如何利用零点、因式与余数求多项式参数a、b值及解P(x)≤0不等式？

Alright, let's work through this problem step by step to find the values of a and b, then solve the inequality P(x) ≤ 0.

Step 1: Calculate a and b using given conditions #

We have two key pieces of information to leverage here:

• x=1 is a zero of P(x): By definition, if x=1 is a root, then P(1) = 0. Substitute x=1 into the polynomial:

  P(1) = a(1)³ + 3(1)² + b(1) + 3 = 0
  

  Simplify to get our first equation:

  a + 3 + b + 3 = 0 → a + b = -6  --- (1)
  

• Dividing P(x) by (x-2) leaves a remainder of 15: Using the Remainder Theorem (which states the remainder when dividing by (x-c) equals P(c)), we know P(2) = 15. Substitute x=2:

  P(2) = a(2)³ + 3(2)² + b(2) + 3 = 15
  

  Simplify to get our second equation:

  8a + 12 + 2b + 3 = 15 → 8a + 2b = 0 → 4a + b = 0  --- (2)
  

Now solve the system of equations:
Subtract equation (1) from equation (2):

(4a + b) - (a + b) = 0 - (-6) → 3a = 6 → a = 2

Plug a=2 back into equation (1):

2 + b = -6 → b = -8

So the polynomial becomes:

P(x) = 2x³ + 3x² - 8x + 3

Step 2: Factorize P(x) to find all roots #

We already know x=1 is a root. Using the Rational Root Theorem (possible roots are ±1, ±3, ±1/2, ±3/2), we can test other values:

• x=-3: P(-3) = 2(-3)³ + 3(-3)² -8(-3)+3 = -54+27+24+3=0 → valid root
• x=1/2: P(1/2)=2(1/2)³+3(1/2)²-8(1/2)+3=1/4+3/4-4+3=0 → valid root

Now write P(x) in factored form (eliminating fractions for clarity):

P(x) = (x + 3)(2x - 1)(x - 1)

Step 3: Solve the inequality P(x) ≤ 0 #

The leading coefficient of P(x) is positive (2), so the cubic tends to -∞ as x→-∞ and +∞ as x→+∞. The roots in order are -3 < 1/2 < 1, splitting the number line into four intervals. Test the sign of P(x) in each:

• (-∞, -3): All factors are negative → negative × negative × negative = negative → P(x) < 0
• (-3, 1/2): x+3 positive, others negative → positive × negative × negative = positive → P(x) > 0
• (1/2, 1): x+3 and 2x-1 positive, x-1 negative → positive × positive × negative = negative → P(x) < 0
• (1, +∞): All factors positive → positive × positive × positive = positive → P(x) > 0

We want P(x) ≤ 0, so include intervals where P(x) is negative plus the roots where P(x)=0. The solution is:

x ≤ -3  or  1/2 ≤ x ≤ 1

内容的提问来源于stack exchange，提问作者natojato