Search
You can find the results of your search below.
Fulltext results:
- Question 5 and 6, Exercise 5.2
- (1)^{2} + 3(1) - 5 \\ &= 1 + 1 + 3 - 5 \\ &= 0. \end{align*} By the factor theorem, \( t - 1 \) is a ... \\ & & 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{align*} Thus, the factorization is: \begin{align*
- Question 3 and 4, Exercise 5.2
- ) + 5(4) + 18 - 18 \\ &= -16 + 20 + 18 - 18 = 0. \end{align*} By the factor theorem, \( x + 2 \) is a ... & -4 & -2 & 22 \\ \hline & 2 & 1 & -11 & 0 \\ \end{array} \] This gives: \begin{align*} f(x) &= (x + 2)(2x^{2} + x - 9). \end{align*} Thus, we can factor \( 2x^{2} + x - 9 \) ... x + 3) - 3(x + 3)] \\ &= (x + 2)(2x - 3)(x + 3). \end{align*} The solution set is $$f(x)=(x + 2)(2x - 3
- Question 1 and 2, Exercise 5.2
- lign*} f(-1)&=(-1)^{3}-7 (-1)-6 \\ &= -1+7-6 =0. \end{align*} By factor theorem, $y+1$ is factor of $f(... w & -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+2y-6) \\ & = (y+1)(y(y-3)+2(y-3)) \\ & = (y+1)(y-3)(y+2). \end{align*} GOOD =====Question 2===== Factorize
- Question 7 and 8, Exercise 5.2
- & 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^{2} - 14x + 20)... 2)(x - 5)\\ &=\left(2x - 1\right) (x - 2)(x - 5) \end{align*} Finally, the complete factorization is: ... gn*} f(x) &= \left(2x - 1\right) (x - 2)(x - 5). \end{align*} =====Question 8===== If $h(x)=4 x^{3}
- Question 8 and 9, Exercise 5.1
- (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 usin... row & 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(x+3)+1(x+3) \\ = & (x+3)(2x+1) \end{align*} i.e. $x+3=0$ or $2x+1=0$ $\implies$ $x=-3
- Question 6 and 7, Exercise 5.1
- - 3(2) - m \\ & = 16 + 12 - 6 - m \\ & = 22 - m. \end{align*} Given that the remainder is 16, so \begin... ign*} 22 - m & = 16 \\ m & = 22 - 16 \\ m & = 6. \end{align*} Hence, the value of \( m \) is 6. GOOD ... \begin{align*} p(1)&=(1)^3-7(1)+6\\ &=1-7+6\\ &=0\end{align*} Hence $1$ is the zero of p(x). \\ Similar... } p(-2)&=(-2)^3-7(-2)+6\\ &=-8+14+6\\ &=12 \neq 0\end{align*} This gives -2 is not zero of $p(x)$.\\ He
- Question 4 & 5, Review Exercise
- en \begin{align*}3y-2&=0\\ 3y&=2\\ y&=\frac{2}{3}\end{align*} Suppose \begin{align*} f(y) &= 6y^{3} - ... frac{16 - 4 - 30 + 18}{9} \\ &= \frac{0}{9} = 0. \end{align*} Hence by the factor theorem, \( 3y - 2 \... \frac{1}{5}\left(5x^3 - 13x^2 - 34x + 24\right) \end{align*} Thus, the required polynomial is: \[ f(x
- Question 1, Exercise 5.1
- }+3 (-2)^{2}-4 (-2)+1 \\ & = -16+12+8+1 \\ &= 5. \end{align*} Hence remiander = 5. =====Question 1(ii)=... + 2(2) + 3 \\ & = 16 + 16 - 4 + 4 + 3 \\ & = 35. \end{align*} Hence, the remainder is 35. ====Go t
- Question 2 and 3, Exercise 5.1
- )&=3^3-2(3)^2-5(3)+6 \\ & = 27-18-15+6 \\ & = 0. \end{align*} Hence, $x-3$ is factor of $p(x)$. GOOD =... 2(3)^2-5(3)+1 \\ & = 27-18-15+1 \\ & = -5 \neq 0 \end{align*} Hence, $x-3$ is not factor of $p(x)$. GOO
- Question 10, Exercise 5.1
- -1 & -10 & -24 \\ \hline & 1 & 10 & 24 & 0 \\ \end{array}\end{align} This gives $$ p(x) = (x+1)(x^2+10x+24)$$ We have volume of room = area of floor $\
- Question 1, Exercise 5.3
- +10x+30)-120=0\\ \implies & x^3+13x^2+30x-120=0. \end{align*} Consider $$p(x)=x^3+13x^2+30x-120$$ Now ... 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
- Question 3, Exercise 5.3
- implies & 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
- Question 4, Exercise 5.3
- x^2-x-6)-2475=0 \\ \implies & 2x^3-x^2-6x-2475=0 \end{align*} Suppose $$p(x)=2x^3-x^2-6x-2475.$$ Since... (11)^3-11^2-6(11)-2475 \\ &=2662-121-66-2475 = 0 \end{align*} This gives $x=11$ is the zeros of $p(x)$.
- Question 5, Exercise 5.3
- 4)+7(x+4)) \\ =& 2(x+4)(3x+7) \\ =& (2x+8)(3x+7) \end{align*} Now \begin{align*} & Length \times Width... 8) = (2x+8)(3x+7) \\ \implies & Length = 3x+7 \\ \end{align*} Hence length of rectangle $ACED$ = $3x+4
- Question 6, Exercise 5.3
- w & 2 & 0 & -2 \\ \hline & 1 & 0 & -1 & 0 \\ \end{array} \] Thus \begin{align*} p(y) & = (y-2)(y^2-1) \\ & = (y-2)(y+1)(y-1) \end{align*} Hence the length of remaining two sides a