Question

hat is the remainder of the quotient of (4x^4+8x^3−18x^2−9x+6)÷(2x2+6x−2) ?(1 point)

Answer 1

To find the remainder of the polynomial division \( (4x^4 + 8x^3 - 18x^2 - 9x + 6) \div (2x^2 + 6x - 2) \), we can use polynomial long division or synthetic division. However, given that we're only interested in the remainder, we can use the Remainder Theorem.

The divisor is a quadratic polynomial \( 2x^2 + 6x - 2 \), which can produce a remainder that is a linear polynomial of the form \( Ax + B \).

We can perform polynomial long division to find the quotient and remainder.

1. **Divide \( 4x^4 \) by \( 2x^2 \)** to find the first term of the quotient:

\[
4x^4 \div 2x^2 = 2x^2
\]

2. **Multiply** \( 2x^2 \) by \( 2x^2 + 6x - 2 \):

\[
2x^2(2x^2 + 6x - 2) = 4x^4 + 12x^3 - 4x^2
\]

3. **Subtract** this result from the original polynomial:

\[
(4x^4 + 8x^3 - 18x^2 - 9x + 6) - (4x^4 + 12x^3 - 4x^2) = -4x^3 - 14x^2 - 9x + 6
\]

4. **Repeat the process**. Now divide \( -4x^3 \) by \( 2x^2 \):

\[
-4x^3 \div 2x^2 = -2x
\]

5. **Multiply** \( -2x \) by \( 2x^2 + 6x - 2 \):

\[
-2x(2x^2 + 6x - 2) = -4x^3 - 12x^2 + 4x
\]

6. **Subtract again**:

\[
(-4x^3 - 14x^2 - 9x + 6) - (-4x^3 - 12x^2 + 4x) = -2x^2 - 13x + 6
\]

7. Now divide \( -2x^2 \) by \( 2x^2 \):

\[
-2x^2 \div 2x^2 = -1
\]

8. **Multiply** by the divisor:

\[
-1(2x^2 + 6x - 2) = -2x^2 - 6x + 2
\]

9. **Subtract again**:

\[
(-2x^2 - 13x + 6) - (-2x^2 - 6x + 2) = -7x + 4
\]

The result is the polynomial division result of quotient plus remainder. The remainder is therefore \( -7x + 4 \).

Thus, the remainder of the division \( (4x^4 + 8x^3 - 18x^2 - 9x + 6) \div (2x^2 + 6x - 2) \) is:

\[
\boxed{-7x + 4}
\]