S Chand CLASS 10 Chapter 7 Factor Theorems- Factorization Exercise 7

  Exercise 7

Question 1

Ans: 

(i) 

Let

 $\begin{aligned} f(x) &=3 x^{3}+8 x^{2}-6 x+1 \\ \text { and } x &+3=0 \\ x &=-3 \text {. } \end{aligned}$

$\therefore$ Remainder $=f(-3)$.

$3 x(-3)^{3}+8 x(-3)^{2}-6 x(-3)+1 .$

$3 x(-27)+8 x 9+18+1 .$

$-81+72+19$

$-9+19$

10

(ii) Let $f(x)=5 x^{3}-8 x^{2}+3 x-y$

and $x=1=0$

$x=1$

Remainder= f(1)

=$5 \times(1)^{3}-8 \times(1)^{2}+3 \times 1-4$

=$5 \times 1-8 \times 1+3-4$

=$5-8-1$

=$-3-1$

=$-4$

(iii)

let $f(x)=x^{3}+3 x^{2}-1$. and

$\begin{aligned}&3 x+2=0 \\&3 x=-2 \\&x=\frac{-2}{3}\end{aligned}$

$\therefore$ Remainder $=f\left(-\frac{2}{3}\right)$

$\quad=\left(-\frac{2}{3}\right)^{3}+3 \times\left(-\frac{2}{3}\right)^{2}-1 .$

$\frac{-8}{27}+3 \times \frac{4}{9}-1$

$=\frac{-8}{27}+\frac{4}{3}-1 .$

$=\frac{-8+36-27}{27}$

$=\frac{28-27}{27}$

$=\frac{1}{27} .$

(iv) 

$\begin{aligned} \text { Let } f(x)=& 4 x^{3}-12 x^{2}+14 x-3 . \\ \text { and } 2 x-1 &=0 \\ 2 x &=1 \\ x &=\frac{1}{2} \end{aligned}$

$\therefore$ Remainder $=f\left(\frac{1}{2}\right)$

$=4 \times\left(\frac{1}{2}\right)^{3}-12 \times\left(\frac{1}{2}\right)^{2}+14 \times \frac{1}{2}-3$

$=4 \times \frac{1}{8}-12 \times \frac{1}{4}+7-3$

$=\frac{1}{2}-3+4$

$=\frac{1-6+8}{2}$

$=-\frac{5+8}{2}$

$=\frac{3}{2}$

Question 2

Ans:  Let $f(x)=x^{3}+3 x^{2}-k x+4$.

and $\begin{aligned} x &-2=0 \\ x &=2 . \end{aligned}$

$\therefore$ Remainder $=f(2)$.

$K=(2)^{3}+3 x(2)^{2}-k x 2+4$

$K=8+3 x y-2 k+21$

$K=8+12-2 k+81$

$K=24-2 k$

$K+2 k=2 4$

$3 K=2 4$

$K=\frac{2 4}{3}$

$K=8$

Question 3

Ans:

Given,

Remainder $=5$

Let $f(x)=9 x^{3}+9 x^{2}+4 x-10$

and $x+3=0$

$x=-3$.

So remainder = f(-3)

$5=a \times(-3)^{3}+9 \times(-3)^{2}+4 \times(-3)-10$

$5=a \dot{x}(-27)+9 \times 9-12-10$

$5=-27 a+81-22$

$5=-279+59 .$

$27 a=-5+59$

$27 a=55 \quad a=\frac{54}{27}$

$a=2 .$

Hence, the value of a is 2

Question 4

Ans: Let $f(x)=9 x^{3}+4 x^{2}+3 x-4$. and $q(x)=x^{3}-4 x-9$.

and $x-2=0$.

$x=2$.

