1.2 is the same as 6/5, right? So, 1.2 = 6/5.So, let's rewrite each term using fractions:First term: ( (6/5)x^3 )Second term: ( (6/5) times (-6x^2) = (-36/5)x^2 )Third term: ( (6/5) times 11x = (66/5)x )Fourth term: ( (6/5) times (-4) = (-24/5) )So, writing it all out:( Y'_B(x) = frac{6}{5}x^3 - frac{36}{5}x^2 + frac{66}{5}x - frac{24}{5} )Alternatively, I can factor out the 6/5 if I want, but I think either form is acceptable. The problem didn't specify, so maybe decimals are fine. But sometimes, fractions are preferred in polynomial equations because they are exact. Hmm, I think I'll go with the fractional form for precision.Let me double-check my calculations to make sure I didn't make any mistakes.Starting with ( Y_B(x) = x^3 - 6x^2 + 11x - 4 ). Multiplying each term by 1.2:- ( x^3 times 1.2 = 1.2x^3 )- ( -6x^2 times 1.2 = -7.2x^2 )- ( 11x times 1.2 = 13.2x )- ( -4 times 1.2 = -4.8 )Yes, that's correct. So, converting to fractions:1.2 = 6/5, so:- ( 6/5 x^3 )- ( -36/5 x^2 )- ( 66/5 x )- ( -24/5 )Yes, that's accurate. So, both forms are correct, but fractions might be better here.Wait, another thought: is the 20% increase applied to the original Farm A's yield or to Farm B's yield? The problem says "both farms increased their yield by 20%", so I think it's applied to each farm's current yield. Since we're only asked about Farm B's new yield, we only need to apply the 20% increase to ( Y_B(x) ), which we did. So, that should be correct.I think I'm confident with these answers. Let me recap:1. ( Y_B(x) = x^3 - 6x^2 + 11x - 4 )2. ( Y'_B(x) = frac{6}{5}x^3 - frac{36}{5}x^2 + frac{66}{5}x - frac{24}{5} )Alternatively, if decimals are preferred, it's ( 1.2x^3 - 7.2x^2 + 13.2x - 4.8 ). But since the original polynomial had integer coefficients, using fractions might be more consistent.Wait, actually, looking back, the original ( Y_A(x) ) had integer coefficients, and adding 2 to get ( Y_B(x) ) also resulted in integer coefficients. Then, when we multiplied by 1.2, we got decimals. So, maybe the problem expects the answer in decimal form? Or perhaps as a polynomial with fractions? Hmm.I think either is acceptable, but since 1.2 is a decimal, maybe they expect decimals. Alternatively, if I write it as fractions, it's more precise. I think I'll present both, but probably the fractional form is better because it's exact.Wait, no, actually, 1.2 is exact as a decimal, but in fraction form, it's 6/5, which is also exact. So, both are exact. So, perhaps writing it in fractions is better because it's more precise in terms of polynomial coefficients.But, just to make sure, let me check the multiplication again:- 1.2 * x^3 = 1.2x^3- 1.2 * (-6x^2) = -7.2x^2- 1.2 * 11x = 13.2x- 1.2 * (-4) = -4.8Yes, that's correct. So, the decimal form is accurate. Alternatively, in fractions:- 6/5 x^3- -36/5 x^2- 66/5 x- -24/5Yes, that's correct as well. So, both forms are correct. Maybe I'll present both, but I think the problem expects the polynomial in terms of x, so either form is acceptable. Since the original polynomial was in integers, adding 2 kept it in integers, but multiplying by 1.2 introduced decimals. So, perhaps the answer is expected in decimals.But, to be thorough, I can present both. However, since the question didn't specify, I think either is fine. Maybe I'll go with the decimal form because it's more straightforward after the multiplication.Wait, but in the first part, we had to add 2, which was straightforward. In the second part, it's a percentage increase, which when applied to a polynomial, is a scalar multiplication. So, yes, multiplying each term by 1.2 is correct.I think I'm confident with my answers now.**Final Answer**1. The polynomial equation for Farm B's yield is boxed{x^3 - 6x^2 + 11x - 4}.2. The new polynomial equation for Farm B's yield after the increase is boxed{1.2x^3 - 7.2x^2 + 13.2x - 4.8}.

