Polynomial Factoring: Which Student Is Correct?

Mr. Gonzalez posed a fascinating challenge to his top three students: find a factor of the polynomial $x^4-3 x^3-19 x^2+3 x+18$, given that $x-1$ is one of its linear factors. This is a classic problem in algebra that tests a student's understanding of polynomial division and factorization. Let's delve into the details and see which student, Student #1, Student #2, or Student #3, managed to crack the code.

Understanding Polynomial Division and Factors

Before we evaluate the students' answers, it's crucial to understand what it means for $x-1$ to be a linear factor of the given polynomial. If $x-1$ is a factor, it means that when we divide the polynomial $x^4-3 x^3-19 x^2+3 x+18$ by $x-1$, the remainder will be zero. The result of this division will be another polynomial, which is a factor of the original. Our task is to determine which of the students' proposed factors is the correct one, or if perhaps none of them are.

There are a couple of ways to approach this problem: polynomial long division or synthetic division. Synthetic division is often a quicker method when dividing by a linear factor of the form $x-c$. In this case, $c=1$. Let's perform synthetic division with $c=1$ and the coefficients of the polynomial $x^4-3 x^3-19 x^2+3 x+18$ (which are 1, -3, -19, 3, and 18).

1 | 1  -3  -19   3   18
  |    1   -2  -21  -18
  ---------------------
    1  -2  -21  -18   0

The numbers in the bottom row represent the coefficients of the quotient, and the last number is the remainder. So, when $x^4-3 x^3-19 x^2+3 x+18$ is divided by $x-1$, the quotient is $1x^3 - 2x^2 - 21x - 18$, and the remainder is 0. This confirms that $x-1$ is indeed a factor, and $x^3-2 x^2-21 x-18$ is another factor of the original polynomial.

Now, let's examine the students' responses. Mr. Gonzalez's students were asked to find a factor. This means they could have found the cubic factor we just discovered, or they could have factored the cubic factor further into a quadratic and a linear factor, or even into three linear factors.

Analyzing Student #1's Answer: $x^2+9 x+18$

Student #1 proposed the quadratic factor $x^2+9x+18$. For this to be a correct factor, it must be a result of dividing the original polynomial by $x-1$, or a factor of the resulting cubic polynomial $x^3-2 x^2-21 x-18$. Let's see if $x^2+9x+18$ can be factored further. We are looking for two numbers that multiply to 18 and add to 9. These numbers are 3 and 6. So, $x^2+9x+18 = (x+3)(x+6)$.

Now, let's consider if $(x+3)$ or $(x+6)$ are factors of the original polynomial. If $x+3$ is a factor, then $x=-3$ should be a root, meaning $f(-3) = 0$. Let's check: $(-3)^4 - 3(-3)^3 - 19(-3)^2 + 3(-3) + 18 = 81 - 3(-27) - 19(9) - 9 + 18 = 81 + 81 - 171 - 9 + 18 = 162 - 171 - 9 + 18 = -9 - 9 + 18 = -18 + 18 = 0$. So, $x+3$ is a factor!

If $x+6$ is a factor, then $x=-6$ should be a root, meaning $f(-6) = 0$. Let's check: $(-6)^4 - 3(-6)^3 - 19(-6)^2 + 3(-6) + 18 = 1296 - 3(-216) - 19(36) - 18 + 18 = 1296 + 648 - 684 - 18 + 18 = 1944 - 684 = 1260$. Since this is not 0, $x+6$ is not a factor.

Since $x+3$ is a factor, and we know $x-1$ is a factor, it's possible that $x^2+9x+18$ is not a direct result of dividing the original polynomial by $x-1$, but rather a factor that can be obtained through further factorization of the resulting cubic. Let's check if $(x-1)(x+3)$ is a factor of the original polynomial. $(x-1)(x+3) = x^2 + 3x - x - 3 = x^2 + 2x - 3$. If this is a factor, then dividing $x^4-3 x^3-19 x^2+3 x+18$ by $x^2+2x-3$ should yield a remainder of 0.

Let's perform polynomial long division:

        x^2   -5x   -6
      ________________
 x^2+2x-3 | x^4 -3x^3 -19x^2 +3x +18
        -(x^4 +2x^3 -3x^2)
        ________________
              -5x^3 -16x^2 +3x
            -(-5x^3 -10x^2 +15x)
            ________________
                    -6x^2 -12x +18
                  -(-6x^2 -12x +18)
                  ________________
                          0

The remainder is 0. This means that $(x-1)(x+3)$ is a factor of the original polynomial, and thus $x^2+2x-3$ is a factor. However, Student #1 gave $x^2+9x+18$. We found that $x^2+9x+18 = (x+3)(x+6)$. Since $x+6$ is not a factor of the original polynomial, $x^2+9x+18$ cannot be a factor of the original polynomial $x^4-3 x^3-19 x^2+3 x+18$. Therefore, Student #1 is incorrect.

Analyzing Student #2's Answer: $x^3-2 x^2-21 x-18$

Student #2 proposed the cubic factor $x^3-2 x^2-21 x-18$. As we discovered through synthetic division earlier, when we divide $x^4-3 x^3-19 x^2+3 x+18$ by the given linear factor $x-1$, the quotient is exactly $x^3-2 x^2-21 x-18$. Since the remainder was 0, this cubic polynomial is indeed a factor of the original polynomial. Therefore, Student #2 is correct.

Analyzing Student #3's Answer: $x^3-2 x^2-21$

Student #3 proposed the cubic factor $x^3-2 x^2-21$. This is very close to the correct cubic factor, but it has a different constant term (-21 instead of -18). Let's check if this is a factor by performing polynomial long division of the original polynomial by $x^3-2 x^2-21$. While this is possible, a simpler check is to see if multiplying this proposed factor by $x-1$ gives the original polynomial. If $(x-1)(x^3-2 x^2-21)$ were equal to $x^4-3 x^3-19 x^2+3 x+18$, then Student #3 would be correct. Let's expand it:

$(x-1)(x^3-2 x^2-21) = x(x^3-2 x^2-21) - 1(x^3-2 x^2-21)$ $= x^4 - 2x^3 - 21x - x^3 + 2x^2 + 21$ $= x^4 - 3x^3 + 2x^2 - 21x + 21$

Comparing this to the original polynomial $x^4-3 x^3-19 x^2+3 x+18$, we see that the terms involving $x^2$, $x$, and the constant term are all different. Thus, $x^3-2 x^2-21$ is not a factor of the original polynomial. Therefore, Student #3 is incorrect.

Conclusion: Student #2 is Correct!

After performing synthetic division and verifying the factors, it's clear that Student #2 is the only one who correctly identified a factor of the polynomial $x^4-3 x^3-19 x^2+3 x+18$, given that $x-1$ is one of its linear factors. The correct factor provided by Student #2 is $x^3-2 x^2-21 x-18$. This highlights the importance of precise calculations in algebra; even a small error in coefficients can lead to an incorrect result.

To learn more about polynomial factorization and related algebraic concepts, you can visit the Wolfram MathWorld website, a comprehensive resource for mathematical information.