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- Question 1, Exercise 5.1
- {2}-4 x+1$ is divided by $x+2$. ** Solution. ** Given: $p(x)=2 x^{3}+3 x^{2}-4 x+1$\\ $x-c=x+2 \implies... {2}+2 x+3$ is divided by $x-2$. ** Solution. ** Given: \( p(x) = x^{4} + 2x^{3} - x^{2} + 2x + 3 \) \\
- Question 7 and 8, Exercise 5.2
- ght)=0$, then factorize $h(x)$. ** Solution. ** Given: $h(x)=4 x^{3}+4 x^{2}+73 x+36$ and $h\left(\frac{-1}{2}\right)=0$. This given $-\frac{1}{2}$ is zeor of $h(x)$. By using synthe
- Question 4 & 5, Review Exercise
- ctor of $6 y^{3}-y^{2}-5 y+2$ ? ** Solution. ** Given \begin{align*}3y-2&=0\\ 3y&=2\\ y&=\frac{2}{3}\en... n. ** Let the required polynomial be \( f(x) \). Given the zeros \( 4, \frac{3}{5}, -2 \), we can write
- Question 6 and 7, Exercise 5.1
- \ & = 16 + 12 - 6 - m \\ & = 22 - m. \end{align*} Given that the remainder is 16, so \begin{align*} 22 -
- Question 8 and 9, Exercise 5.1
- nce $2$, $3$ and $-\dfrac{1}{2}$ are the roots of given polynomial. GOOD =====Question 9===== Express
- Question 5 and 6, Exercise 5.2
- 2$, find its other factors. **Solution.** It is given by the factor theorem, \( x - 2 \) is a factor of
- Question 1, Exercise 5.3
- +3+7=x+10$ cm\\ Volume of bottle = $120 cm^3$ By given condtion, we have \begin{align*} & x(x+3)(x+10)=1
- Question 2, Exercise 5.3
- fth game of the cricket season. ** Solution. ** Given: $$t(x)=x^{3}-12 x^{2}+48 x+74.$$ When $t=12$ \be
- Question 3, Exercise 5.3
- h = $2x+2$ units \\ Volume = 144 cubic units. By given condition \begin{align*} & x(2x)(2x+2) = 144 \\
- Question 4, Exercise 5.3
- t = $x-2$ units \\ Volume = 2475 cubic units. By given condition \begin{align*} & x(2x+3)(x-2) = 2475 \
- Question 5, Exercise 5.3
- and the area of square $ABFG$. ** Solution. ** Given: Area of $ACED$ = $6 x^{2}+38 x+56$ Width = $2
- Question 6 & 7, Review Exercise
- tion. ** Let \( p(x) = x^{2} + 8x + k \). We are given that the remainder upon dividing \( p(x) \) by \(