Quadratic Functions (IGCSE Old Questions)

1. $f(x)=x^{2}+6 x+8$
   
   Given that $f(x)$ can be expressed in the form $(x+A)^{2}+B$ where $A$ and $B$ are constants,
   1. find the value of $A$ and the value of $B$.
   2. Hence, or otherwise, find
      1. the value of $x$ for which $f(x)$ has its least value
      2. the least value of $f(x)$
   The curve $C$ has equation $y=x^{2}+6 x+8$ The line $l$, with equation $y=2-x$, intersects $C$ at two points.
   3. Find the $x$-coordinate of each of these two points.
   4. Find the $x$-coordinate of the points where $C$ crosses the $x$-axis.



$\begin{aligned} f(x) &=x^{2}+6 x+8 \\\\ &=x^{2}+2(3 x)+9-1 \\\\ &=\left(x^{2}+3\right)^{2}-1\\\\ f(x) &=x^{2}+6 x+8 \\\\ &=x^{2}+2(3 x)+9-1 \\\\ &=\left(x^{2}+3\right)^{2}-1 \end{aligned}$ (a) $A=3$ and $B=-1$ (b) (i) $f(x)$ has the least value when $x=-3$. $\quad$(ii) The least value of $f(x)=-1$. $\begin{aligned} & C: \quad y=x^{2}+6 x+8 \\\\ & l: \quad y=2-x \end{aligned}$ (c) At the points of intersection of $C$ and $l$, $\begin{aligned} & x^{2}+6 x+8=2-x \\\\ &\therefore x^{2}+7 x+6=0 \\\\ &(x+6)(x+1)=0 \\\\ &\therefore x=-6 \text { or } x=-1 \\\\ &\text { At } x=-6, y=2-(-6)=8 \\\\ &\text { At } x=-1, y=2-(-1)=3 \end{aligned}$ $\therefore\quad$ The point of intersection of $C$ and $l$ are $(-6,8)$ and $(-1,3)$. (d) When $C$ cross $x$-axis, $\begin{aligned} & x^{2}+6 x+8=0 \\\\ & (x+2)(x+4)=0 \end{aligned}$ $\therefore\quad$ The curve $C$ cross $x$-axis at $(-2,0)$ and $(-4,0)$.




2. $f(x)=3 x^{2}+6 x+7$
   
   Given that $\mathrm{f}(x)$ can be written in the form $A(x+B)^{2}+C$, where $A, B$ and $C$ are rational numbers,
   1. find the value of $A$, the value of $B$ and the value of $C$.
   2. Hence, or otherwise, find
      1. the value of $x$ for which $\dfrac{1}{f(x)}$ is a maximum,
      2. the maximum value of $\dfrac{1}{f(x)}$.



$\begin{aligned} f(x) &=3 x^{2}+6 x+7 \\\\ &=3 x^{2}+6 x+3+4 \\\\ &=3\left(x^{2}+2 x+1\right)+4 \\\\ &=3(x+1)^{2}+4 \end{aligned}$ (a) $A=3, B=1, C=4$ (b) (i) $\dfrac{1}{f(x)}$ is maximum when $f(x)$ is minimun. $\quad \quad f(x)$ is minimun when $x=-1.$ $\therefore \quad\dfrac{1}{f(x)}$ is maximum when $x=-1.$ $\quad$ (ii) The minimum value of $f(x)$ is 4. $\therefore \quad$ The maximum value of $f(x)$ is $\dfrac{1}{4}$.




3. $f(x)=2 x^{2}-8 x+5$
   
   Given that $f(x)$ can be written in the form $a(x-b)^{2}+c$
   1. find the value of $a$, the value of $b$ and the value of $c$.
   2. Write down
      1. the minimum value of $f(x)$
      2. the value of $x$ at which this minimum occurs.



$\begin{aligned} f(x) &=2 x^{2}-8 x+5 \\\\ &=2\left(x^{2}-4 x\right)+5 \\\\ &=2\left(x^{2}-4 x+4\right)+5-8 \\\\ &=2(x-2)^{2}-3 \end{aligned}$ (a) $a=2, b=2$ and $c=-3$ (b) (i) the minimun value of $f(x)=-3$ $\quad$ (ii) $f(x)$ is minimum when $x=2$.




4. $f(x)=6+5 x-2 x^{2}$
   
   Given that $f(x)$ can be written in the form $p(x+q)^{2}+r$, where $p, q$ and $r$ are rational numbers,
   1. find the value of $p$, the value of $q$ and the value of $r$.
   2. Hence, or otherwise, find
      1. the maximum value of $f(x)$
      2. the value of $x$ for which this maximum occurs.
   $g(x)=6+5 x^{3}-2 x^{6}$
   
   
   3. Write down
      1. the maximum value of $g(x)$,
      2. the exact value of $x$ for which this maximum occurs.



$\begin{aligned} f(x) &=6+5 x-2 x^{2} \\\\ &=-2\left(x^{2}-\dfrac{5}{2} x\right)+6 \\\\ &=-2\left(x^{2}-2\left(\dfrac{5}{4}\right) x+\left(\dfrac{5}{4}\right)^{2}\right)+6+2\left(\dfrac{5}{4}\right)^{2} \\\\ &=-2\left(x-\dfrac{5}{4}\right)^{2}+6+\dfrac{25}{8} \\\\ &=-2\left(x-\dfrac{5}{4}\right)^{2}+\dfrac{73}{8} \end{aligned}$ (a) $P=-2, \quad q=-\dfrac{5}{4}, r=\dfrac{73}{8}$ (b) (i) the maximum value of $f(x)=\dfrac{73}{8}$ $\quad$ (ii) $f(x)$ is maximun when $x=\dfrac{5}{4}$ $\quad\quad g(x)=6+5 x^{3}-2 x^{6}$ $\quad\quad\quad\quad=-2\left(x^{3}-\dfrac{5}{4}\right)^{2}+\dfrac{73}{8}$ (c) the maximum value of $g(x)=\dfrac{73}{8}$ $\quad g(x)$ has maximum value when $\begin{aligned} \quad &x^{3}-\dfrac{5}{4}=0 \\ \quad &x=\sqrt[3]{\dfrac{5}{4}} \end{aligned}$




5. The roots of the equation $x^{2}+6 x+2=0$ are $\alpha$ and $\beta$, where $\alpha>\beta$. Without solving the equation,
   1. find
      1. the value of $\alpha^{2}+\beta^{2}$,
      2. the value of $\alpha^{4}+\beta^{4}$.
   2. Show that $\alpha-\beta=2 \sqrt{7}$.
   3. Factorise completely $\alpha^{4}-\beta^{4}$.
   4. Hence find the exact value of $\alpha^{4}-\beta^{4}$.
   5. Given that $\beta^{4}=A+B \sqrt{7}$ where $A$ and $B$ are positive constants find the value of $A$ and the value of $B$.



