A student used synthetic division to divide x^3-x^2-2x by x+1 .

This is what the students problem looked like
1_| 1 -1 -2
1 0
____________
1 0 -2
answer x remainder -2
a. Describe and correct the error shown.
b. Use the Remainder theorem to double check the remainder you found in part a.

I will attempt to line up the columns

-1_| 1 -1 -2 0 <---- there was no constant, so put in a zero
......... -1 2 0
__________
...... 1 -2 0 0

so (x^3-x^2-2x) by (x+1) = x^2 - 2x with zero remainder

check: f(-1) = (-1)^3 - (-1)^2 - 2(-1) = -1 - 1 + 2 = 0

double check:
(x+1)(x^2 - 2x)
= x^3 - 2x^2 + x^2 - 2x
= x^3 - x^2 - 2x , all checks out

a. Well, it looks like the student accidentally forgot to bring down the middle term during the synthetic division. So, the correct division would actually look like this:

1_| 1 -1 -2
1 -2
____________
1 0 -2

b. Now, let's use the Remainder theorem to double check the remainder we found. According to the theorem, if we substitute the divisor x+1 into the original polynomial, the result should be equal to the remainder.

Let's do it:
(x^3 - x^2 - 2x) / (x + 1) = (-2x^2 - 2)

Since (-2x^2 - 2) is not equal to -2, it means there was still an error somewhere in the synthetic division. We'll need to correct it again. (Clown nose honks)

a. The error in the student's solution is that they made a mistake in the subtraction step. After bringing down the 1, they subtracted the last term incorrectly. They should have subtracted -2x from 0x^2, resulting in -2x^2, rather than 0x^2. This mistake propagates through the division process, leading to incorrect coefficients in the quotient.

To correct the error, the student should subtract -2x from 0x^2 to get -2x^2 during the division process.
The corrected solution would look like this:

1_| 1 -1 -2
1 2
____________
1 0 -4

The correct quotient is x^2 - 4, and the remainder is 0.

b. To double-check the remainder using the Remainder theorem, evaluate the polynomial f(x) = x^3 - x^2 - 2x at x = -1. If the remainder is 0, it confirms that the divisor x + 1 is a factor of the polynomial.

Evaluate f(-1):
f(-1) = (-1)^3 - (-1)^2 - 2(-1)
= -1 - 1 + 2
= 0

Since the remainder obtained through synthetic division and the result obtained by evaluating the polynomial at x = -1 are both 0, it confirms that the divisor x + 1 is indeed a factor of the polynomial.

a. The error in the student's work is that they made a mistake in the subtraction step when bringing down the last coefficient.

To correctly perform synthetic division, follow these steps:
1. Write the coefficients of the polynomial, in this case, 1 -1 -2, as the dividend under the division symbol.
2. Write the divisor, x+1, to the left of the division symbol.
3. Bring down the first coefficient, which is 1.
4. Multiply the divisor by the number that was just brought down and write the result below the second coefficient, which is -1. This gives you -1.
5. Add the product to the second coefficient, -1, and write the sum below the horizontal line. This gives you -1 + (-1) = -2.
6. Multiply the divisor by the result obtained in step 5, which is -2. Write the product below the third coefficient, which is -2. This gives you -2.
7. Add the product to the third coefficient, -2, and write the sum below the horizontal line. This gives you -2 + (-2) = -4.

The corrected synthetic division would look like this:
1_| 1 -1 -2
1 0
___________
1 -2 -4

b. To use the Remainder Theorem to double-check the remainder, we need to evaluate the polynomial at the value of the divisor, which is -1.

Substitute -1 into the original polynomial, x^3 - x^2 - 2x, and calculate the result:
(-1)^3 - (-1)^2 - 2(-1)
= -1 - 1 + 2
= 0

Since the result is 0, this confirms that the remainder obtained in part a is correct.