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- Question 2, Exercise 5.3
- ld during the match can be modeled by $t(x)=x^{3}-12 x^{2}+48 x+74$, where $x$ is the number of games ... ket season. ** Solution. ** Given: $$t(x)=x^{3}-12 x^{2}+48 x+74.$$ When $t=12$ \begin{align*} t(12)&=(12)^3-12(12)^2+48(12)+74 \\ &=650. \end{align*} During the 12th game 650 tick
- Question 3 and 4, Exercise 5.2
- ray}{r|rrrr} 3 & 3 & -5 & 0 & -36 \\ & & 9 & 12 & 36 \\ \hline & 3 & 4 & 12 & 0 \\ \end{array} \] This gives: \begin{align*} f(x) &= (x - 3)(3x^{2} + 4x + 12). \end{align*} The final factorization is: \begin{align*} f(x) &= (x - 3)(3x^{2} + 4x + 12). \end{align*} ====Go to ==== <text al
- Question 1, Review Exercise
- If $f(x)$ is divided by $x-2$, then remainder is 12 . What is $f(2)$ ?\\ * (a) $-12$ * (b) $\quad f(-2)$ * %%(c)%% $12$ * (d) zero\\ <btn type="link" collapse="a8">See Ans... /btn><collapse id="a8" collapsed="true">%%(c)%%: $12$</collapse> ====Go to ==== <text align="right
- Question 5 and 6, Exercise 5.2
- )$ is one of the factor of $2 x^{3}-15 x^{2}+16 x+12$, find its other factors. **Solution.** It is g... n{align} \begin{array}{r|rrrr} 2 & 2 & -15 & 16 & 12 \\ & & 4 & -22 & -12 \\ \hline & 2 & -11 & -6 & 0 \\ \end{array} \end{align} This gives: \begin{al
- Question 4 and 5, Exercise 5.1
- n*} &(-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:
- Question 6 and 7, Exercise 5.1
- (2) - m \\ & = 2(8) + 3(4) - 3(2) - m \\ & = 16 + 12 - 6 - m \\ & = 22 - m. \end{align*} Given that th... gin{align*} p(-2)&=(-2)^3-7(-2)+6\\ &=-8+14+6\\ &=12 \neq 0\end{align*} This gives -2 is not zero of $
- Question 8 and 9, Exercise 5.1
- egin{align} p(2) &= 2(2)^3+3(2)^2-11(2)-6 \\ &=16+12-22-6 = 0 \end{align} Hence 2 is zero of $p(x)$. \... rr} 4 & 1 & -1 & -14 & 11 \\ & \downarrow & 4 & 12 & -8 \\ \hline & 1 & 3 & -2 & |\,\, 3 \\ \end{arr
- Question 4 & 5, Review Exercise
- 5x - 3)(x + 2)\\ &=\frac{1}{5} (5x^2 - 3x - 20x + 12 )(x + 2)\\ &=\frac{1}{5} (5x^2 - 23x + 12 )(x + 2)\\ &= \frac{1}{5}\left(5x^3 + 10x^2 - 23x^2 - 46x +
- Question 1, Exercise 5.1
- ) \\ & = 2(-2)^{3}+3 (-2)^{2}-4 (-2)+1 \\ & = -16+12+8+1 \\ &= 5. \end{align*} Hence remiander = 5. ==
- Question 1, Exercise 5.3
- n 1===== The volume of a drinking water bottle is 120 cubic centimeters. The bottle is 7 centimeters l... of bottle = $x+3+7=x+10$ cm\\ Volume of bottle = $120 cm^3$ 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