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Algebra · real student question

Find the quotient (6x^2 - 8x - 6) divided by (3x + 2).

Question

Find the quotient:

(6x2−8x−6)÷(3x+2)(6x^2-8x-6)\div(3x+2)

Step-by-step solution

  1. Recognise the non-monic divisor. The divisor 3x+23x+2 leads with 3x3x, not xx, so synthetic division does not apply directly and each quotient term comes from dividing by 3x3x rather than by xx. Long division handles this without any extra machinery.

  2. First step. 6x23x=2x\dfrac{6x^2}{3x}=2x. Multiply back: 2x(3x+2)=6x2+4x2x(3x+2)=6x^2+4x. Subtract: (6x2−8x)−(6x2+4x)=−12x(6x^2-8x)-(6x^2+4x)=-12x, then bring down −6-6 to get −12x−6-12x-6.

  3. Second step. −12x3x=−4\dfrac{-12x}{3x}=-4. Multiply back: −4(3x+2)=−12x−8-4(3x+2)=-12x-8. Subtract: (−12x−6)−(−12x−8)=2(-12x-6)-(-12x-8)=2.

  4. Stop when the degree drops. The leftover 22 is a constant, lower in degree than the linear divisor, so the process terminates with quotient 2x−42x-4 and remainder 22.

  5. Express the answer. 6x2−8x−63x+2=2x−4+23x+2\dfrac{6x^2-8x-6}{3x+2}=2x-4+\dfrac{2}{3x+2}.

  6. Check by multiplying back and by substitution. (3x+2)(2x−4)+2=6x2−12x+4x−8+2=6x2−8x−6(3x+2)(2x-4)+2=6x^2-12x+4x-8+2=6x^2-8x-6. As a second check, the divisor vanishes at x=−23x=-\tfrac23, and the dividend there is 6⋅49+163−6=83+163−6=8−6=26\cdot\tfrac49+\tfrac{16}{3}-6=\tfrac83+\tfrac{16}{3}-6=8-6=2, which is the remainder.

Answer

2x−4+23x+2(quotient 2x−4, remainder 2)2x-4+\frac{2}{3x+2}\quad\text{(quotient }2x-4\text{, remainder }2\text{)}

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