Solution: We analyze the cubic polynomial $ R(x) = x^3 - 3x + 2 $. To find the number of real roots, we first attempt rational root theorem: possible rational roots are $ \pm1, \pm2 $. Testing:

["Solution to Analyze the Cubic Polynomial $ R(x) = x^3 - 3x + 2 $: Finding the Number of Real Roots", "When studying cubic polynomials, determining the exact number of real roots is essential for understanding the function’s behavior and applications in modeling. In this article, we analyze the cubic polynomial:", "$$\nR(x) = x^3 - 3x + 2\n$$", "We aim to find how many real roots $ R(x) $ has by applying the Rational Root Theorem, testing possible rational roots, and using polynomial factorization along with calculus to confirm the total number of real solutions.", "---", "### Step 1: Apply the Rational Root Theorem", "The Rational Root Theorem states that any possible rational root $ \frac{p}{q} $ of the polynomial must have $ p $ dividing the constant term and $ q $ dividing the leading coefficient.", "For $ R(x) = x^3 - 3x + 2 $:", "- Constant term: $ 2 $ → possible $ p = \pm1, \pm2 $\n- Leading coefficient: $ 1 $ → possible $ q = \pm1 $", "Thus, the possible rational roots are:\n$$\n\pm1, \pm2\n$$", "---", "### Step 2: Test Possible Rational Roots", "Evaluate $ R(x) $ at each candidate:", "1. $ R(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0 $\n👉 So, $ x = 1 $ is a rational root", "2. $ R(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4 <br/>\ne 0 $", "3. $ R(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4 <br/>\ne 0 $", "4. $ R(-2) = (-2)^3 - 3(-2) + 2 = -8 + 6 + 2 = 0 $\n👉 So, $ x = -2 $ is also a rational root", "We have found two rational roots: $ x = 1 $ and $ x = -2 $", "---", "### Step 3: Factor the Polynomial", "Since $ x = 1 $ and $ x = -2 $ are roots, $ (x - 1) $ and $ (x + 2) $ are factors of $ R(x) $. We perform polynomial division or use factorization.", "Let’s factor $ R(x) $ using these roots:", "We suspect:\n$$\nR(x) = (x - 1)(x + 2)Q(x)\n$$", "Multiply $ (x - 1)(x + 2) $:\n$$\n(x - 1)(x + 2) = x^2 + 2x - x - 2 = x^2 + x - 2\n$$", "Now divide $ x^3 - 3x + 2 $ by $ x^2 + x - 2 $:", "Using polynomial long division:", "- $ x^3 \div x^2 = x $, so multiply $ x(x^2 + x - 2) = x^3 + x^2 - 2x $\n- Subtract: $ (x^3 + 0x^2 - 3x + 2) - (x^3 + x^2 - 2x) = -x^2 - x + 2 $\n- Next term: $ -x^2 \div x^2 = -1 $, so $ -1(x^2 + x - 2) = -x^2 - x + 2 $\n- Subtract: $ (-x^2 - x + 2) - (-x^2 - x + 2) = 0 $", "Thus:\n$$\nR(x) = (x - 1)(x + 2)(x - 1) = (x - 1)^2(x + 2)\n$$", "---", "### Step 4: Determine the Number of Real Roots", "From the factorization:\n$$\nR(x) = (x - 1)^2(x + 2)\n$$", "The roots are:\n- $ x = 1 $ (with multiplicity 2)\n- $ x = -2 $ (with multiplicity 1)", "All three roots are real. However, counting distinct real roots, we have two: $ x = 1 $ and $ x = -2 $.", "But since we are analyzing the number of real roots counting multiplicity, there are three real roots (one repeated).", "To determine the number of distinct real roots, we conclude there are two.", "---", "### Step 5: Confirm with Calculus (Optional but Insightful)", "For deeper understanding, consider the derivative:\n$$\nR'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1)\n$$", "Critical points at $ x = 1 $ and $ x = -1 $", "Evaluate $ R(x) $ at turning points:", "- $ R(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4 > 0 $\n- $ R(1) = 0 $ (as found)", "Behavior at infinity:\n- $ \lim_{x \ o -\infty} R(x) = -\infty $\n- $ \lim_{x \ o \infty} R(x) = \infty $", "Since the function goes from $ -\infty $, increases to a local max at $ x = -1 $, $ R(-1) = 4 > 0 $, then decreases to a double root at $ x = 1 $, and finally increases to $ \infty $, it must cross the x-axis exactly three times:", "- Once between $ -\infty $ and $ -1 $: since $ R(-2) = 0 $, and $ R(-1) = 4 $, but $ R(x) $ approaches $ -\infty $, there must be a root left of $ -2 $ — but wait: $ R(-2) = 0 $, so actually the crossing includes $ x = -2 $, then another between $ -2 $ and $ 1 $? Let's check sign changes.", "Wait — correction: $ R(-2) = 0 $, $ R(-1) = 4 $, $ R(1) = 0 $", "Since it starts at $ -\infty $, reaches local max at $ x = -1 $, $ R(-1) = 4 > 0 $, then drops to $ R(1) = 0 $, and finally increases to $ \infty $, the sign changes show:", "- From $ -\infty $ to $ -2 $: $ R $ goes from $ -\infty $ to $ 0 $ — derivative negative → decreasing\n- But between $ -2 $ and $ 1 $: $ R(-2) = 0 $, increases to $ R(-1) = 4 $, then decreases to $ R(1) = 0 $ — so no crossing in $ (-2,1) $\n- Then from $ x = 1 $ to $ \infty $: $ R $ increases from 0 to $ \infty $", "But earlier we found $ R(x) = (x - 1)^2(x + 2) $, so the only roots are $ x = -2 $ and $ x = 1 $ (double).", "Waj, this implies only two distinct real roots: $ x = -2 $, $ x = 1 $", "But wait — from the graph:", "- At $ x = -3 $: $ R(-3) = (-27) - 3(-3) + 2 = -27 + 9 + 2 = -16 < 0 $\n- $ R(-2) = 0 $\n- $ R(-1) = 4 > 0 $ → sign change → root at $ x = -2 $\n- $ R(0) = 0 + 0 + 2 = 2 > 0 $\n- $ R(1) = 0 $\n- $ R(2) = 8 - 6 + 2 = 4 > 0 $", "But $ R(1) = 0 $, yet the function goes from $ R(-1) = 4 > 0 $ to $ R(1) = 0 $, so no sign change — just touches zero.", "So:\n- One root at $ x = -2 $\n- One root at $ x = 1 $ (double)\n- No other real roots", "Thus, only two real roots (one repeated)\nThree real roots counting multiplicity", "---", "### Final Answer: Number of Real Roots", "The cubic polynomial $ R(x) = x^3 - 3x + 2 $ has:", "- Two distinct real roots: $ x = -2 $ and $ x = 1 $ (since $ x = 1 $ has multiplicity 2)\n- Three real roots counting multiplicity, but the number of distinct real roots is 2", "This analysis confirms that cubic polynomials can have one, two, or three real roots — in this case, two distinct, three total with multiplicity.", "Understanding this helps in physics, economics, and engineering models where polynomial equations describe behavior or equilibrium points.", "For further analysis, derivative tests and graph behavior confirm the root structure without needing numerical methods.", "---", "Keywords: cubic polynomial, real roots, rational root theorem, $ R(x) = x^3 - 3x + 2 $, factorization, multiplicity, calculus analysis, polynomial roots, distinct real roots."]