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- Question 3 and 4, Exercise 5.2
- sing factor theorem: $2 x^{3}+5 x^{2}-9 x-18$ ** Solution. ** Suppose \( f(x) = 2x^{3} + 5x^{2} - 9x - 18 ... )] \\ &= (x + 2)(2x - 3)(x + 3). \end{align*} The solution set is $$f(x)=(x + 2)(2x - 3)(x + 3).$$ =====Que... by using factor theorem: $3 x^{3}-5 x^{2}-36$ **Solution.** Suppose \( f(x) = 3x^{3} - 5x^{2} - 36 \). \
- Question 1, Exercise 5.1
- $2 x^{3}+3 x^{2}-4 x+1$ is divided by $x+2$. ** Solution. ** Given: $p(x)=2 x^{3}+3 x^{2}-4 x+1$\\ $x-c=x... {4}+2 x^{3}-x^{2}+2 x+3$ is divided by $x-2$. ** Solution. ** Given: \( p(x) = x^{4} + 2x^{3} - x^{2} + 2x
- Question 2 and 3, Exercise 5.1
- t $x-3$ is a factor of $x^{3}-2 x^{2}-5 x+6$. ** Solution. ** Let $p(x)=x^{3}-2 x^{2}-5 x+6$ and $x-c=x-3$... is a factor of $x^{3}-2 x^{2}-5 x+1$ or not. ** Solution. ** Let $p(x)=x^{3}-2 x^{2}-5 x+1$ and $x-c=x-3$
- Question 4 and 5, Exercise 5.1
- is $4 y^{2}-8 y+10$, then find other factor. ** Solution. ** =====Question 5===== Find the value of... x^{2}-7 x+6$ is exactly divisible by $(x+1)$. ** Solution. ** Let $p(x)=x^{3}+q x^{2}-7 x+6$ and $x-c=x+1$
- Question 6 and 7, Exercise 5.1
- divided by $x-2$ gives the remainder of 16 . ** Solution. ** Let $p(x)=2 x^{3}+3 x^{2}-3 x-m$ and $x-c=x-... ther 1 and -2 are the zeros of $x^{3}-7 x+6$. ** Solution. ** Suppose $p(x)=x^3-7x+6$.\\ $1$ will be zero o
- Question 8 and 9, Exercise 5.1
- s of the polynomial $2 x^{3}+3 x^{2}-11 x-6$. ** Solution. ** Suppose $p(x)=2x^3+3x^2-11x-6$. \\ Since \be... )+r$, where $a=4$. (:!: statement corrected). ** Solution. ** By synthetic division \begin{align} \begin{a
- Question 1 and 2, Exercise 5.2
- torize by using factor theorem: $y^{3}-7 y-6$ ** Solution. ** Suppose $f(y)=y^{3}-7 y-6$. \begin{align*} ... y using factor theorem: $2 x^{3}-x^{2}-2 x+1$ ** Solution. ** \begin{align*} f(x) &= 2x^{3} - x^{2} - 2x +
- Question 5 and 6, Exercise 5.2
- by using factor theorem: $t^{3}+t^{2}+3 t-5$ ** Solution. ** Suppose \( f(t) = t^{3} + t^{2} + 3t - 5 \).... {3}-15 x^{2}+16 x+12$, find its other factors. **Solution.** It is given by the factor theorem, \( x - 2 \
- Question 7 and 8, Exercise 5.2
- 10$ if ' $\dfrac{1}{2}$ ' is one of its zero. ** Solution. ** Using synthetic division to divide \( f(x) \... frac{-1}{2}\right)=0$, then factorize $h(x)$. ** Solution. ** Given: $h(x)=4 x^{3}+4 x^{2}+73 x+36$ and $h
- Question 2 & 3, Review Exercise
- $\left(64 y^{3}-8\right) \div(4 y-2) \quad$ ** Solution. ** \begin{align*} \frac{(64 y^{3}-8)}{(4 y-2)}&... 3===== $\left(125 y^{3}-8\right) \div(5 y-2)$ ** Solution. ** \begin{align*} \frac{(125 y^{3}-8)}{(5 y-2)}
- Question 4 & 5, Review Exercise
- s $3 y-2$ a factor of $6 y^{3}-y^{2}-5 y+2$ ? ** Solution. ** Given \begin{align*}3y-2&=0\\ 3y&=2\\ y&=\fr... are $4, \frac{3}{5},-2$, find the polynomial. ** Solution. ** Let the required polynomial be \( f(x) \). G
- Question 6 & 7, Review Exercise
- \left(x^{2}+8 x+k\right)$ by $(x-4)$ is zero. ** Solution. ** Let \( p(x) = x^{2} + 8x + k \). We are give... -x+32-\frac{121}{x+4}$. What is the dividend? ** Solution. ** :!: The question doesn't seem to solvable.
- Question 10, Exercise 5.1
- is $(x+1)$ feet. Find the area of its floor. ** Solution. ** Suppose $p(x)=x^{3}+11 x^{2}+34 x+24$. By u
- Question 1, Exercise 5.3
- he dimensions of the bottle. (:!: Correction) ** Solution. ** Consider length of bottle = $x$ cm\\ Width o
- Question 2, Exercise 5.3
- uring the twelfth game of the cricket season. ** Solution. ** Given: $$t(x)=x^{3}-12 x^{2}+48 x+74.$$ When