$x^{2}+6 x+2=0$ $\alpha$ and $\beta$ are the roots of the equation. $\begin{aligned} \therefore \quad x^{2}+6 x+2 &=(x-\alpha)(x-\beta) \\\\ &=x^{2}-(\alpha+\beta) x+\alpha \beta \\\\ \therefore \quad \alpha+\beta=-6 & \text { and } \alpha \beta=2 \end{aligned}$ $\text{(a) (i) Since we have } (\alpha+\beta)^{2}=\alpha^{2}+2 \alpha \beta+\beta^{2}$, $\begin{aligned} \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=36-4 \\\\ &=32 \end{aligned}$ $\begin{aligned} \text{(ii)}\qquad \alpha^{4}+\beta^{4} & \\\\ \left(\alpha^{2}\right)^{2}+\left(\beta^{2}\right)^{2}&=\left(\alpha^{2}+\beta^{2}\right)^{2}-2\left(\alpha^{2} \beta^{2}\right) \\\\ &=\left(\alpha^{2}+\beta^{2}\right)^{2}-2(\alpha \beta)^{2} \\\\ &=(32)^{2}-2(2)^{2} \\\\ &=1016 \end{aligned}$ $ \begin{aligned} \text{(b) Since }(\alpha-\beta)^{2} &=\alpha^{2}-2 \alpha \beta+\beta^{2} \\\\ &=\alpha^{2}+\beta^{2}-2(\alpha \beta) \\\\ &=32-2(2) \\\\ &=28 \\\\ \alpha-\beta &=\sqrt{28} \\\\ &=2 \sqrt{7} \end{aligned}$ $\begin{aligned} \text{(c) } \alpha^{4}-\beta^{4} &=\left(\alpha^{2}-\beta^{2}\right)\left(\alpha^{2}+\beta^{2}\right) \\\\ &=(\alpha-\beta)(\alpha+\beta)\left(\alpha^{2}+\beta^{2}\right) \end{aligned}$




6. The equation $x^{2}+m x+15=0$ has roots $\alpha$ and $\beta$ and the equation $x^{2}+h x+k=0$ has roots $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$
   1. Write down the value of $k$.
   2. Find an expression for $h$ in terms of $m$.
   
   Given that $\beta=2 \alpha+1$,
   
   
   3. find the two possible values of $\alpha$.
   4. Hence find the two possible values of $m$.



$\alpha$ and $\beta$ are the roots of the equation $x^{2}+m x+15=0$ $\begin{aligned} \therefore \quad & x^{2}+m x+15=(x-\alpha)(x-\beta) \\\\ & x^{2}+m x+15=x^{2}-(\alpha+\beta) x+\alpha \beta \end{aligned}$ $\therefore \alpha+\beta=-m$ and $\alpha \beta=15$ $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ are the rodts of the equation $x^{2}+h x+k=0$ $\therefore x^{2}+h x+k=\left(x-\dfrac{\alpha}{\beta}\right)\left(x-\dfrac{\beta}{\alpha}\right)$ $\therefore x^{2}+h x+k=x^{2}-\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right) x+1$ $\begin{aligned} \text {(a) } k&=1\\\\ \text { (b) }h &=-\left(\dfrac{\alpha}{\beta}+\dfrac{B}{\alpha}\right) \\\\ &=-\dfrac{\alpha^{2}+\beta^{2}}{\alpha \beta} \\\\ &=-\dfrac{(\alpha+\beta)^{2}-2 \alpha \beta}{d \beta} \\\\ &=-\dfrac{m^{2}-30}{15} \\\\ &=\dfrac{30-m^{2}}{15} \end{aligned}$ $\begin{aligned} \text { (c) }&\beta=2 \alpha+1 \\\\ &\alpha \beta=15 \\\\ & \alpha(2 d+1)=15 \\\\ &2 \alpha^{2}+\alpha-15=0 \\\\ &(\alpha+3)(2 \alpha-5)=0 \\\\ &\alpha=-3 \text { or } \alpha=\dfrac{5}{2} \end{aligned}$ $\begin{aligned} \text { (d) } \alpha+\beta &=-m \\\\ m &=-\alpha-\beta \\\\ &=-\alpha-2 \alpha-1 \\\\ &=-(3 \alpha+1) \\\\ \text { when } \alpha &=-3, m=8 \\\\ \text { when } \alpha &=\dfrac{5}{2}, m=-\dfrac{17}{2} \end{aligned}$




7. The equation $2 x^{2}-7 x+4=0$ has roots $\alpha$ and $\beta$ Without solving this equation, form a quadratic equation with integer coefficients which has roots $\alpha+\dfrac{1}{\beta}$ and $\beta+\dfrac{1}{\alpha}$.



The roots of the equation $2 x^{2}-7 x+4=0$ are $\alpha$ and $\beta$. $\begin{aligned} \therefore 2 x^{2}-7 x+4 &=2(x-\alpha)(x-\beta) \\\\ &=2 x^{2}-2(\alpha+\beta) x+2 \alpha \beta \\\\ \therefore \alpha+\beta=\dfrac{7}{2},\ & \alpha \beta=2 \end{aligned}$ The quadratic equation which has roots $\alpha+\dfrac{1}{\beta}$ and $\beta+\dfrac{1}{\alpha}$ is $\begin{aligned} &\left[x-\left(\alpha+\dfrac{1}{\beta}\right)\right]\left[x-\left(\beta+\dfrac{1}{\alpha}\right)\right]=0 \\\\ \therefore\quad & x^{2}-\left(\alpha+\beta+\dfrac{1}{\alpha}+\dfrac{1}{\beta}\right) x+\left(\alpha \beta+\dfrac{1}{\alpha \beta}+2\right)=0 \end{aligned}$ $\begin{aligned} \quad &x^{2}-\left(\alpha+\beta+\dfrac{\alpha+\beta}{\alpha \beta}\right) x+\left(\alpha \beta+\dfrac{1}{\alpha \beta}+2\right)=0 \\\\ \quad &x^{2}-\left(\dfrac{7}{2}+\dfrac{7}{4}\right) x+\left(2+\dfrac{1}{2}+2\right)=0 \\\\ \quad &4 x^{2}-(14+7) x+(8+2+8)=0 \\\\ \quad &4 x^{2}-21 x+18=0 \end{aligned}$




