Question

Divide the polynomials using polynomial long division. (6x^3-16x^2+11x-2/(3x-2)








A
2x2+4x−12x^2+4x-12x
2
+4x−1

B
2x2−4x2x^2-4x2x
2
−4x

C
2x2−4x+12x^2-4x+12x
2
−4x+1

D
2x2−4x−12x^2-4x-12x
2
−4x−1

Answer 1

To divide the polynomial \( 6x^3 - 16x^2 + 11x - 2 \) by \( 3x - 2 \) using polynomial long division, follow these steps:

### Step 1: Set up the long division

Divide \( 6x^3 - 16x^2 + 11x - 2 \) by \( 3x - 2 \).

### Step 2: Divide the leading terms

Take the leading term of the dividend \( 6x^3 \) and divide it by the leading term of the divisor \( 3x \):
\[
\frac{6x^3}{3x} = 2x^2
\]
So, the first term of the quotient is \( 2x^2 \).

### Step 3: Multiply and subtract

Multiply \( 2x^2 \) by the entire divisor \( 3x - 2 \):
\[
2x^2 \cdot (3x - 2) = 6x^3 - 4x^2
\]
Now subtract this result from the original polynomial:
\[
(6x^3 - 16x^2 + 11x - 2) - (6x^3 - 4x^2) = -16x^2 + 4x^2 + 11x - 2 = -12x^2 + 11x - 2
\]

### Step 4: Repeat the process

Now divide the new leading term \( -12x^2 \) by the leading term of the divisor \( 3x \):
\[
\frac{-12x^2}{3x} = -4x
\]
The next term in the quotient is \( -4x \).

Now multiply \( -4x \) by \( 3x - 2 \):
\[
-4x \cdot (3x - 2) = -12x^2 + 8x
\]
Subtract this result:
\[
(-12x^2 + 11x - 2) - (-12x^2 + 8x) = 11x - 8x - 2 = 3x - 2
\]

### Step 5: One more round

Now divide the new leading term \( 3x \) by \( 3x \):
\[
\frac{3x}{3x} = 1
\]
The next term in the quotient is \( 1 \).

Multiply \( 1 \) by \( 3x - 2 \):
\[
1 \cdot (3x - 2) = 3x - 2
\]
Subtract this:
\[
(3x - 2) - (3x - 2) = 0
\]

Since the remainder is 0, we have finished the division.

### Final Result

The quotient of the division is:
\[
2x^2 - 4x + 1
\]

Thus, the final answer is:
\[
\text{C) } 2x^2 - 4x + 1
\]