Divide the polynomial \$2x^4 + 6x^3 - x^2 - 3x + 6\$ by the binomial (x + 3).

The given polynomial is

\$2x^4 + 6x^3 - x^2 - 3x + 6\$ (x + 3)

First, we’ll set up the synthetic division:

\$-3 | 2 6 - 1 - 3 6\$

Now, let’s perform the synthetic division: + 1. Bring down the leading coefficient, which is

2

2.Multiply the divisor, -3, by the result from step 1, which gives us

-6

3.Add the coefficient from the polynomial in the first column to the result from step 2. So

6 + (-6) = 0

4.Multiply the divisor, -3, by the result from step 3, which gives us

0

5.Add the coefficient from the polynomial in the second column to the result from step 4. So

-1 + 0 = - 1

6.Multiply the divisor, -3, by the result from step 5, which gives us

\$ -3 times -1\$ = 3

7.Add the coefficient from the polynomial in the third column to the result from step 6. So

-3 + 3 = 0

8.Multiply the divisor, -3, by the result from step 7, which gives us

0

9.Add the coefficient from the polynomial in the fourth column to the result from step 8. So

6 + 0 = 6

So, the quotient is \$2x^3 - x -1\$ and the remainder is 6.

Therefore, \$2x^4 + 6x^3− x^2−3x+6\$ divided by ( x + 3) equals \$2x^3 − x−1\$ with a remainder of 6.

It incorrectly states the quotient as \$2x^3 + x − 1\$, but the correct one, shown in the division, is \$2x^3 − x − 1\$. Hence, option A is wrong

Quotient is \$2x^3 + x -1\$ and the remainder is 6

This statement is accurate, as it yields the correct quotient \$2x^3 − x − 1\$ and leaves a remainder of 6

Quotient is \$2x^3 - x -1\$ and the remainder is 6

The quotient obtained through polynomial division \$- 2x^3 - x -1\$ is incorrect

Quotient is \$- 2x^3 - x -1\$ and the remainder is 6

Because the sign of the x variable and constant term in the quotient is incorrect

Quotient is \$- 2x^3 + x +1\$ and the remainder is 6

Option

B

Can you summarize what you’ve understood in the above steps?