8. $f(x)=2 x^{2}-5 x+1$
   
   The equation $f(x)=0$ has roots $\alpha$ and $\beta$. Without solving the equation
   1. find the value of $\alpha^{2}+\beta^{2}$
   2. show that $\alpha^{4}+\beta^{4}=\dfrac{433}{16}$
   3. form a quadratic equation with integer coefficients which has roots $\left(\alpha^{2}+\dfrac{1}{\alpha^{2}}\right)$ and $\left(\beta^{2}+\dfrac{1}{\beta^{2}}\right)$



$f(x)=2 x^{2}-5 x+1$ The roots of the equation $f(x)=0$ are $\alpha$ and $\beta$ $\begin{aligned} &2 x^{2}-5 x+1=2(x-\alpha)(x-\beta) \\\\ &2 x^{2}-5 x+1=2 x^{2}-2(\alpha+\beta) x+22 \beta \end{aligned}$ $\therefore \alpha+\beta=\dfrac{5}{2}$ and $\alpha \beta=\dfrac{1}{2}$ $ \begin{aligned} \text { (a) } \alpha^{2}+\beta^{2} &=\left(\alpha+\beta^{2}\right)-2 \alpha \beta \\\\ &=\dfrac{25}{4}-1 \\\\ &=\dfrac{21}{4}\\\\ \text { (b) } \alpha^{4}+\beta^{4} &=\left(\alpha^{2}+\beta^{2}\right)^{2}-2(\alpha \beta)^{2} \\\\ &=\left(\dfrac{21}{4}\right)^{2}-2\left(\dfrac{1}{2}\right)^{2} \\\\ &=\dfrac{433}{16} \end{aligned}$ $\text { (c) }$ The quadratic equation which has roots $\left(\alpha^{2}+\dfrac{1}{\alpha^{2}}\right)$ and $\left(\beta^{2}+\dfrac{1}{\beta^{2}}\right)$ is $\begin{aligned} &{\left[x-\left(\alpha^{2}+\dfrac{1}{\alpha^{2}}\right)\right]\left[x-\left(\beta^{2}+\dfrac{1}{\beta^{2}}\right)\right]=0} \\\\ &x^{2}-\left(a^{2}+\beta^{2}+\dfrac{1}{\alpha^{2}}+\dfrac{1}{\beta^{2}}\right) x+\left(\alpha^{2} \beta^{2}+\dfrac{\alpha^{2}}{\beta^{2}}+\dfrac{\beta^{2}}{\alpha^{2}}+\dfrac{1}{\alpha^{2} \beta^{2}}\right)=0 \\\\ &x^{2}-\left[\alpha^{2}+\beta^{2}+\dfrac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}}\right] x+\left((a \beta)^{2}+\dfrac{\alpha^{4}+\beta^{4}}{(\alpha \beta)^{2}}+\dfrac{1}{(\alpha \beta)^{2}}\right)=0 \\\\ &x^{2}-\left(\dfrac{21}{4}+\dfrac{21}{4} \times 4\right) x+\left(\dfrac{1}{4}+\dfrac{433}{16}(4)+4\right)=0 \\\\ &4 x^{2}-105 x+450=0 \end{aligned}$




9. The equation $x^{2}+p x+1=0$ has roots $\alpha$ and $\beta$
   1. Find, in terms of $p$, an expression for
      1. $\alpha+\beta$.
      2. $\alpha^{2}+\beta^{2}$.
      3. $\alpha^{3}+\beta^{3}$.
   2. Find a quadratic equation, with coefficients expressed in terms of $p$, which has roots $a^{3}$ and $\beta^{3}$.



$\therefore \quad x^{2}+p x+1=(x-\alpha)(x-\beta)$ $x^{2}+p x+1=x^{2}-(\alpha+\beta) x+\alpha \beta$ $\begin{aligned} \text{(a) } &\alpha+\beta=-p, \alpha \beta=1\\\\ \text{(i) } & \therefore\ \alpha+\beta=-p\\\\ \text{(ii) }& \alpha^{2}+\beta^{2}\\\\ & =(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=p^{2}-1\\\\ \text{(iii) } &\text { Since }(\alpha+\beta)^{3}=\alpha^{3}+3 \alpha^{2} \beta+3 \alpha \beta^{2}+\beta^{3} \\\\ &=\alpha^{3}+\beta^{3}+3 \alpha \beta(\alpha+\beta) \\\\ &\alpha^{3}+\beta^{3}=\left(\alpha+\beta^{3}\right)-3 \alpha \beta(\alpha+\beta)\\\\ &\alpha^{3}+\beta^{3}=-p^{3}-3(-p)=3 p-p^{3} \end{aligned}$ $\text{(b)}$ The guadratic equation Which has roots $\alpha^{3}$ and $\beta^{3}$ is $\begin{aligned} &\left(x-\alpha^{3}\right)\left(x-\beta^{3}\right)=0 \\\\ &x^{2}-\left(\alpha^{3}+\beta^{3}\right) x+(\alpha \beta)^{3}=0 \\\\ &x^{2}-\left(3 p-p^{3}\right) x+1=0 \\\\ &x^{2}+\left(p^{3}-3 p\right) x+1=0 \end{aligned}$




10. 1. Show that $(\alpha+\beta)\left(\alpha^{2}-\alpha \beta+\beta^{2}\right)=\alpha^{3}+\beta^{3}$.
       The roots of the equation $2 x^{2}+6 x-7=0$ are $\alpha$ and $\beta$ where $\alpha>\beta$.
       Without solving the equation,
    2. find the value of $\alpha^{3}+\beta^{3}$.
    3. show that $\alpha-\beta=\sqrt{23}$.
    4. Hence find the exact value of $\alpha^{3}-\beta^{3}$.



