Polynomials, Cubic Equations and Partial Fractions practice questions

• $\text{Perform long division of } 2x^{3} + 2x^{2} - 3x - 1 \text{ by } (x + 1)$
• $2x^{3} \div x = 2x^{2}$. Multiply: $2x^{2}(x + 1) = 2x^{3} + 2x^{2}$. Subtract to get $-3x - 1$.
• $-3x \div x = -3$. Multiply: $-3(x + 1) = -3x - 3$. Subtract to get remainder $2$.
• $\text{Quotient} = 2x^{2} - 3$
• $\text{Remainder} = 2$
• $\therefore 2x^{3} + 2x^{2} - 3x - 1 = (x + 1)(2x^{2} - 3) + 2$

Answer: ${"q2":"2","q1":"0","q0":"-3","remainder":"2"}$

• $\text{Since } (x + 2) \text{ is a factor, } P(-2) = 0 \quad \scriptsize\textit{Factor Theorem}$
• $(-2)^3 + 4(-2)^2 + k(-2) - 4 = 0$
• $-8 + 16 - 2k - 4 = 0$
• $-2k = -4$
• $k = 2$

Answer: $2$

• $\text{Let } f(x) = x^{3} + x^{2} - 16x - 16$
• Test $x = -1$, a factor of the constant term $-16$:
• $f(-1) = (-1)^{3} + (-1)^{2} - 16(-1) - 16$
• $= -1 + 1 + 16 - 16$
• $= 0$
• $\therefore (x + 1) \text{ is a factor} \quad \scriptsize\textit{Factor Theorem}$
• $f(x) = (x + 1)(x^2 + bx + c)$
• Equating constant terms: $c = -16$

Answer: $-4, -1, 4$