$\therefore$ Remainder f $(2)=q(2) .$

So 

$\begin{aligned} f(2) &=9 \times(2)^{3}+4 \times(2)^{2}+3 \times 2-4 \\ &=8 a+16+6-4 . \\ &=8 a+16+2 . \\ &=84+18 . \end{aligned}$

$\begin{aligned} q(2) &=(2)^{3}-4 \times 2-9 . \\ &=8-8-9 . \\ &=-a \end{aligned}$

$\begin{aligned} \therefore 8 a+18 &=-a \\ 8 a+a &=-18 \\ 9 a &=-18 \\ a &=\frac{-18}{9} \\ a &=-2 . \end{aligned}$

Hence the value of a is -2

Question 5

Ans: 

(i) 

$\begin{aligned} f(x) &=x^{3}-6 x^{2}+11 x^{-6} \\ \text { Let } x &=3=0 \\ x &=3 \end{aligned}$

Remainder $f(3)=(3)^{3}-6 \times(3)^{2}+11 \times 3-6$

$=27-6 \times 9+33-6$

$=27-54+27$

$=54-54 .$

$=0 .$

ஃ Remainder is zero 

So x-3 is a factor of f(x)

(ii)$f(x)=2 x^{3}-9 x^{2}+x+12$

Let $x+1=0$

x=(-1)

Remainder =f(-1)

$=2 x(-1)^{3}-9 x(-1)^{2}+(-1)+12$

$=-2-9 \times 1-1+12$

$=-2-9+11$

$=-11+11$

$=0$

Remainder =0

So x +1 is a factor of f(x)

(iii)  $f(x)=7 x^{2}-2 \sqrt{8} x-6$

Let $x-\sqrt{2}=0 .$

$x=\sqrt{2} .$

$\begin{aligned} \text { Remainder } &=f(\sqrt{2}) \\ &=7 \times(\sqrt{2})^{2}-2 \sqrt{8} \times \sqrt{2}-6 . \\ &=7 \times 2-2 \sqrt{16}-6 \\ &=14-2 \sqrt{16}-6 \\ &=8-2 \times 4 \\ &=8-8 \\ &=0 \end{aligned}$

Remainder =0  

So , x - $(\sqrt{2})$ is a factor of f(x)

(iv)

$\begin{aligned} f(x) &=3 x^{3}+x^{2}-20 x+12 \\ \text { let } & 3 x-2=0 \\ & x=\frac{2}{3} \\ \therefore \text { Remainder } &=f\left(\frac{2}{3}\right) \\ &=3 \times\left(\frac{2}{3}\right)^{3}+\left(\frac{2}{3}\right)^{2}-20 \times \frac{2}{3}+12 \\ &=\frac{1}{3} \times \frac{8}{27}+\frac{4}{9}-\frac{40}{3}+12 \\ &=\frac{8}{9}+\frac{4}{9}-\frac{40}{3}+12 \\ &=\frac{8+4-120+108}{9} \\ &=12-12 \\ &=0 \end{aligned}$

Remainder =0

So , 3x-2 is a factor of f(x)

Question 6

Ans:  Let $f(x)=x^{3}+a x+2 a-2$

and $x+1=0$

$x=-1 .$

$\therefore$ Remainder $=f(-1)$

$=(-1)^{3}+a x(-1)+2 \times a-2$

$=-1-9+2 a-2$

$\therefore-3+a$

$\therefore \quad x+1$ is a factor of $f(x)$

so, Remainder $=0 .$

$-3+a=0$

a=3 

Question 7

Ans:   

$\begin{aligned}&\text { Lit } f(x)=\dot{x}^{3}+10 x^{2}+a x+b . \\&\text { and } \begin{aligned} & x-1=0 \\ & x=1 . \end{aligned} \\&\therefore \text { Remainder }=f(1) \\&=(1)^{3}+10 \times(1)^{2}+a \times 1+b . \\&= & 1+10+a+b \\&= & 11+a+b . \\&\therefore x-1 \text { is the factor of } f(x)\end{aligned}$

So, remainder =0 

11+ a +b =0 

a+ b = -11.............(i)

Again x-2=0

x=2

Remainder =f(2)