$\begin{aligned} \text{(a) }\quad &(\alpha+\beta)\left(\alpha^{2}-\alpha \beta+\beta^{2}\right) \\\\ &=\alpha^{3}-\alpha^{2} \beta+\alpha \beta^{2}+\alpha^{2} \beta-\alpha \beta^{2}+\beta^{3} \\\\ &=\alpha^{3}+\beta^{3} \end{aligned}$ The roots of the equation $2 x^{2}+6 x-7=0$ are $\alpha$ and $\beta$. $\begin{aligned} \therefore\quad 2 x^{2}+6 x-7 &=2(x-\alpha)(x-\beta) \\\\ &=2 x^{2}-2(\alpha+\beta)+2 \alpha \beta \end{aligned}$ $\therefore \quad a+\beta=-3$ and $\alpha \beta=-\dfrac{7}{2}$ $\begin{aligned} \therefore\quad\quad \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=9+7 \\\\ &=16 \\\\ \text{(b) }\quad \alpha^{3}+\beta^{3} &=(\alpha+\beta)\left(\alpha^{2}-\alpha \beta+\beta^{2}\right) \\\\ &=(-3)\left(16+\frac{7}{2}\right) \\\\ &=-\frac{117}{2} \end{aligned}$ $\begin{aligned} \text{(c) }\quad (a-\beta)^{2} &=a^{2}-2 \alpha \beta+\beta^{2} \\\\ &=16+7 \\\\ &=23 \\\\ \therefore \alpha-\beta &=\sqrt{23} \end{aligned}$ $\begin{aligned} \text{(d)}\quad \text{Since}(\alpha-\beta)^{3} &=\alpha^{3}-3 \alpha^{2} \beta+3 \alpha \beta^{2}-\beta^{3} \\\\ &=\alpha^{3}-\beta^{3}-3 \alpha \beta(\alpha-\beta) \\\\ \alpha^{3}-\beta^{3} &=(\alpha-\beta)^{3}+3 \alpha \beta(\alpha-\beta) \\\\ &=23 \sqrt{23}+3\left(-\frac{7}{2}\right) \sqrt{23} \\\\ &=\frac{25}{2} \sqrt{23} \end{aligned}$




11. $f(x)=x^{2}+(k-3) x+4$
    
    The roots of the equation $f(x)=0$ are $\alpha$ and $\beta$.
    1. Find, in terms of $k$, the value of $\alpha^{2}+\beta^{2}$.
    
    Given that $4\left(\alpha^{2}+\beta^{2}\right)=7 \alpha^{2} \beta^{2}$,
    
    
    2. without solving the equation $f(x)=0$, form a quadratic equation, with integer coefficients, which has roots $\dfrac{1}{\alpha^{2}}$ and $\dfrac{1}{\beta^{2}}$
    3. find the possible values of $k$.



$f(x)=x^{2}+(k-3) x+4$ The roots of $f(x)=0$ are $\alpha$ and $\beta$. $\begin{aligned} &\therefore x^{2}+(k-3) x+4=(x-\alpha)(x-\beta) \\\\ &x^{2}+(k-3) x+4=x^{2}-(\alpha+\beta)+\alpha \beta \\\\ &\therefore-(\alpha+\beta)=k-3 \\\\ &\alpha+\beta=3-k \\\\ &\alpha \beta=4 \end{aligned}$ $\begin{aligned} \text { (a) }\quad\quad \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=(3-k)^{2}-8 \\\\ &=k^{2}-6 k+1 \\\\ \alpha\left(\alpha^{2}+\beta^{2}\right) &=7 \alpha^{2} \beta^{2} \\\\ \dfrac{\alpha^{2}+\beta^{2}}{d^{2} \beta^{2}} &=\dfrac{7}{4} \\\\ \dfrac{1}{\alpha^{2}}+\dfrac{1}{\beta^{2}} &=\dfrac{7}{4} \end{aligned}$ $\text { (b) }$ The quadratic equation Which has roots $\dfrac{1}{\alpha^{2}}$ and $\dfrac{1}{\beta^{2}}$ is $\begin{aligned} &\left(x-\dfrac{1}{\alpha^{2}}\right)\left(x-\dfrac{1}{\beta^{2}}\right)=0 \\\\ &x^{2}-\left(\dfrac{1}{\alpha^{2}}+\dfrac{1}{\beta^{2}}\right) x+\dfrac{1}{(\alpha \beta)^{2}}=0 \\\\ &x^{2}-\dfrac{7}{4} x+\dfrac{1}{16}=0 \\\\ &16 x^{2}-28+1=0\\\\ \text { (c) } &\alpha\left(\alpha^{2}+\beta^{2}\right)=7 \alpha^{2} \beta^{2} \\\\ & 4\left(k^{2}-6 k+1\right)=7(4)^{2} \\\\ & k^{2}-6 k+1=28 \\\\ & k^{2}-6 k-27=0 \\\\ & (k+3)(k-9)=0 \\\\ & k=-3 \text { or } k=9 \end{aligned}$




12. $f(x)=x^{2}+p x+7 \quad p \in \mathbb{R}$.
    
    The roots of the equation $\mathrm{f}(x)=0$ are $\alpha$ and $\beta$.
    1. Find, in terms of $p$ where necessary,
       1. $\alpha^{2}+\beta^{2}$,
       2. $\alpha^{2} \beta^{2}$.
    
    Given that $7\left(\alpha^{2}+\beta^{2}\right)=5 \alpha^{2} \beta^{2}$.
    
    
    2. find the possible values of $p$.
    
    Using the positive value of $p$ found in part (b) and without solving the equation $f(x)=0$,
    
    
    3. form a quadratic equation with roots $\dfrac{2 p}{\alpha^{2}}$ and $\dfrac{2 p}{\beta^{2}}$.



$f(x)=x^{2}+p x+7, p \in R$ The roots of $f(x)=0$ are $\alpha$ and $\beta$. $\begin{aligned} \therefore\quad x^{2}+p x+7 &=(x-\alpha)(x-\beta) \\\\ x^{2}+p x+7 &=x^{2}-(\alpha+\beta) x+\alpha \beta\\\\ \therefore\quad \alpha+\beta=-p\ & \text{ and }\ \alpha \beta=7\\\\ \text { (i) } \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=p^{2}-14\\\\ \text { (ii) }\alpha^{2} \beta^{2} &=(\alpha \beta)^{2}=49\\\\ 7\left(a^{2}+\beta^{2}\right) &=5 \alpha^{2} \beta^{2} \\\\ 7\left(p^{2}-14\right) &=5(49) \\\\ p^{2}-14 &=35 \\\\ p^{2} &=49 \\\\ p &=\pm 7 \end{aligned}$ The quadratic equations which has roots $\dfrac{2 p}{\alpha^{2}}$ and $\dfrac{2 p}{\beta^{2}}$ where $p>0$ is $\left(x-\dfrac{2 p}{\alpha^{2}}\right)\left(x-\dfrac{2 p}{\beta^{2}}\right)=0$ $\begin{aligned} &x^{2}-\left(\dfrac{2 p}{\alpha^{2}}+\dfrac{2 p}{\beta^{2}}\right) x+\dfrac{4 p^{2}}{\alpha^{2} \beta^{2}}=0 \\\\ &x^{2}-2 p\left(\dfrac{\alpha^{2}+\beta^{2}}{\alpha^{2} \beta^{2}}\right) x+\dfrac{4 p^{2}}{\alpha^{2} \beta^{2}}=0 \\\\ &x^{2}-14\left(\dfrac{49-14}{49}\right) x+\dfrac{4(49)}{49}=0 \\\\ &x^{2}-10 x+4=0 \end{aligned}$




13. The equation $3 x^{2}-5 x+4=0$ has roots $\alpha$ and $\beta$. Without solving this equation, form a quadratic equation with integer coefficients that has roots $\alpha+\dfrac{1}{2 \beta}$ and $\beta+\dfrac{1}{2 \alpha}$.



