What is the quotient of ( 6x^2 - 7x - 5 ) ÷ (2x + 1 ) for all values of x where the expression is defined?
A.) 3x - 2 - 7/2x + 1
B.) 3x +5
C.) 3x - 5
D.) 3x -2 - 3/2x + 1
To find the quotient of the polynomial division \((6x^2 - 7x - 5) ÷ (2x + 1)\), we can use polynomial long division or synthetic division. Let's do polynomial long division:
1. Divide the first term of the dividend \(6x^2\) by the first term of the divisor \(2x\), to get the first term of the quotient, which is \(3x\).
2. Multiply the divisor \((2x + 1)\) by the first term of the quotient \(3x\), which gives you \(6x^2 + 3x\).
3. Subtract this from the dividend:
\(6x^2 - 7x - 5\)
\(- (6x^2 + 3x)\)
4. This leaves us with \(-10x - 5\) after subtraction.
5. Now, divide the term \(-10x\) by the first term of the divisor \(2x\), which gives \(-5\).
6. Multiply the divisor \((2x + 1)\) by \(-5\), giving us \(-10x - 5\).
7. Subtract this from the current remainder:
\(-10x - 5\)
\(- (-10x - 5)\)
8. This leaves us with \(0\), which means the remainder is nothing and thus the polynomial division is exact.
From the steps above, the quotient of \((6x^2 - 7x - 5) ÷ (2x + 1)\) is \(3x - 5\), which is option C.
The answer is:
C.) \(3x - 5\)