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- Question 1, Exercise 5.1
- \ $x-c=x+2 \implies c=-2$. By Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c) = p... - 2 \implies c = 2 \). By the Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c) = p
- Question 1, Review Exercise
- $, then $x+1$ will be its:\\ * (a) divisor as well as factor\\ * (b) dividend\\ * %%(c)%%... collapse id="a4" collapsed="true">(a): divisor as well as factor </collapse> v. If 2 is a zero of the
- Question 6 & 7, Review Exercise
- ** Solution. ** Let \( p(x) = x^{2} + 8x + k \). We are given that the remainder upon dividing \( p(x... e remainder is \( p(4) \). Since \( p(4) = 0 \), we have: \begin{align*} p(4) &= (4)^2 + 8(4) + k \\
- Question 6 and 7, Exercise 5.1
- x-2$ $\implies c=2$. \\ By the Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c) = p
- Question 10, Exercise 5.1
- align} This gives $$ p(x) = (x+1)(x^2+10x+24)$$ We have volume of room = area of floor $\times$ heig
- Question 3 and 4, Exercise 5.2
- x) &= (x + 2)(2x^{2} + x - 9). \end{align*} Thus, we can factor \( 2x^{2} + x - 9 \) as: \begin{align*
- Question 1, Exercise 5.3
- Volume of bottle = $120 cm^3$ By given condtion, we have \begin{align*} & x(x+3)(x+10)=120 \\ \implie
- Question 2, Exercise 5.3
- 650. \end{align*} During the 12th game 650 ticket were sold. GOOD ====Go to ==== <text align
- Question 5, Exercise 5.3
- a of $ACED$ = $6 x^{2}+38 x+56$ Width = $2 x+8$ We have \begin{align*} & 6 x^{2}+38 x+56 \\ = & 2(3x
- Question 4 & 5, Review Exercise
- (x) \). Given the zeros \( 4, \frac{3}{5}, -2 \), we can write the polynomial as: \[ f(x) = (x - 4)\l