The equation $3 x^{2}-5 x+4=0$ has roots $\alpha$ and $\beta$. $\therefore \quad 3 x^{2}-5 x+4=3(x-\alpha)(x-\beta)$ $\quad\quad 3 x^{2}-5 x+4=3 x^{2}-3(\alpha+\beta) x+3 \alpha \beta$ $\therefore \quad \alpha+\beta=\dfrac{5}{3}$ and $\alpha \beta=\dfrac{4}{3}$ The quadratic equation which has roots $\alpha+\dfrac{1}{2 \beta}$ and $\beta+\dfrac{1}{2 \alpha}$ is $\begin{aligned} &{\left[x-\left(\alpha+\dfrac{1}{2 \beta}\right)\right]\left[x-\left(\beta+\dfrac{1}{2 \alpha}\right)\right]=0} \\\\ &x^{2}-\left(\alpha+\beta+\dfrac{1}{2 \alpha}+\dfrac{1}{2 \beta}\right) x+\left(\alpha \beta+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{4 \alpha \beta}\right)=0 \end{aligned}$ $x^{2}-\left(\alpha+\beta+\dfrac{\alpha+\beta}{2 \alpha \beta}\right) x+\left(1+\alpha \beta+\dfrac{1}{4 \alpha \beta}\right)=0$ $x^{2}-\left(\dfrac{5}{3}+\dfrac{5 / 3}{8 / 3}\right) x+\left(1+\dfrac{4}{3}+\dfrac{1}{16 / 3}\right)=0$ $x^{2}-\dfrac{5}{24} x+\dfrac{121}{48}=0$ $48 x^{2}-110 x+121=0$




14. The roots of the equation $x^{2}+3 x-5=0$ are $\alpha$ and $\beta$.
    1. Without solving the equation, find
       1. the value of $\alpha^{2}+\beta^{2}$.
       2. the value of $\alpha^{4}+\beta^{4}$.
    
    Given that $\alpha>\beta$ and without solving the equation,
    
    
    2. show that $\alpha-\beta=\sqrt{29}$.
    3. Factorise $\alpha^{4}-\beta^{4}$ completely.
    4. Hence find the exact value of $\alpha^{4}-\beta^{4}$.
    
    Given that $\beta^{4}=p+q \sqrt{29}$ where $p$ and $q$ are positive constants,
    
    
    5. find the value of $p$ and the value of $q$.



The roots of the equation $x^{2}+3 x-5=0$ are $\alpha$ and $\beta$. $\begin{aligned} \therefore \quad x^{2}+3 x-5&=(x-\alpha)(x-\beta) \\\\ x^{2}+3 x-5 & =x^{2}-(\alpha+\beta) x+\alpha \beta \\\\ \therefore\quad \alpha +\beta=-3 & \text { and } \alpha \beta=-5 \end{aligned}$ $\begin{aligned} \text { (a) (i) }\qquad \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 d \beta \\\\ &=9-2(-5) \\\\ &=19\\\\ \quad \text {(ii) }\qquad \alpha^{4}+\beta^{4} &=\left(\alpha^{2}\right)^{2}+\left(\beta^{2}\right)^{2} \\\\ &=\left(\alpha^{2}+\beta^{2}\right)^{2}-2 \alpha^{2} \beta^{2} \\\\ &=19^{2}-2(-5)^{2} \\\\ &=311\\\\ \text { (b) }\qquad\quad (\alpha-\beta)^{2} &=\alpha^{2}+\beta^{2}-2 \alpha \beta \\\\ &=19-2(-5) \\\\ &=29 \\\\ \therefore\quad \alpha-\beta &=\sqrt{29}\\\\ \text { (c) }\qquad\quad \alpha^{4}-\beta^{4} &=\left(\alpha^{2}+\beta^{2}\right)\left(\alpha^{2}-\beta^{2}\right) \\\\ &=\left(\alpha^{2}+\beta^{2}\right)(\alpha+\beta)(\alpha-\beta) \\\\ \text { (d) }\qquad\quad \alpha^{4}-\beta^{4} &=(19)(-3)(\sqrt{29}) \\\\ &=-57 \sqrt{29} \end{aligned}$ $\begin{aligned} \text { (e) } \alpha^{4}+\beta^{4}-\left(\alpha^{4}-\beta^{4}\right)&=311+57 \sqrt{29} \\\\ 2 \beta^{4}&=311+57 \sqrt{29} \\\\ \beta^{4}&=\dfrac{311}{2}+\dfrac{57}{2} \sqrt{29}\\\\ \beta^{4}&=p+q \sqrt{29}\ \text{( given )}\\\\ \therefore\quad p=\dfrac{311}{2},\ & q=\dfrac{57}{2} \end{aligned}$




15. It is given that $\alpha$ and $\beta$ are such that $a+\beta=-\dfrac{5}{2}$ and $\alpha \beta=-5$.
    1. Form a quadratic equation with integer coefficients that has roots $\alpha$ and $\beta$
    
    Without solving the equation found in part (a),
    
    
    2. find the value of
       1. $\alpha^{2}+\beta^{2}$.
       2. $a^{3}+\beta^{3}$.
    3. Hence form a quadratic equation with integer coefficients that has roots $\left(\alpha-\dfrac{1}{\alpha^{2}}\right)$ and $\left(\beta-\dfrac{1}{\beta^{2}}\right)$.
    
