Write the quotient and remainder when we divide :
`(6x^(2) - 31 x + 47)` by (2x - 5)

Answer

To find the quotient and remainder when dividing \(6x^2 - 31x + 47\) by \(2x - 5\), we can use polynomial long division. Here’s the step-by-step solution:

### Step 1: Set up the division
We will divide \(6x^2 - 31x + 47\) by \(2x - 5\).

### Step 2: Divide the leading terms
Divide the leading term of the dividend \(6x^2\) by the leading term of the divisor \(2x\):
\[ \frac{6x^2}{2x} = 3x \]
This means our first term in the quotient is \(3x\).

### Step 3: Multiply and subtract
Now, multiply \(3x\) by the entire divisor \(2x - 5\):
\[ 3x \cdot (2x - 5) = 6x^2 - 15x \]
Now subtract this from the original polynomial:
\[ (6x^2 - 31x + 47) - (6x^2 - 15x) = -31x + 15x + 47 = -16x + 47 \]

### Step 4: Repeat the process
Now, we will divide the new leading term \(-16x\) by the leading term of the divisor \(2x\):
\[ \frac{-16x}{2x} = -8 \]
So, the next term in the quotient is \(-8\).

### Step 5: Multiply and subtract again
Multiply \(-8\) by the entire divisor \(2x - 5\):
\[ -8 \cdot (2x - 5) = -16x + 40 \]
Now subtract this from \(-16x + 47\):
\[ (-16x + 47) - (-16x + 40) = 47 - 40 = 7 \]

### Step 6: Conclusion
At this point, we cannot divide further since the degree of the remainder \(7\) is less than the degree of the divisor \(2x - 5\).

Thus, the quotient is:
\[ 3x - 8 \]
And the remainder is:
\[ 7 \]

### Final Answer
**Quotient:** \(3x - 8\)
**Remainder:** \(7\)

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