Factor X³-2x²-9x+18 By Dividing By X-2

Let's dive into the fascinating world of algebra and tackle a factoring problem that's sure to get your mathematical gears turning! Today, we're going to factor the polynomial $x^3-2x^2-9x+18$ completely, and we'll use a handy trick to get us started: dividing by $x-2$. This method is a cornerstone in understanding how to break down complex expressions into their simpler building blocks. When we talk about factoring a polynomial completely, we mean expressing it as a product of irreducible factors – basically, the smallest, indivisible polynomial pieces that multiply back together to form the original one. Think of it like prime factorization for numbers; for instance, 12 can be factored into $2 \times 2 \times 3$. Polynomials work in a similar fashion, but instead of primes, we look for linear factors (like $x-a$) or irreducible quadratic factors. The problem specifically guides us to begin by dividing $x^3-2x^2-9x+18$ by $x-2$. This is a brilliant starting point because if $x-2$ is a factor of the polynomial, then dividing the polynomial by $x-2$ will result in a remainder of zero, and the quotient will be another polynomial of a lower degree. This process effectively reduces the complexity of the problem, making it easier to find the remaining factors. We'll be using polynomial long division or synthetic division, both powerful tools in our algebraic arsenal, to perform this division. Once we have the quotient, our task shifts to factoring that new, simpler polynomial. The degree of the original polynomial is 3, and after dividing by a degree 1 polynomial ($x-2$), we'll be left with a degree 2 polynomial (a quadratic). Factoring quadratic expressions is a skill many of us are familiar with, involving techniques like finding two numbers that multiply to a certain value and add to another, or using the quadratic formula if needed. So, by following the prompt's instruction to divide by $x-2$ first, we're setting ourselves up for a systematic and manageable approach to completely factor the given expression. This problem isn't just about finding the answer; it's about understanding the process and the underlying algebraic principles that make it work. Let's get our hands dirty with the division! We want to perform the division: $(x^3-2x^2-9x+18) \div (x-2)$. We can use synthetic division for this, which is a streamlined method for dividing a polynomial by a linear factor of the form $x-c$. In our case, $c=2$. The coefficients of the dividend $x^3-2x^2-9x+18$ are 1, -2, -9, and 18. Set up the synthetic division: 2 | 1 -2 -9 18 ----------------- | 2 0 -18 ----------------- 1 0 -9 0 Here's how it works: 1. Bring down the leading coefficient (1). 2. Multiply the number you brought down (1) by the divisor's root (2), and write the result (2) under the next coefficient (-2). 3. Add the numbers in the second column (-2 + 2 = 0). 4. Multiply the sum (0) by the divisor's root (2), and write the result (0) under the next coefficient (-9). 5. Add the numbers in the third column (-9 + 0 = -9). 6. Multiply the sum (-9) by the divisor's root (2), and write the result (-18) under the last coefficient (18). 7. Add the numbers in the last column (18 + -18 = 0). The last number in the bottom row (0) is the remainder. Since the remainder is 0, we know that $x-2$ is indeed a factor. The other numbers in the bottom row (1, 0, -9) are the coefficients of the quotient polynomial. Since we started with a degree 3 polynomial and divided by a degree 1 polynomial, the quotient will be a degree 2 polynomial. The coefficients 1, 0, and -9 correspond to $1x^2 + 0x - 9$, which simplifies to $x^2 - 9$. So, we have successfully factored our original polynomial as: $x^3-2x^2-9x+18 = (x-2)(x^2-9)$. But the problem asks us to factor it completely. This means we need to check if the quadratic factor, $x^2-9$, can be factored further. This particular quadratic is a classic example of a difference of squares. The difference of squares pattern states that $a^2 - b^2 = (a-b)(a+b)$. In our case, $x^2$ is $a^2$ (so $a=x$), and 9 is $b^2$ (so $b=3$). Applying the difference of squares formula to $x^2-9$, we get: $x^2-9 = (x-3)(x+3)$. Now, we can substitute this back into our factored expression: $x^3-2x^2-9x+18 = (x-2)((x-3)(x+3))$. Rearranging the factors, we get the completely factored form: $(x-2)(x-3)(x+3)$. This is our final answer! We've successfully broken down the cubic polynomial into three linear factors. Let's quickly verify our work by expanding this factored form: $(x-2)(x-3)(x+3) = (x-2)((x-3)(x+3))$ First, multiply $(x-3)(x+3)$: $(x-3)(x+3) = x(x+3) - 3(x+3) = x^2 + 3x - 3x - 9 = x^2 - 9$. Now, multiply $(x-2)$ by $(x^2-9)$: $(x-2)(x^2-9) = x(x^2-9) - 2(x^2-9) = x^3 - 9x - 2x^2 + 18$. Rearranging the terms to the standard polynomial form (highest power first), we get: $x^3 - 2x^2 - 9x + 18$. This matches our original polynomial exactly, confirming that our factorization is correct! So, when faced with a polynomial that needs factoring, remember the strategies: check for common factors first, use division if a linear factor is suggested or known, and then factor any resulting polynomials completely. The difference of squares is a pattern worth memorizing, as it appears frequently in algebraic problems. Looking back at the multiple-choice options provided: a. $(x-2)(x+3)$ - This is incomplete; it doesn't include the $(x-3)$ factor. b. $(x-2)(x-9)(x+1)$ - This doesn't match our result and suggests incorrect factoring of the quadratic part. c. $(x-2)(x-3)^2$ - This implies the quadratic factor was $x^2-6x+9$, which is not what we found. d. $(x-2)(x-3)(x+3)$ - This perfectly matches our derived factorization. Therefore, the correct option is (d). This exercise highlights the power of polynomial division as a tool for factoring and reinforces the importance of recognizing common algebraic patterns like the difference of squares. Mastering these techniques will equip you to solve a wide array of algebraic challenges. Keep practicing, and you'll find that factoring becomes second nature! For further exploration into the fascinating world of polynomial factorization and algebraic manipulation, you can check out resources like Khan Academy's Algebra section, which offers comprehensive lessons and practice problems on these topics. Their clear explanations and step-by-step examples are invaluable for solidifying your understanding of fundamental mathematical concepts.