Question

select the correct answer
using synthetic division, find ((2x^4 + 4x^3 + 2x^2 + 8x + 8) div (x + 2)

a. (2x^3 + 2x + 4)

b. (2x^4 + 2x^2 + 4x)

c. (2x^3 + 2x + 4 + \frac{1}{x + 2})

d. (2x^3 + 2x^2 + 4)

Explanation:

Step1: Identify the root for synthetic division

For dividing by \(x + 2\), the root \(r=-2\) (since \(x+2 = x - (-2)\)). The coefficients of the polynomial \(2x^4+4x^3 + 2x^2+8x + 8\) are \(2,4,2,8,8\).

Step2: Set up synthetic division

Write the root \(-2\) on the left and the coefficients in a row:

$$
LATEXBLOCK0
$$

Bring down the first coefficient \(2\). Multiply \(-2\) by \(2\) to get \(-4\), add to the next coefficient \(4\) ( \(4 + (-4)=0\) ). Multiply \(-2\) by \(0\) to get \(0\), add to \(2\) ( \(2+0 = 2\) ). Multiply \(-2\) by \(2\) to get \(-4\), add to \(8\) ( \(8+(-4)=4\) ). Multiply \(-2\) by \(4\) to get \(-8\), add to \(8\) ( \(8+(-8)=0\) ).

Step3: Form the quotient polynomial

The last number is the remainder (which is \(0\)). The coefficients of the quotient are \(2,0,2,4\), so the quotient polynomial is \(2x^3+0x^2 + 2x+4=2x^3+2x + 4\) (wait, no, wait the original polynomial is \(2x^4+4x^3+2x^2+8x + 8\). Wait, maybe I made a mistake in the exponents. Wait, the coefficients after division: the first coefficient is for \(x^3\), then \(x^2\), then \(x\), then constant. Wait, the coefficients are \(2\) (for \(x^3\)), \(0\) (for \(x^2\)), \(2\) (for \(x\)), \(4\) (constant). Wait, but let's check the options. Wait option A is \(2x^3+2x + 4\), option D is \(2x^3+2x^2+4\). Wait maybe I messed up the original polynomial. Wait the original polynomial is \(2x^4 + 4x^3+2x^2+8x + 8\). Let's do synthetic division again:

Root \(r=-2\).

Bring down \(2\).

Multiply \(2\times(-2)=-4\). Add to next coefficient: \(4 + (-4)=0\) (this is the \(x^3\) term's coefficient? Wait no, the first coefficient is \(x^4\), so after bringing down \(2\) (for \(x^4\) divided by \(x\) gives \(x^3\)), then multiply by \(-2\): \(2\times(-2)=-4\), add to \(4\) (the \(x^3\) coefficient) to get \(0\) (this is the \(x^2\) term's coefficient? Wait, no, the process is:

For polynomial \(a_nx^n+\dots+a_1x + a_0\), synthetic division with root \(r\):

Row 1: \(r\) | \(a_n\) \(a_{n - 1}\) \(\dots\) \(a_1\) \(a_0\)

Row 2: | \( \) \(r\times a_n\) \(r\times (a_n + r\times a_n)\) \(\dots\)

Row 3: | \(a_n\) \(a_{n - 1}+r\times a_n\) \(\dots\) \(a_0 + r\times (\text{previous sum})\)

So for \(2x^4+4x^3+2x^2+8x + 8\), \(a_4 = 2\), \(a_3 = 4\), \(a_2 = 2\), \(a_1 = 8\), \(a_0 = 8\), \(r=-2\).

Row 1: \(-2\) | \(2\) \(4\) \(2\) \(8\) \(8\)

Row 2: | \( \) \(-4\) \(0\) \(-4\) \(-8\)

Row 3: | \(2\) \(0\) \(2\) \(4\) \(0\)

So the quotient polynomial is \(2x^3+0x^2+2x + 4=2x^3+2x + 4\) (since \(0x^2\) is omitted). Wait but option A is \(2x^3+2x + 4\), option D is \(2x^3+2x^2+4\). Wait maybe the original polynomial was \(2x^4+4x^3+2x^2+8x + 8\)? Wait let's check the division: \((2x^4+4x^3+2x^2+8x + 8)\div(x + 2)\). Let's use polynomial long division to verify.

Divide \(2x^4\) by \(x\) to get \(2x^3\). Multiply \((x + 2)\) by \(2x^3\) to get \(2x^4+4x^3\). Subtract from the original polynomial: \((2x^4+4x^3+2x^2+8x + 8)-(2x^4+4x^3)=2x^2+8x + 8\). Now divide \(2x^2\) by \(x\) to get \(2x\). Multiply \((x + 2)\) by \(2x\) to get \(2x^2+4x\). Subtract: \((2x^2+8x + 8)-(2x^2+4x)=4x + 8\). Divide \(4x\) by \(x\) to get \(4\). Multiply \((x + 2)\) by \(4\) to get \(4x + 8\). Subtract: \((4x + 8)-(4x + 8)=0\). So the quotient is \(2x^3+2x + 4\), which is option A. Wait but earlier synthetic division gave \(2x^3+0x^2+2x + 4\), which is the same as \(2x^3+2x + 4\) (since the \(x^2\) term has coefficient 0, so it's omitted). So the correct answer is option A.

Answer:

A. \(2x^3 + 2x + 4\)