answer：Okay, so I have this problem about an irrigation system for a public park. The flow rate is given by the function ( f(t) = -2t^2 + 8t ), where ( t ) is time in hours and ( f(t) ) is the flow rate in cubic meters per hour. There are two questions here. Let me tackle them one by one.Starting with question 1: Determine the time intervals during which the flow rate is at least 6 cubic meters per hour. Hmm, so I need to find when ( f(t) geq 6 ). That means I need to solve the inequality ( -2t^2 + 8t geq 6 ).First, I'll rewrite the inequality to make it easier to solve. Subtract 6 from both sides:( -2t^2 + 8t - 6 geq 0 )Now, I have a quadratic inequality. To solve this, I should find the roots of the quadratic equation ( -2t^2 + 8t - 6 = 0 ). Let me write that down:( -2t^2 + 8t - 6 = 0 )I can simplify this equation by dividing all terms by -2 to make the coefficients smaller. Remember, dividing by a negative number will reverse the inequality sign later, but since I'm just solving the equation right now, let's do that:Divide by -2:( t^2 - 4t + 3 = 0 )Wait, hold on, actually, if I divide each term by -2, the equation becomes:( t^2 - 4t + 3 = 0 )Wait, is that correct? Let me check:-2t^2 divided by -2 is t^2.8t divided by -2 is -4t.-6 divided by -2 is 3.Yes, that's correct. So now, the equation is ( t^2 - 4t + 3 = 0 ).Now, let's factor this quadratic. Looking for two numbers that multiply to 3 and add up to -4. Hmm, factors of 3 are 1 and 3. Since the middle term is -4t, both numbers should be negative. So, -1 and -3 multiply to 3 and add up to -4. So, the factoring is:( (t - 1)(t - 3) = 0 )Wait, hold on, if I factor ( t^2 - 4t + 3 ), it's (t - 1)(t - 3) because 1*3=3 and 1+3=4, but since the middle term is -4t, it should be (t - 1)(t - 3). Let me check:( (t - 1)(t - 3) = t^2 - 3t - t + 3 = t^2 - 4t + 3 ). Yes, that's correct.So, the roots are t = 1 and t = 3.Now, going back to the inequality ( t^2 - 4t + 3 geq 0 ). Wait, no, actually, I had divided the original inequality by -2, which would reverse the inequality sign. Let me clarify.Wait, no, actually, when I divided the equation by -2, I was solving for the equality, not the inequality. So, the inequality after division would be:( t^2 - 4t + 3 leq 0 )Because when I divide both sides of an inequality by a negative number, the direction of the inequality flips. So, originally, the inequality was ( -2t^2 + 8t - 6 geq 0 ). After dividing by -2, it becomes ( t^2 - 4t + 3 leq 0 ).So, now, I need to solve ( t^2 - 4t + 3 leq 0 ). Since the quadratic opens upwards (because the coefficient of t^2 is positive), the expression will be less than or equal to zero between its roots.So, the roots are t = 1 and t = 3, so the quadratic is negative or zero between t = 1 and t = 3. Therefore, the solution to the inequality is [1, 3].But wait, let me double-check. Let me pick a test point in each interval to see where the inequality holds.The intervals are:1. t < 12. 1 < t < 33. t > 3Let's pick t = 0 for the first interval:( t^2 - 4t + 3 = 0 - 0 + 3 = 3 ), which is positive. So, the inequality ( t^2 - 4t + 3 leq 0 ) is not satisfied here.Next, t = 2:( 4 - 8 + 3 = -1 ), which is negative. So, the inequality holds here.Lastly, t = 4:( 16 - 16 + 3 = 3 ), which is positive. So, the inequality doesn't hold here.Therefore, the solution is indeed [1, 3].So, the flow rate is at least 6 cubic meters per hour between t = 1 and t = 3 hours. Therefore, the time interval is [1, 3].Moving on to question 2: If the irrigation system needs to operate for a continuous period where the flow rate is at least 6 cubic meters per hour, calculate the maximum possible duration of this period in hours.From question 1, we found that the flow rate is at least 6 cubic meters per hour between t = 1 and t = 3. So, the duration is 3 - 1 = 2 hours.Wait, is that the maximum possible? Let me think. The flow rate function is a quadratic, which is a parabola opening downward because the coefficient of t^2 is negative (-2). So, the vertex is the maximum point.The vertex occurs at t = -b/(2a). For the original function ( f(t) = -2t^2 + 8t ), a = -2, b = 8.So, t = -8/(2*(-2)) = -8/(-4) = 2.So, the maximum flow rate occurs at t = 2 hours, which is the vertex. The maximum flow rate is f(2) = -2*(4) + 8*(2) = -8 + 16 = 8 cubic meters per hour.So, the flow rate starts at t = 0, increases to a maximum at t = 2, then decreases. So, the flow rate crosses 6 cubic meters per hour at t = 1 and t = 3, as we found earlier.Therefore, the continuous period where the flow rate is at least 6 is from t = 1 to t = 3, which is 2 hours. So, the maximum possible duration is 2 hours.Wait, but just to make sure, is there any other interval where the flow rate is above 6? Since it's a downward opening parabola, it only crosses the line y=6 at two points, t=1 and t=3. So, the only interval where the flow rate is above 6 is between t=1 and t=3. Therefore, the maximum duration is indeed 2 hours.So, summarizing:1. The time intervals are [1, 3].2. The maximum duration is 2 hours.I think that's it. Let me just recap to ensure I didn't make any mistakes.For question 1, I set up the inequality correctly, solved for the roots, tested intervals, and concluded the interval [1, 3]. For question 2, I recognized that the parabola peaks at t=2, so the flow rate is above 6 only between t=1 and t=3, giving a duration of 2 hours. Seems solid.**Final Answer**1. The time intervals are boxed{[1, 3]}.2. The maximum possible duration is boxed{2} hours.