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- Question 5 and 6, Exercise 5.2
- * Suppose \( f(t) = t^{3} + t^{2} + 3t - 5 \). \begin{align*} f(1) &= (1)^{3} + (1)^{2} + 3(1) - 5 \\ &... actor of \( f(t) \). Using synthetic division: \begin{align} \begin{array}{r|rrrr} 1 & 1 & 1 & 3 & -5 \\ & & 1 & 2 & 5 \\ \hline & 1 & 2 & 5 & 0 \\ \end{array} \end{align} This gives: \begin{align*} f(t) &= (t - 1)(t^{2} + 2t + 5). \end{ali
- Question 3 and 4, Exercise 5.2
- Suppose \( f(x) = 2x^{3} + 5x^{2} - 9x - 18 \). \begin{align*} f(-2) &= 2(-2)^{3} + 5(-2)^{2} - 9(-2) - ... tor of \( f(x) \). Using synthetic division: \[ \begin{array}{r|rrrr} -2 & 2 & 5 & -9 & -18 \\ & & -4... 2 & 1 & -11 & 0 \\ \end{array} \] This gives: \begin{align*} f(x) &= (x + 2)(2x^{2} + x - 9). \end{ali... n*} Thus, we can factor \( 2x^{2} + x - 9 \) as: \begin{align*} (x + 2)(2x^{2} + x - 9) &=(x + 2)( 2x^{2}
- Question 1 and 2, Exercise 5.2
- $ ** Solution. ** Suppose $f(y)=y^{3}-7 y-6$. \begin{align*} f(-1)&=(-1)^{3}-7 (-1)-6 \\ &= -1+7-6 =0.... is factor of $f(y)$. Using synthetic division: \begin{align} \begin{array}{r|rrrr} -1 & 1 & 0 & -7 & -6 \\ & \downarrow & -1 & 1 & 6 \\ \hline & 1 & -1 & -6 & 0 \\ \end{array}\end{align} This gives \begin{align*} f(y)& =(y+1)(y^2-y-6) \\ & = (y+1)(y^2-3y
- Question 7 and 8, Exercise 5.2
- to divide \( f(x) \) by \( x - \frac{1}{2} \): \begin{align} \begin{array}{r|rrrr} \frac{1}{2} & 2 & -15 & 27 & -10 \\ & & 1 & -7 & 10 \\ \hline & 2 & -14... & 20 & 0 \\ \end{array} \end{align} This gives: \begin{align*} f(x) &= \left(x - \frac{1}{2}\right)(2x^{... align*} Finally, the complete factorization is: \begin{align*} f(x) &= \left(2x - 1\right) (x - 2)(x - 5
- Question 8 and 9, Exercise 5.1
- on. ** Suppose $p(x)=2x^3+3x^2-11x-6$. \\ Since \begin{align} p(2) &= 2(2)^3+3(2)^2-11(2)-6 \\ &=16+12-2... of $p(x)$. \\ Then by using synthetic division: \begin{align} \begin{array}{r|rrrr} 2 & 2 & 3 & -11 & -6 \\ & \downarrow & 4 & 14 & 6 \\ \hline & 2 & 7 & 3 & 0 \\ \end{array}\end{align} Now \begin{align*} & 2x^2+7x+3 \\ = & 2x^2+6x+x+3 \\ = & 2x(
- Question 6 and 7, Exercise 5.1
- plies c=2$. \\ By the Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c) = p(2) \\ & = 2... \end{align*} Given that the remainder is 16, so \begin{align*} 22 - m & = 16 \\ m & = 22 - 16 \\ m & = 6... .\\ $1$ will be zero of $p(x)$ if $p(1)=0$. Thus \begin{align*} p(1)&=(1)^3-7(1)+6\\ &=1-7+6\\ &=0\end{al... rly, $-2$ will be zero of $p(x)$ if f $p(-2)=0$. \begin{align*} p(-2)&=(-2)^3-7(-2)+6\\ &=-8+14+6\\ &=12
- Question 4 & 5, Review Exercise
- $6 y^{3}-y^{2}-5 y+2$ ? ** Solution. ** Given \begin{align*}3y-2&=0\\ 3y&=2\\ y&=\frac{2}{3}\end{align*} Suppose \begin{align*} f(y) &= 6y^{3} - y^{2} - 5y + 2\\ f\left(... {5}\right)(x + 2). \] Multiplying the factors: \begin{align*} f(x) &= (x - 4) \left( \frac{5x - 3}{5} \
- Question 1, Exercise 5.1
- 2 \implies c=-2$. By Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c) = p(-2) \\ & = ... ies c = 2 \). By the Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c) = p(2) \\ & = (
- Question 2 and 3, Exercise 5.1
- em $x-3$ is factor of $p(x)$ iff $p(3)=0$. Now \begin{align*} p(3)&=3^3-2(3)^2-5(3)+6 \\ & = 27-18-15+6... em $x-3$ is factor of $p(x)$ iff $p(3)=0$. Now \begin{align*} p(3)&=3^3-2(3)^2-5(3)+1 \\ & = 27-18-15+1
- Question 10, Exercise 5.1
- 11 x^{2}+34 x+24$. By using synthetic division: \begin{align} \begin{array}{r|rrrr} -1 & 1 & 11 & 34 & 24 \\ & \downarrow & -1 & -10 & -24 \\ \hline & 1 & 10
- Question 1, Exercise 5.3
- bottle = $120 cm^3$ By given condtion, we have \begin{align*} & x(x+3)(x+10)=120 \\ \implies & x(x^2+3... align*} Consider $$p(x)=x^3+13x^2+30x-120$$ Now \begin{align*} p(2)&=2^3+13(2)^2+30(2)-120 \\ &=8+52+60-
- Question 3, Exercise 5.3
- \ Volume = 144 cubic units. By given condition \begin{align*} & x(2x)(2x+2) = 144 \\ \implies & 4x^2(x+... end{align*} Suppose $$p(x)=x^3+x^2-36.$$ Since \begin{align*} p(3)&=3^3+3^2-36 \\ &=27+9-36 = 0 \end{al
- Question 4, Exercise 5.3
- Volume = 2475 cubic units. By given condition \begin{align*} & x(2x+3)(x-2) = 2475 \\ \implies & x(2x^... ign*} Suppose $$p(x)=2x^3-x^2-6x-2475.$$ Since \begin{align*} p(11)&=2(11)^3-11^2-6(11)-2475 \\ &=2662-
- Question 5, Exercise 5.3
- D$ = $6 x^{2}+38 x+56$ Width = $2 x+8$ We have \begin{align*} & 6 x^{2}+38 x+56 \\ = & 2(3x^2+19x+28) \... (x+4)(3x+7) \\ =& (2x+8)(3x+7) \end{align*} Now \begin{align*} & Length \times Width = Area\\ \implies &
- Question 6, Exercise 5.3
- is zeros of $p(y)$ Using synthetic division: \[ \begin{array}{r|rrrr} 2 & 1 & -2 & -1 & 2 \\ & \downarro... hline & 1 & 0 & -1 & 0 \\ \end{array} \] Thus \begin{align*} p(y) & = (y-2)(y^2-1) \\ & = (y-2)(y+1)(y