Exercise 3.7 — Book-Back Answer Key
Exercise 3.7 — Book-Back Answer Key
Each answer below was solved independently; all ten agree with the standard key. The letter in brackets matches the workbook (A = option 1 … D = option 4).
- Q1 — Option 4, \(-4\) [D]. \(x^{3}=-64\Rightarrow x=-4\); the other zeros are \(2\pm 2\sqrt{3}\,i\).
- Q2 — Option 1, \(mn\) [A]. \(\deg f\big(g(x)\big)=\deg f\times\deg g\).
- Q3 — Option 3, exactly \(n\) complex roots [C]. Fundamental Theorem of Algebra, counting multiplicity.
- Q4 — Option 1, \(-\dfrac{q}{r}\) [A]. \(\sum\frac{1}{\alpha}=\dfrac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\dfrac{q}{-r}\).
- Q5 — Option 3, \(\dfrac{4}{5}\) [C]. \(p\) must divide \(5\) and \(q\) divide \(4\); \(\frac{4}{5}\) fits neither.
- Q6 — Option 4, \(|k|\ge 6\) [D]. \(x^{2}-kx+9\) is real-rooted iff \(k^{2}-36\ge 0\).
- Q7 — Option 1, \(2\) [A]. \((\sin^{2}x-1)^{2}=0\Rightarrow\sin x=\pm1\Rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}\).
- Q8 — Option 3, \(a\lt 0\) [C]. Descartes’ Rule: a positive zero is possible only when the \(x\)-coefficient is negative.
- Q9 — Option 1, one negative and two imaginary zeros [A]. \(x^{3}+2x+3=(x+1)\left(x^{2}-x+3\right)\), and \(x^{2}-x+3\) has \(\Delta=-11\).
- Q10 — Option 2, \(n\) [B]. The sum equals \((1-x)^{n}\), whose only zero is \(x=1\) with multiplicity \(n\).