Search
You can find the results of your search below.
Fulltext results:
- 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(2) \
- Question 6 and 7, Exercise 5.1
- $ $\implies c=2$. \\ By the Remainder Theorem, we have \begin{align*} \text{Remainder} & = p(c) = p(2) \
- Question 10, Exercise 5.1
- gn} This gives $$ p(x) = (x+1)(x^2+10x+24)$$ We have volume of room = area of floor $\times$ height.
- Question 1, Exercise 5.3
- ume of bottle = $120 cm^3$ By given condtion, we have \begin{align*} & x(x+3)(x+10)=120 \\ \implies &
- Question 5, Exercise 5.3
- f $ACED$ = $6 x^{2}+38 x+56$ Width = $2 x+8$ We have \begin{align*} & 6 x^{2}+38 x+56 \\ = & 2(3x^2+19
- Question 6 & 7, Review Exercise
- emainder is \( p(4) \). Since \( p(4) = 0 \), we have: \begin{align*} p(4) &= (4)^2 + 8(4) + k \\ &= 16