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

The given expressions are

\$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:

Step1: Bring down the leading coefficient, which is

2

Step2: Multiply the divisor, -3, by the result from step 1, which gives us

(- 6)

Step3: Add the coefficient from the polynomial in the first column to the result from step 2. So

6 + (- 6) = 0

Step4: Multiply the divisor, -3, by the result from step 3, which gives us

0

Step5: Add the coefficient from the polynomial in the second column to the result from step 4. So

(- 1) + 0 = - 1

Step6: Multiply the divisor, -3, by the result from step 5, which gives us

\$ -3 times -1 = 3\$

Step7: Add the coefficient from the polynomial in the third column to the result from step 6. So

(-3) + 3 = 0

Step8: Multiply the divisor, -3, by the result from step 7, which gives us

0

Step9: 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\$ and the remainder is 6.

Therefore, \$2x^4 + 6x^3− x^2−3x+6\$ divided by ( x + 3) equals \$2x^3 − x\$ 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 \$. Hence, option A is wrong

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

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

Quotient is \$2x^3 - x \$ 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?