Solution: Factor the divisor: $ x^2 - 3x + 2 = (x-1)(x-2) $. Use the Remainder Theorem: Let $ R(x) = ax + b $ be the remainder. Evaluate $ f(1) = 1 - 2 + 1 + 1 = 1 = a(1) + b $. Evaluate $ f(2) = 32 - 16 + 4 + 1 = 21 = a(2) + b $. Solve: $ a + b = 1 $, $ 2a + b = 21 $. Subtract: $ a = 20 $, then $ b = -19 $. Remainder: $ 20x - 19 $. \boxed{20x - 19}

["# Factor the Divisor: $ x^2 - 3x + 2 = (x-1)(x-2) $ Using the Remainder Theorem", "Understanding how to factor quadratic expressions efficiently is a cornerstone of algebra. One powerful technique relies on the Remainder Theorem, which helps not only verify roots of polynomials but also break them down into linear factors.", "In this article, we explore how to factor $ x^2 - 3x + 2 $ into $ (x-1)(x-2) $ using the Remainder Theorem — a method ideal for confirming possible divisors and solving polynomial factorization problems.", "---", "## Step 1: Recognize the goal", "We aim to factor the quadratic $ x^2 - 3x + 2 $. A standard approach is to express it as a product of two linear binomials:\n$$\nx^2 - 3x + 2 = (x - a)(x - b)\n$$\nwhich expands to:\n$$\nx^2 - (a+b)x + ab\n$$\nMatching coefficients, we find $ a + b = 3 $ and $ ab = 2 $. The integers $ a = 1 $, $ b = 2 $ satisfy both conditions. Thus,\n$$\nx^2 - 3x + 2 = (x - 1)(x - 2)\n$$", "But how can we confirm these factors using algebraic tools? This is where the Remainder Theorem shines.", "---", "## Step 2: Apply the Remainder Theorem", "The Remainder Theorem states that if $ f(x) $ is divided by $ x - c $, the remainder is $ f(c) $.", "Let $ f(x) = x^2 - 3x + 2 $. We want to factor $ f(x) $, and start by testing potential roots — conveniently, the constant term 2 gives small integer candidates: $ \pm 1, \pm 2 $.", "### Evaluate $ f(1) $\n$$\nf(1) = (1)^2 - 3(1) + 2 = 1 - 3 + 2 = 0\n$$\nSince the remainder is zero, $ x - 1 $ is a factor of $ f(x) $.", "### Evaluate $ f(2) $\n$$\nf(2) = (2)^2 - 3(2) + 2 = 4 - 6 + 2 = 0\n$$\nAgain, remainder is zero — so $ x - 2 $ is also a factor.", "Thus, $ x^2 - 3x + 2 $ is divisible by both $ x - 1 $ and $ x - 2 $. This confirms:\n$$\nx^2 - 3x + 2 = (x - 1)(x - 2)\n$$", "---", "## Step 3: Use the Remainder Theorem explicitly to compute the remainder $ R(x) $", "Suppose we don’t yet know the factorization, but suspect $ x - 1 $ and $ x - 2 $ are roots. Let the full remainder when dividing by a general linear factor be linear: $ R(x) = ax + b $. Then:", "$$\nf(x) = (x - 1)(x - 2)Q(x) + ax + b\n$$", "But since $ x = 1 $ and $ x = 2 $ are roots, $ f(1) = 0 $ and $ f(2) = 0 $, so plugging in confirms $ R(1) = 0 $ and $ R(2) = 0 $.", "Set $ R(x) = ax + b $. Then:", "- $ R(1) = a(1) + b = a + b = f(1) = 1 $\n- $ R(2) = a(2) + b = 2a + b = f(2) = 21 $", "---", "## Step 4: Solve the system of equations", "We solve:\n$$\n\begin{cases}\na + b = 1 \\n2a + b = 21\n\end{cases}\n$$", "Subtract the first equation from the second:\n$$\n(2a + b) - (a + b) = 21 - 1 \Rightarrow a = 20\n$$", "Substitute $ a = 20 $ into $ a + b = 1 $:\n$$\n20 + b = 1 \Rightarrow b = -19\n$$", "Thus, the remainder is:\n$$\nR(x) = 20x - 19\n$$", "---", "## Step 5: Interpret the result", "While we already factored $ x^2 - 3x + 2 $ directly, the Remainder Theorem gives a rigorous way to confirm the linear factors and even find the remainder when divided by binomials. This method works broadly for any degree polynomial when seeking linear factors.", "---", "### Final Answer:\n$$\n\boxed{20x - 19}\n$$", "This remainder confirms that:\n$$\nx^2 - 3x + 2 = (x - 1)(x - 2)\n$$\nusing the algebraic power of the Remainder Theorem.", "---", "### Why This Matters in Real Life", "Efficient polynomial factoring and remainder analysis support fields like control theory, signal processing, cryptography, and computer algebra systems — where breaking down complex expressions into simpler parts is essential for computation and modeling.", "---", "Keywords: factor quadratic, use Remainder Theorem, solve $ f(1) $, solve $ f(2) $, polynomial division, factor $ x^2 - 3x + 2 $, remainder 20x - 19, algebraic root finding, divide by linear factor, polynomial remainder theorem."]