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- Question 5 and 6, Exercise 5.2
- pose \( f(t) = t^{3} + t^{2} + 3t - 5 \). \begin{align*} f(1) &= (1)^{3} + (1)^{2} + 3(1) - 5 \\ &= 1 + 1 + 3 - 5 \\ &= 0. \end{align*} By the factor theorem, \( t - 1 \) is a factor... of \( f(t) \). Using synthetic division: \begin{align} \begin{array}{r|rrrr} 1 & 1 & 1 & 3 & -5 \\ & ... & 5 \\ \hline & 1 & 2 & 5 & 0 \\ \end{array} \end{align} This gives: \begin{align*} f(t) &= (t - 1)(t^{2
- Question 3 and 4, Exercise 5.2
- e \( f(x) = 2x^{3} + 5x^{2} - 9x - 18 \). \begin{align*} f(-2) &= 2(-2)^{3} + 5(-2)^{2} - 9(-2) - 18 \\ ... 5(4) + 18 - 18 \\ &= -16 + 20 + 18 - 18 = 0. \end{align*} By the factor theorem, \( x + 2 \) is a factor... & -11 & 0 \\ \end{array} \] This gives: \begin{align*} f(x) &= (x + 2)(2x^{2} + x - 9). \end{align*} Thus, we can factor \( 2x^{2} + x - 9 \) as: \begin{alig
- 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. \end{align*} By factor theorem, $y+1$ is factor of $f(y)$. Using synthetic division: \begin{align} \begin{array}{r|rrrr} -1 & 1 & 0 & -7 & -6 \\ & ... 6 \\ \hline & 1 & -1 & -6 & 0 \\ \end{array}\end{align} This gives \begin{align*} f(y)& =(y+1)(y^2-y-6)
- Question 7 and 8, Exercise 5.2
- vide \( f(x) \) by \( x - \frac{1}{2} \): \begin{align} \begin{array}{r|rrrr} \frac{1}{2} & 2 & -15 & 27... \\ \hline & 2 & -14 & 20 & 0 \\ \end{array} \end{align} This gives: \begin{align*} f(x) &= \left(x - \frac{1}{2}\right)(2x^{2} - 14x + 20)\\ &=\left(x - \frac... - 5)\\ &=\left(2x - 1\right) (x - 2)(x - 5) \end{align*} Finally, the complete factorization is: \begin
- Question 6 and 7, Exercise 5.1
- c=2$. \\ By the Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c) = p(2) \\ & = 2(2)^{3... 2) - m \\ & = 16 + 12 - 6 - m \\ & = 22 - m. \end{align*} Given that the remainder is 16, so \begin{align*} 22 - m & = 16 \\ m & = 22 - 16 \\ m & = 6. \end{align*} Hence, the value of \( m \) is 6. GOOD ===
- Question 8 and 9, Exercise 5.1
- Suppose $p(x)=2x^3+3x^2-11x-6$. \\ Since \begin{align} p(2) &= 2(2)^3+3(2)^2-11(2)-6 \\ &=16+12-22-6 = 0 \end{align} Hence 2 is zero of $p(x)$. \\ Then by using synthetic division: \begin{align} \begin{array}{r|rrrr} 2 & 2 & 3 & -11 & -6 \\ & ... & 6 \\ \hline & 2 & 7 & 3 & 0 \\ \end{array}\end{align} Now \begin{align*} & 2x^2+7x+3 \\ = & 2x^2+6x+x+
- Question 4 & 5, Review Exercise
- {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(\frac{2}{3}\right) &= 6\left(\frac{2}... {16 - 4 - 30 + 18}{9} \\ &= \frac{0}{9} = 0. \end{align*} Hence by the factor theorem, \( 3y - 2 \) is a
- Question 2 and 3, Exercise 5.1
- 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 \\ & = 0. \end{align*} Hence, $x-3$ is factor of $p(x)$. GOOD =====Qu... 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 \\ & = -5 \neq 0 \end{align*} Hence, $x-3$ is not factor of $p(x)$. GOOD ====
- Question 3, Exercise 5.3
- me = 144 cubic units. By given condition \begin{align*} & x(2x)(2x+2) = 144 \\ \implies & 4x^2(x+1)=144... ies & x^2(x+1)=36 \\ \implies & x^3+x^2-36=0 \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{align*} This gives $x=3$ is zeros of $p(x)$. Thus width = $2(3
- Question 4, Exercise 5.3
- e = 2475 cubic units. By given condition \begin{align*} & x(2x+3)(x-2) = 2475 \\ \implies & x(2x^2+3x-4... x-6)-2475=0 \\ \implies & 2x^3-x^2-6x-2475=0 \end{align*} Suppose $$p(x)=2x^3-x^2-6x-2475.$$ Since \begin{align*} p(11)&=2(11)^3-11^2-6(11)-2475 \\ &=2662-121-66-2475 = 0 \end{align*} This gives $x=11$ is the zeros of $p(x)$. Thus
- Question 5, Exercise 5.3
- 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)) \\ =& 2(x+4)(3x+7) \\ =& (2x+8)(3x+7) \end{align*} Now \begin{align*} & Length \times Width = Area\\ \implies & Length \times (2x+8) = 6 x^{2}+38 x+56 \... (2x+8)(3x+7) \\ \implies & Length = 3x+7 \\ \end{align*} Hence length of rectangle $ACED$ = $3x+4$ Now
- Question 2 & 3, Review Exercise
- ght) \div(4 y-2) \quad$ ** Solution. ** \begin{align*} \frac{(64 y^{3}-8)}{(4 y-2)}&= \frac{(4y - 2)(1... } + 8y + 4)}{4y - 2}\\ & = 16y^{2} + 8y + 4 .\end{align*} =====Question 3===== $\left(125 y^{3}-8\right) \div(5 y-2)$ ** Solution. ** \begin{align*} \frac{(125 y^{3}-8)}{(5 y-2)} &= \frac{(5y - 2)... 4\right)}{5y - 2} \\ & = 25y^{2} + 10y + 4. \end{align*} ====Go to ==== <text align="left"><bt
- Question 1, Exercise 5.1
- lies c=-2$. By Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c) = p(-2) \\ & = 2(-2)^... (-2)^{2}-4 (-2)+1 \\ & = -16+12+8+1 \\ &= 5. \end{align*} Hence remiander = 5. =====Question 1(ii)===== F... = 2 \). By the Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c) = p(2) \\ & = (2)^{4}... 2) + 3 \\ & = 16 + 16 - 4 + 4 + 3 \\ & = 35. \end{align*} Hence, the remainder is 35. ====Go to ====
- Question 1, Exercise 5.3
- e = $120 cm^3$ By given condtion, we have \begin{align*} & x(x+3)(x+10)=120 \\ \implies & x(x^2+3x+10x+30)-120=0\\ \implies & x^3+13x^2+30x-120=0. \end{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+60-120 =0 \end{align*} This gives $x=2$ is zero of $p(x)$. Hence Leng
- Question 4 and 5, Exercise 5.1
- actor of $p(x)$ iff $p(-1)=0$. This gives \begin{align*} &(-1)^3+q(-1)^2-7(-1)+6=0 \\ -&1+q+7+6=0\\ &q+12=0\\ &q=-12 \end{align*} ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit05:ex5-1-p2|< Question 2 & 3 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:uni