(d) \( x^{3}-2 x^{2}-5 x+0 \) Find the quotient and the remainder if \( f(x) \) is divisible by \( x \) (a) \( f(x)=x^{3}+3 x^{2}-x-2 ; x-2 \) (b) \( f(x)=-2 x^{4}=3 x^{2}+7 x-26 ; x+3 \) (c) \( f(x)=4 x^{3}-2 x^{2}-5 x+1 ; x-2 / 3 \)

**(a) Divide \( f(x)=x^{3}+3x^{2}-x-2 \) by \( x-2 \):** Write the polynomial with coefficients: \(1\) (for \(x^3\)), \(3\) (for \(x^2\)), \(-1\) (for \(x\)), and \(-2\) (constant). Using synthetic division with \( r=2 \) (since \( x-2=0 \) gives \( x=2 \)): 1. Bring down the first coefficient: \(1\). 2. Multiply by \(2\): \(1 \times 2 = 2\). 3. Add to the next coefficient: \(3 + 2 = 5\). 4. Multiply: \(5 \times 2 = 10\). 5. Add to the next coefficient: \(-1 + 10 = 9\). 6. Multiply: \(9 \times 2 = 18\). 7. Add to the constant: \(-2 + 18 = 16\). Thus, the quotient is \[ x^2+5x+9 \] and the remainder is \[ 16. \] --- **(b) Divide \( f(x)=-2x^4+3x^2+7x-26 \) by \( x+3 \):** First, write the polynomial in full degree order (note the missing \( x^3 \) term): \[ -2x^4+0x^3+3x^2+7x-26. \] We use synthetic division with \( r=-3 \) (since \( x+3=0 \) gives \( x=-3 \)): Coefficients: \(-2\), \(0\), \(3\), \(7\), \(-26\). 1. Bring down \(-2\). 2. Multiply: \(-2 \times (-3) = 6\). 3. Add: \(0 + 6 = 6\). 4. Multiply: \(6 \times (-3) = -18\). 5. Add: \(3 + (-18) = -15\). 6. Multiply: \(-15 \times (-3) = 45\). 7. Add: \(7 + 45 = 52\). 8. Multiply: \(52 \times (-3) = -156\). 9. Add: \(-26 + (-156) = -182\). The quotient becomes \[ -2x^3+6x^2-15x+52 \] and the remainder is \[ -182. \] --- **(c) Divide \( f(x)=4x^3-2x^2-5x+1 \) by \( x-\frac{2}{3} \):** We use synthetic division with \( r=\frac{2}{3} \) (since \( x-\frac{2}{3}=0 \) gives \( x=\frac{2}{3} \)). Coefficients: \(4\), \(-2\), \(-5\), \(1\). 1. Bring down \(4\). 2. Multiply: \(4 \times \frac{2}{3} = \frac{8}{3}\). 3. Add: \(-2 + \frac{8}{3} = -\frac{6}{3}+\frac{8}{3} = \frac{2}{3}\). 4. Multiply: \(\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}\). 5. Add: \(-5 + \frac{4}{9} = -\frac{45}{9}+\frac{4}{9} = -\frac{41}{9}\). 6. Multiply: \(-\frac{41}{9} \times \frac{2}{3} = -\frac{82}{27}\). 7. Add: \(1 + \left(-\frac{82}{27}\right) = \frac{27}{27}-\frac{82}{27} = -\frac{55}{27}\). Thus, the quotient is \[ 4x^2 + \frac{2}{3}x - \frac{41}{9} \] and the remainder is \[ -\frac{55}{27}. \] --- **(d) Divide \( f(x)=x^3-2x^2-5x \) by \( x \):** Since the divisor is \( x \), divide each term by \( x \): \[ \frac{x^3}{x} = x^2,\quad \frac{-2x^2}{x} = -2x,\quad \frac{-5x}{x} = -5. \] Thus, we have \[ x^3-2x^2-5x = x\left(x^2-2x-5\right). \] The quotient is \[ x^2-2x-5, \] and since the division is exact, the remainder is \[ 0. \]