    
    
    

    $\alpha+\beta=-\dfrac{5}{2},\ \alpha \beta=-5$ (a) The quadratic equation which has roots $\alpha$ and $\beta$ is $\begin{aligned} &(x-\alpha)(x-\beta)=0 \\\\ &x^{2}-(\alpha+\beta) x+\alpha \beta=0 \\\\ &x^{2}-\left(-\dfrac{5}{2}\right) x-5=0 \\\\ &2 x^{2}+5 x-10=0 \end{aligned}$ $\begin{aligned} \text {(b) (i) } \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=\dfrac{25}{4}-2(-5)=\dfrac{65}{4}\\\\ \quad \text { (ii) } (\alpha+\beta)^{3} &=\alpha^{3}+3 \alpha^{2} \beta+3 \alpha \beta^{2}+\beta^{3} \\\\ &=\alpha^{3}+\beta^{3}+3 \alpha \beta(\alpha+\beta) \\\\ \therefore\quad \alpha^{3}+\beta^{3} &=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta) \\\\ &=-\dfrac{125}{8}+15\left(-\dfrac{5}{2}\right) \\\\ &=-\dfrac{425}{8} \end{aligned}$ $\text {(c)}$ The quadratic equation which has roots $\alpha-\dfrac{1}{\alpha^{2}}$ and $\beta-\dfrac{1}{\beta^{2}}$ is $\begin{aligned} &{\left[x-\left(\alpha-\dfrac{1}{\alpha^{2}}\right)\right]\left[x-\left(\beta-\dfrac{1}{\beta^{2}}\right)\right]=0} \\\\ &x^{2}-\left(\alpha-\dfrac{1}{\alpha^{2}}+\beta-\dfrac{1}{\beta^{2}}\right) x+\left(\alpha \beta-\dfrac{\alpha}{\beta^{2}}-\dfrac{\beta}{\alpha^{2}}+\dfrac{1}{\alpha^{2} \beta^{2}}\right)=0 \\\\ &x^{2}-\left(\alpha+\beta-\dfrac{\alpha^{2}+\beta^{2}}{\alpha^{2} \beta^{2}}\right) x+\left(\alpha \beta-\dfrac{\alpha^{3}+\beta^{3}}{d^{2} \beta^{2}}+\dfrac{1}{\alpha^{2} \beta^{2}}\right)=0 \\\\ &x^{2}-\left(-\dfrac{5}{2}-\dfrac{65 / 4}{25}\right) x+\left(-5-\dfrac{-425 / 8}{25}+\dfrac{1}{25}\right)=0 \\\\ &200 x^{2}+630 x-567=0 \end{aligned}$

    
    
    

Factorial Expression : Exercise

Factorials

We define $\boldsymbol{n} !=\boldsymbol{n}(\boldsymbol{n}-\mathbf{1})(\boldsymbol{n}-\mathbf{2}) \cdots \mathbf{3} \cdot \mathbf{2} \cdot \mathbf{1}$ if $n$ is a nonnegative integer. An empty product is normally defined to be 1 . With this convention, $0 !=1$ An alternative is to define $\boldsymbol{n} !$ recursively on the nonnegative integers. $\boldsymbol{n} != \begin{cases}1 & \text { if } \boldsymbol{n}=\mathbf{0} \\ \boldsymbol{n}(\boldsymbol{n}-\mathbf{1}) ! & \text { if } \boldsymbol{n} \geq \mathbf{1}\end{cases}$

Exercise

1. Evaluate
   $\begin{array}{lll} \text{(a)}\ 2 !& \text{(b)}\ 3 !& \text{(c)}\ 4 !\\\\ \text{(e)}\ 5 !& \text{(f)}\ 6 !& \text{(g)}\ 10 ! \end{array}$





$\begin{aligned} \text{(a)}\ &2 !=2 \times 1=2 \\\\ \text{(b)}\ &3 !=3 \times 2 \times 1=6 \\\\ \text{(c)}\ &4 !=4 \times 3 \times 2 \times 1=24 \\\\ \text{(d)}\ &5 !=5 \times 4 \times 3 \times 2 \times 1=120 \\\\ \text{(e)}\ &6 !=6 \times 5 \times 4 \times 3 \times 2 \times 1=720 \\\\ \text{(f)}\ &10 !=10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=362880 \end{aligned}$



2. Express in factorial form:
   $\begin{array}{l} \text{(a)}\ 4 \times 3 \times 2 \times 1 \\\\ \text{(b)}\ 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \quad \\\\ \text{(c)}\ 6 \times 5 \\\\ \text{(d)}\ 8 \times 7 \times 6\\\\ \text{(e)}\ 10 \times 9 \times 8 \times 7 \\\\ \text{(f)}\ 15 \times 14 \times 13 \times 12\\\\ \text{(g)}\ \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} \\\\ \text{(h)}\ \dfrac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\\\\ \text{(i)}\ \dfrac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \end{array}$





$\begin{aligned} \text{(a)}\ &\quad 4 \times 3 \times 2 \times 1\\\\ &=4 !\\\\ \text{(b)}\ &\quad 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\\\\ &=7 !\\\\ \text{(c)}\ &\quad 6 \times 5\\\\ &=\dfrac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{6 !}{4 !}\\\\ \text{(d)}\ &\quad 8 \times 7 \times 6\\\\ &=\dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{8 !}{5 !}\\\\ \text{(e)}\ &\quad 10 \times 9 \times 8 \times 7\\\\ &=\dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{10 !}{6 !}\\\\ \text{(f)}\ &\quad 15 \times 14 \times 13 \times 12\\\\ &=\dfrac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{15 !}{11 !}\\\\ \text{(g)}\ &\quad \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1}\\\\ &=\dfrac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(6 \times 5 \times 4 \times 3 \times 2 \times 1)}\\\\ &=\dfrac{9 !}{3 ! 6 !}\\\\ \text{(h)}\ &\quad \dfrac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\\\\\ &=\dfrac{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1)(9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}\\\\ &=\dfrac{13 !}{4 ! 9 !}\\\\ \text{(i)}\ &\quad \dfrac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1)( 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}\\\\ &=\dfrac{15 !}{5 ! 10 !} \end{aligned}$ $\begin{aligned} &\textbf{Alternative Method }\\\\ \text{(a)}\ &\quad 4 \times 3 \times 2 \times 1\\\\ &=4 !\\\\ \text{(b)}\ &\quad 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\\\\ &=7 !\\\\ \text{(c)}\ &\quad 6 \times 5=\dfrac{6 \times 5 \times 4 !}{4 !}\\\\ &=\dfrac{6 !}{4 !}\\\\ \text{(d)}\ &\quad 8 \times 7 \times 6\\\\ &=\dfrac{8 \times 7 \times 6 \times 5 !}{5 !}\\\\ &=\dfrac{8 !}{5 !}\\\\ \text{(e)}\ &\quad 10 \times 9 \times 8 \times 7\\\\ &=\dfrac{10 \times 9 \times 8 \times 7 \times 6 !}{6 !}\\\\ &=\dfrac{10 !}{6 !}\\\\ \text{(f)}\ &\quad 15 \times 14 \times 13 \times 12=\dfrac{15 \times 14 \times 13 \times 12 \times 11 !}{11 !}\\\\ &=\dfrac{15 !}{11 !}\\\\ \text{(g)}\ &\quad \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1}\\\\ &=\dfrac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}\\\\ &=\dfrac{9 !}{3 ! 6 !}\\\\ \text{(h)}\ &\quad \dfrac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{13 \times 12 \times 11 \times 10 \times 9 !}{4 \times 3 \times 2 \times 1 \times 9 !}\\\\ &=\dfrac{13 !}{4 ! 9 !} \\\\ \text{(i)}\ &\quad \dfrac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{15 \times 14 \times 13 \times 12 \times 11 \times 10 !}{5 \times 4 \times 3 \times 2 \times 1 \times 10 !}\\\\ &=\dfrac{15 !}{5 ! 10 !} \end{aligned}$



