Divide. \[ \left(10 x^{3}-21 x^{2}+14 x-4\right)=(5 x-3) \] Your answer should give the quotient and the remainder. Quotient: Remainder: \( \square \)

To divide the polynomial \(10x^{3} - 21x^{2} + 14x - 4\) by \(5x - 3\), we can use polynomial long division. 1. Divide the leading term of the dividend \(10x^{3}\) by the leading term of the divisor \(5x\) to get \(2x^{2}\). 2. Multiply \(2x^{2}\) by \(5x - 3\) which gives \(10x^{3} - 6x^{2}\). 3. Subtract this from the original polynomial: \[ (10x^{3} - 21x^{2} + 14x - 4) - (10x^{3} - 6x^{2}) = -15x^{2} + 14x - 4 \] 4. Now, repeat the process. Divide \(-15x^{2}\) by \(5x\) to get \(-3x\). 5. Multiply \(-3x\) by \(5x - 3\) to get \(-15x^{2} + 9x\). 6. Subtract: \[ (-15x^{2} + 14x - 4) - (-15x^{2} + 9x) = 5x - 4 \] 7. Next, divide \(5x\) by \(5x\) which gives \(1\). 8. Multiply \(1\) by \(5x - 3\) yielding \(5x - 3\). 9. Subtract: \[ (5x - 4) - (5x - 3) = -1 \] Now we have reached the end of our division. Quotient: \(2x^{2} - 3x + 1\) Remainder: \(-1\)