$=x^{3}+10 x^{2}+a x+b$

$=(2)^{3}+10 \times(2)^{2}+a \times 2+b$

$=8+10 \times 4+2 a+b$

$=8+40+2 a+b$

$=48+2 a+b$

∴ x -2 is a factor of f(x)

So, remainder =0

48+2a+b=0

2a +b =-48............(ii)

From eq (i) and (ii)

$\begin{aligned} 9+b &=-11 \\ 2 a+b &=-48 \end{aligned}$

Multiply by 1 in both 

$\begin{aligned} a+b &=-11 \\-2 a+b &=748 \\-a &=37 \\ a &=-37 \end{aligned}$

put the value of a in eq (i)

$-37+b=-11$

$b=-11+37$

$b=26$

Hence the value of a and b is -37 and 26 respectively 

Question 8

Ans: $\operatorname{let} f(x)=x^{2}+a x-6$ and

$q(x)=x^{2}-9 x+b$

and let $x-2=0$

$=x=2$

∴ Remainder = f(2)

$=(x)^{2}+a x-6$

$=(2)^{2}+a \times 2-6$

$=4+2 a-6$

=2 a-2 

∴ x -2 is a factor of f(x)

So f(2) = 0

$\begin{aligned} 2 a-2 &=0 \\ 2 a &=2 \\ a &=\frac{2}{2} \\ a &=1 . \end{aligned}$

Again; 

x-2 =0 

x=2

$\begin{aligned} \therefore q(2) &=(2)^{2}-9 \times 2+b \\ &=4-18+b \\ &=-14+b \end{aligned}$

∴  (x-2) is a factor or of q(x)

∴ -14+b = 0

b=14 

Hence, a =1 and b = 14

Question 9

Ans: Let f(x) = $p x^{2}+5 x+r$

and let x-2 =0 

x=2

∴ f(x) = $p(2)^{2}+5 \times 2+r$

$=4 p+10+r$

∴ x - 2 is its factor 

∴ f(2) =0

4p +10+r =0 

4p +r =-10.................(i)

Again let x $-\frac{1}{2}=0$

 $x=\frac{1}{2}$

$\begin{aligned} \therefore f &\left(\frac{1}{2}\right) \\ &=p \times\left(\frac{1}{2}\right)^{2}+5 \times \frac{1}{2}+r \\ &=\frac{p}{4}+\frac{5}{2}+r \\ & \therefore x-\frac{1}{2} \text { is its factor } \end{aligned}$

So , f $\left(\frac{1}{2}\right)=0 .$

$\frac{P}{4}+\frac{5}{2}+r=0$

$\frac{P+10+4 r}{4}=0 .$

$P+10+4 r=0 .$

$P+4 r=-10 .$.........(ii)

From eq (i) and (ii)

$-4 p+r=-10$

$p+4 r=-10$

Multiply by 4 in eq (i)

16p+4r =-40

-p+4r =-10

$15 p=-30$

$p=-\frac{-30}{15}$

$p=-2 .$

Put the value of p = -2 in eq (ii)

$\begin{aligned}-2+4 r &=-10 \\ 4 r &=-10+2 \\ 4 r &=-8 \\ r &=-2 . \end{aligned}$

Hence p = -2 and r=-2

Hence prove 

p=r

-2=-2

Question 10

Ans: Let f(x) = $=x^{3}+a x^{2}+b x+6$

Let 

  $\begin{aligned} & x-2=0 \\ &=x=2 . \end{aligned}$

$\begin{aligned} \therefore f(x) &=(2)^{3}+a \times(2)^{2}+3 \times 2+6 \\ &=8+4 a+2 b+6 . \\ &=14+4 a+2 b . \\ \therefore x-2 & \text { is a factor } \end{aligned}$

So remainder =0

$\begin{aligned} 14+4 a+2 b &=0 \\ 4 a+2 b &=-14 . \\ 2(2 a+b) &=-14 \\ 24+b &=-\frac{14}{2} \\ 2 a+b &=-7 . \end{aligned}$.............(i)