3. Simplify without using a calculator:
   $\begin{array}{ll} \text{(a)}\ \dfrac{7 !}{6 !}& \text{(b)}\ \dfrac{8 !}{6 !}\\\\ \text{(c)}\ \dfrac{12 !}{10 !}& \text{(d)}\ \dfrac{120 !}{119 !}\\\\ \text{(e)}\ \dfrac{10 !}{8 ! \times 2 !}& \text{(f)}\ \dfrac{100 !}{98 ! \times 2 !}\\\\ \text{(g)}\ \dfrac{7 !}{3 !}& \text{(h)}\ \dfrac{8 !}{5 !}\\\\ \text{(i)}\ \dfrac{4 !}{2 ! 2 !}& \text{(j)}\ \dfrac{6 !}{3 ! 2 !}\\\\ \text{(k)}\ \dfrac{6 !}{(3 !)^{2}}& \text{(l)}\ \dfrac{5 !}{3 !} \times \dfrac{7 !}{4 !} \end{array}$





$\begin{array}{l} \text{(a)}\ \dfrac{7 !}{6 !}=\dfrac{7 \times 6 !}{6 !}=7\\\\ \text{(b)}\ \dfrac{8 !}{6 !}=\dfrac{8 \times 7 \times 6 !}{6 !}=56\\\\ \text{(c)}\ \dfrac{12 !}{10 !}=\dfrac{12 \times 11 \times 10 !}{10 !}=132\\\\ \text{(d)}\ \dfrac{120 !}{119 !}=\dfrac{120 \times 119 !}{119 !}=120\\\\ \text{(e)}\ \dfrac{10 !}{8 ! \times 2 !}=\dfrac{10 \times 9 \times 8 !}{8 ! \times(2 \times 1)}=45\\\\ \text{(f)}\ \dfrac{100 !}{98 ! \times 2 !}=\dfrac{100 \times 99 \times 98 !}{98 ! \times 2 \times 1}=4950\\\\ \text{(g)}\ \dfrac{7 !}{3 !}=\dfrac{7 \times 6 \times 5 \times 4 \times 3 !}{3 !}=840\\\\ \text{(h)}\ \dfrac{8 !}{5 !}=\dfrac{8 \times 7 \times 6 \times 5 !}{5 !}=336\\\\ \text{(i)}\ \dfrac{4 !}{2 ! 2 !}=\dfrac{4 \times 3 \times 2 !}{2 !(2 \times 1)}=6\\\\ \text{(j)}\ \dfrac{6 !}{3 ! 2 !}=\dfrac{6 \times 5 \times 4 \times 3 !}{3 ! \times(2 \times 1)}=60\\\\ \text{(k)}\ \dfrac{6 !}{(3 !)^{2}}=\dfrac{6 \times 5 \times 4 \times 3 !}{3 !(3 \times 2 \times 1)}=20\\\\ \text{(l)}\ \dfrac{5 !}{3 !} \times \dfrac{7 !}{4 !}=(5 \times 4) \times(7 \times 6 \times 5)=4200 \end{array}$



4. Simplify:
   $\begin{array}{l} \text{(a)}\ \dfrac{n !}{(n-1) !}\\\\ \text{(b)}\ \dfrac{(n+2) !}{n !}\\\\ \text{(c)}\ \dfrac{(n+1) !}{(n-1) !} \end{array}$





$\begin{aligned} \text{(a)}\ &\dfrac{n !}{(n-1) !}\\\\\ &=\dfrac{n(n-1) !}{(n-1) !}\\\\ &=n \\\\ \text{(b)}\ &\dfrac{(n+2) !}{n !}\\\\ &=\dfrac{(n+2)(n+1) n !}{n !}\\\\ &=n^{2}+3 n+2 \\\\ \text{(c)}\ &\dfrac{(n+1) !}{(n-1) !}\\\\ &=\dfrac{(n+1) n(n-1) !}{(n-1) !}\\\\ &=n^{2}+n \end{aligned}$



5. Rewrite each of the following using factorial notation.
   $\begin{array}{l} \text{(a)}\ n(n-1)(n-2)(n-3)\\\\ \text{(b)}\ n(n-1)(n-2)(n-3)(n-4)(n-5)\\\\ \text{(c)}\ \dfrac{n(n-1)(n-2)}{5 \times 4 \times 3 \times 2 \times 1}\\\\ \text{(d)}\ \dfrac{n(n-1)(n-2)(n-3)(n-4)}{3 \times 2 \times 1} \end{array}$





$\begin{aligned} \text{(a)}\ &n(n-1)(n-2)(n-3)\\\\ &=\frac{n(n-1)(n-2)(n-3)(n-4) !}{(n-4) !}\\\\ &=\frac{n !}{(n-4) !} \\\\ \text{(b)}\ &n(n-1)(n-2)(n-3)(n-4)(n-5)\\\\ &=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) !}{(n-6) !}\\\\ &=\frac{n !}{(n-6) !} \\\\ \text{(c)}\ &\frac{n(n-1)(n-2)}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\frac{n(n-1)(n-2)(n-3) !}{5 !(n-3) !}=\frac{n !}{5 !(n-3) !} \\\\ \text{(d)}\ &\frac{n(n-1)(n-2)(n-3)(n-4)}{3 \times 2 \times 1}\\\\ &=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5) !}{3 !(n-5) !}\\\\ &=\frac{n !}{3 !(n-5) !} \end{aligned}$



6. Express the following as a single factorial notation.
   (a) $n !(n+1)$
   (b) $(n-1) !\left(n^{2}+n\right)$
   (c) $(n+4)(n+5)(n+3) !$
   (d) $n !\left(n^{2}+3 n+2\right)$
   (e) $(n+1)(n+2)(n+3)$
   (f) $(n-3)(n-4)(n-5)$





$\begin{aligned} \text { (a) } & n !(n+1) \\\\ &=(n+1) n ! \\\\ &=(n+1) ! \\\\ \text { (b) } &(n-1) !\left(n^{2}+n\right) \\\\ &=\left(n^{2}+n\right)(n-1) ! \\\\ &=(n+1) n(n-1) ! \\\\ &=(n+1) ! \\\\ &=(n+5)(n+4)(n+3) ! \\\\ &=(n+5) !\\\\ \text { (c) } &(n+4)(n+5)(n+3) ! \\\\ &=(n+5)(n+4)(n+3) ! \\\\ &=(n+5) !\\\\ \text { (d) } & n !\left(n^{2}+3 n+2\right) \\\\ =&(n+2)(n+1) n ! \\\\ =&(n+2) !\\\\ \text { (e) } &(n+1)(n+2)(n+3) \\\\ =& \frac{(n+3)(n+2)(n+1) n !}{n !} \\\\ =& \frac{(n+3) !}{n !}\\\\ \text { (f) } & (n-3)(n-4)(n-5) \\\\ &=\frac{(n-3)(n-4)(n-5)(n-6) !}{(n-6) !} \\\\ &=\frac{(n-3) !}{(n-6) !} \end{aligned}$



7. Write as a product by factorizing:
   (a) $5 !+4 !$
   (b) $11 !-10 !$
   (c) $5 !+7 !$
   (d) $12 !-10 !$
   (e) $9 !+8 !+7 !$
   (f) $7 !-6 !+8 !$
   (g) $12 !-2 \times 11 !$
   (h) $3 \times 9 !+5 \times 8 !$





$\begin{aligned} \text { (a) } & \quad 5 !+4 ! \\\\ &= 5 \times 4 !+4 ! \\\\ &=(5+1) 4 ! \\\\ &= 6 \times 4 ! \\\\ \text { (b) } & \quad 11 !-10 ! \\\\ &=(11-1) 10 ! \\\\ &= 10 \times 10 !\\\\ \text { (c) } & \quad 5 !+7 ! \\\\ &= 5 !+7 \times 6 \times 5 ! \\\\ &=(1+42) 5 ! \\\\ &= 43 \times 5 ! \\\\ \text { (d) } & \quad 12 !-10 ! \\\\ &= 12 \times 11 \times 10 !-10 ! \\\\ &=(132-1) 10 ! \\\\ &= 131 \times 10 !\\\\ \text { (e) } & \quad 9 !+8 !+7 ! \\\\ &= 9 \times 8 \times 7 !+8 \times 7 !+7 ! \\\\ &=(72+8+1) 7 ! \\\\ &= 81 \times 7 ! \\\\ \text { (f) } & \quad 7 !-6 !+8 ! \\\\ &= 7 \times 6 !-6 !+8 \times 7 \times 6 ! \\\\ &=(7-1+56) 6 ! \\\\ &= 62 \times 6 !\\\\ \text { (g) } & \quad 12 !-2 \times 11 ! \\\\ &= 12 \times 11 !-2 \times 11 ! \\\\ &=(12-2) 11 ! \\\\ &= 10 \times 11 ! \\\\ \text { (h) } & \quad 3 \times 9 !+5 \times 8 ! \\\\ &= 3 \times 9 \times 8 !+5 \times 8 ! \\\\ &=(27+5) 8 ! \\\\ &= 32 \times 8 ! \end{aligned}$



8. Simplify by factorizing:
   $\begin{array}{l} \text{(a)}\ \dfrac{12 !-11 !}{11}\\\\ \text{(b)}\ \dfrac{10 !+9 !}{11}\\\\ \text{(c)}\ \dfrac{10 !-8 !}{89}\\\\ \text{(d)}\ \dfrac{10 !-9 !}{9 !}\\\\ \text{(e)}\ \dfrac{6 !+5 !-4 !}{4 !}\\\\ \text{(f)}\ \dfrac{n !+(n-1) !}{(n-1) !}\\\\ \text{(g)}\ \dfrac{n !-(n-1) !}{n-1}\\\\ \text{(h)}\ \dfrac{(n+2) !+(n+1) !}{n+3} \end{array}$





$\begin{aligned} \text { (a) } & \quad \dfrac{12 !-11 !}{11} \\\\ &= \dfrac{12 \times 11 !-11 !}{11} \\\\ &= \dfrac{(12-1) 11 !}{11} \\\\ &= \dfrac{11 \times 11 !}{11} \\\\ &= 11 !\\\\ \text { (b) } & \quad \dfrac{10 !+9 !}{11} \\\\ &= \dfrac{10 \times 9 !+9 !}{11} \\\\ &= \dfrac{(10+1) 9 !}{11} \\\\ &= \dfrac{11 \times 9 !}{11} \\\\ &= 9 !\\\\ \text { (c) } & \quad \dfrac{10 !-8 !}{89} \\\\ &= \dfrac{10 \times 9 \times 8 !-8 !}{89} \\\\ &= \dfrac{(90-1) \times 8 !}{89} \\\\ &= \dfrac{89 \times 8 !}{89} \\\\ &= 8 !\\\\ \text { (d) } & \quad \dfrac{10 !-9 !}{9} \\\\ &= \dfrac{10 \times 9 !-9 !}{9} \\\\ &= \dfrac{(10-1) 9 !}{9} \\\\ &= \dfrac{9 \times 9 !}{9} \\\\ &= 9 !\\\\ \text { (e) } & \quad \dfrac{6 !+5 !-4 !}{4 !} \\\\ &= \dfrac{6 \times 5 \times 4 !+5 \times 4 !-4 !}{4 !} \\\\ &= \dfrac{(30+5-1) 4 !}{4 !} \\\\ &= 34\\\\ \text { (f) } & \quad \dfrac{n !+(n-1) !}{(n-1) !} \\\\ &=\dfrac{n(n-1) !+(n-1) !}{(n-1) !} \\\\ &=\dfrac{(n+1)(n-1) !}{(n-1) !} \\\\ &=n+1\\\\ \text { (g) } & \quad \dfrac{n !-(n-1) !}{n-1} \\\\ &=\dfrac{n(n-1) !-(n-1) !}{n-1} \\\\ &=\dfrac{(n-1)(n-1) !}{n-1} \\\\ &=(n-1) !\\\\ \text { (h) } & \quad \dfrac{(n+2) !+(n+1) !}{n+3} \\\\ &= \dfrac{(n+2)(n+1) !+(n+1) !}{n+3} \\\\ &= \dfrac{(n+2+1)(n+1) !}{n+3} \\\\ &= \dfrac{(n+3)(n+1) !}{n+3} \\\\ &=(n+1) ! \end{aligned}$