 Again let x -3=0 =x=3

$\begin{aligned} \therefore f(3) &=(3)^{3}+a \times(3)^{2}+b \times 3+6 \\ &=27+9 a+3 b+6 \\ &=33+9 a+3 b . \end{aligned}$

Remainder =3

33+9a +3b =3

$\begin{aligned} 9 a+3 b=& 3-33 \\ 3(3 a+b) &=-30 \\ 3 a+b &=\frac{-30}{3} \end{aligned}$

$3 a+b=-10$..........(ii)

On eq (i) and (ii)

$\begin{aligned} 2 a+b &=-7 \\ 3 a+b &=-10 \\-a &=3 \\ a &=-3 . \end{aligned}$

put the value of a in eq (i)

$2 x(-3)+b=-7$

$-6+b=-7$

$b=-7+6$

$b=-1$

Hence the value of a and b is -3 and -1 respectively 

Question 11

Ans: (i) Let $f(x)=x^{3}+13 x^{2}+32 x+20$

Let $x+2=0$

$=x=-2$.

$\begin{aligned} \therefore f(-2) &=(-2)^{3}+13 \times(-2)^{3}+32 \times(-2)+20 \\ &=-8+13 \times 4-64+20 \\ &=-8+52-44 \\ &=44-44 \\ &=0 \end{aligned}$

Remainder =0

So, x+2 is its factor 

Now, dividing f(x) by (x+2)

(To be added)

$f(x)=(x+2)\left(x^{2}+11 x+10\right)$

$=(x+2)\left(x^{2}+x+10(+10)\right.$

$=(x+2)(x(x+1)+10(x+1)) .$

$=(x+y)(x+1)(x+10)$

$=(x+2)(x+1)(x+10) .$

(ii) Let $f(x)=4 x^{3}+20 x^{2}+33 x+18$.

and $2 x+3=0$

$\begin{aligned}&2 x=-3 \\&x=\frac{-3}{2} .\end{aligned}$

$\begin{aligned} \therefore f\left(-\frac{3}{2}\right) &=4 \times\left(\frac{3}{2}\right)^{3}+20 \times\left(-\frac{3}{2}\right)^{2}+33 \times\left(-\frac{3}{2}\right)+18 \\ &=4 \times\left(\frac{-27}{8}\right)+20 \times \frac{9}{4}-\frac{99}{2}+18 \\ &=\frac{-27}{2}+45-\frac{99}{2}+18 \\ &=\frac{-27+90-99+36}{2} \\ &=\frac{63-63}{2} \\ &=\frac{0}{2} \\ &=0 \end{aligned}$

Remainder =0 

So, 2x+3 is its factor 

Now dividing f(x) by 2x+3

(TO BE ADDED)

$\begin{aligned} f(x) &=(2 x+3)\left(2 x^{2}+7 x+6\right) \\ &=(2 x+3)\left\{2 x^{2}+4 x+3 x+6\right\} \\ &=(2 x+3)\{2 x(x+2)+3(x+2)\} \\ &=(2 x+3)(x+2)(2 x+3) \\ &=(2 x+3)^{2}(x+2) \end{aligned}$

Question 12

Ans: (i) Let $f(x)=x^{3}-23 x^{2}+142 x-120$

and $x-10=0$

$\begin{aligned} x &=10 \\ f(10) &=(10)^{3}-23 \times(10)^{2}+142 \times 10-120 \\ &=1000-2300+1420-120 \\ &=-1300+1300 \\ &=0 \end{aligned}$

Remainder =0

Lo, $x-10$ is a factor of $f(x)$,

Now dividing $f(x)$ by $(x-10)$,

(TO BE ADDED)

$\begin{aligned} f(x) &=(x-10) \cdot\left(x^{2}-13 x+12\right) \\ &=(x-10)\left(x^{2}-x-12 x+12\right) \\ &=(x-10)(x(x-1)-12(x-1)) \\ &=(x-10)(x-1)(x-12)) \\ &=(x-10)(x-1)(x-12) \end{aligned}$

(ii) Let f (x) =$9 z^{3}-27 z^{2}-100 z+300$

and 

$\begin{aligned} 3 z &+10=0 \\ 3 z &=-10 \\ z &=\frac{-10}{3} \end{aligned}$

$\begin{aligned} \therefore f\left(\frac{-10}{3}\right) &=9 \times\left(\frac{-10}{3}\right)^{3}-27 \times\left(\frac{-10}{3}\right)^{2}-100 \times\left(\frac{-10}{3}\right)+300 \\ &=9 \times\left(\frac{-1000}{27}\right)-27 \times \frac{100}{9}+\frac{1000}{3}+300 \\ & \end{aligned}$

=0

∴ Remainder =0

∴ 3z +10 is the factor of f (z)

Now dividing f(z) by 3z +10

(TO BE ADDED)

$\begin{aligned} f(z) &=(3 z+10)\left(3 z^{2}-19 z+30\right) \\ &=(3 z+10)\left(3 z^{2}-9 z-10 z+30\right) \\ &=(3 z+10)(3 z(z-3)-10(z-3)) \\ &=(3 z+10)((z-3)(3 z-10)) \\ &=(3 z+10)(z-3)(3 z-10) . \end{aligned}$

Question 13

Ans: Let $f(x)=x^{3}+3 x^{2}+a x+b .$

$let x-2=0$

$x=2$

$\begin{aligned} \therefore f(x) &=(2)^{3}+3 \times(2)^{2}+a \times 2+b \\ &=8+12+2 a+b \\ &=20+2 a+b \end{aligned}$

∴ x-2 is a factor of f(x)

So, remainder =0

20 +2a +b =0

$2 a+b=-20 .$ ............(i)

Again x+1 =0

x=-1

$\begin{aligned} \therefore f(-1) &=(-1)^{3}+3 x(-1)^{2}+9 x(-1)+b . \\ &=-1+3-9+b . \\ &=2-a+b . \end{aligned}$

∴ x+1 is  a factor,

So, remainder =0

$2-a+b=0$

$2=a-b$

$a-b=2$ .............(ii)

From eq (i) and (ii)

$\begin{aligned} 2 a+b &=-20 \\ a-b &=2 . \\ 3 a &=-18 \\ a &=\frac{-18}{3} \\ a &=-6 . \end{aligned}$

Put the value of a= -6 in eq (ii) 

$\begin{aligned}-6-b &=2 \\-b &=2+6 \\-b &=8 \\ b &=-8 \end{aligned}$

$\therefore a=-6$ and $b=-8$

So , f(x) =  $x^{3}+3 x^{2}-6 x-8$

$\therefore ( x-2)$ is a factor of $f(x) .$

Dividing f(x) by x-2

(TO BE ADDED)

$\begin{aligned} f(x) &=(x -2)\left(x^{2}+5 x+4\right) . \\ &=(x-2)\left(x^{2}+x+4 x+4\right) \\ &=(x-2)(x(x+1)+y(x+1)) \\ &=(x-2)(x+1)(x+4) . \end{aligned}$

∴ Remaining factor is (x+4)

Question 14

Ans: Let f(x)  = $x^{3}+a x+b$

and x +2 =0

x =-2

∴  f(-2) = $(-2)^{3}+a \times(-2)+b$

=-8 -2a+b 

∴   x +2 is its factor 

So, the remainder =0

$\begin{aligned}-8-2 a+b &=0 \\-2 a+b &=8 \end{aligned}$ ......(i)

Let x - 3=0

x=3

$\begin{aligned} \therefore f(3) &=3^{3}+a \times 3+b \\ &=27+3a+b \\ \therefore x-3 & \text { is its factor } \end{aligned}$

So, the remainder =0

$27+3 a+b=0$

$3 a+b=-27$ ..........(ii)

From eq(i)  and (ii)

$\begin{aligned}-2 a+b &=8 \\ 3 a+b &=-27 \end{aligned}$

 Multiply by 1 in eq (i)

(TO BE ADDED)

$\begin{aligned}-5 a &=35 \\ a &=\frac{35}{-5} \\ a &=-7 \end{aligned}$

Put the value of a=- in eq (i)

$\begin{aligned}-2 \times(-7) &+b=8 \\ 14+b &=8 \\ b &=8-14 \\ b &=-6 . \end{aligned}$

$a=-7$ and $b=-6$

∴  x +2 is a factor of f(x) =  $x^{3}-7 x-6$ 

So, dividing it by x +2

(TO BE ADDED)

$\begin{aligned} f(x) &=(x+2)\left(x^{2}-2 x-3\right) . \\ &=(x+2)\left\{x^{2}-3 x+x-3\right\} . \\ &=(x+2)\{x(x-3)+1(x-3)\} . \\ &=(x+2)(x-3)(x+1) . \end{aligned}$

Remaining factor of f (x) is (x+1)

Question 15

Ans: (i) Let $f(x)=x^{3}-19 x-30$

Factor of 30 = $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$

Then by trial and error methods 

Let x =-2

So f(-2)= $(-2)^{3}-19 \times (-2)-30$

$=-8+38-30$

$=30-30$

$=0$

x +2 is its factor 

Let x =-3 

, $\begin{aligned} f(-3) &=(-3)^{3}-19 \times(-3)-30 \\ &=-27+57-30 \\ &=30-30 \\ &=0 \end{aligned}$

$\therefore x+2 ; is  its factor

$\text { Let } x=-3$

, $\begin{aligned} f(-3) &=(-3)^{3}-19 \times(-3)-30 \\ &=-27+57-30 \\ &=30-30 \\ &=0 \end{aligned}$

x+3 is its another factor of f(x) 

Let x =5,

, $\begin{aligned} f(5) &=(5)^{3}-19 \times(5)-30 \\ &=125-95-30 \\ &=30-30 \\ &=0 \end{aligned}$

∴ x-5 is its third factor 

Hence,  $x^{3}-19 x-30=(x+2)(x+3)(x-5)$

(ii) Let $f(x)=x^{3}+7 x^{2}-21 x-27$.

Factor of $27=\pm 1, \pm 3, \pm 9, \pm 27$.

 Then , by trial and error method 

Let x =-1 

$\begin{aligned} f(-1) &=(-1)^{3}+7 \times(-1)^{2}-21 \times(-1)-27 \\ &=-1+7+21-27 \\ &=6-6 \\ &=0 \end{aligned}$

∴  x+1 is its factor 

Then dividing f(x) by x+1

(TO BE ADDED)

$\begin{aligned} f(x) &=(x+1)\left(x^{2}+6 x-27\right) \\ &=(x+1)\left\{x^{2}+9 x-3 x-27\right\} \\ &=(x+1)\{x(x+9)-3(x+9)\} \\ &=(x+1)(x+9)(x-3) \end{aligned}$

(iii)  let $f(x)=x^{3}-3 x^{2}-9 x-5$

Factor  of $-5=\pm 1, \pm 5$

Then, by trial and error method, 

Let x =-1

$\begin{aligned} \therefore f(-1) &=(-1)^{3}-3 \times(1)^{2}-9 \times(-1)-5 \\ &=-1-3+9-5 \\ &=-4+4 \\ &=0 \end{aligned}$

∴ x+1 is its factor of f(x)

Then, dividing f(x) by x+1, 

(TO BE ADDED)

$\begin{aligned} f(x) &=(x+1)\left(x^{2}-4 x-5\right) \\ &=(x+1)\left\{x^{2}-5 x+x-5\right\} \\ &=(x+1)\{x(x-5)+1(x-5)\} \\ &=(x+1)(x-5)(x+1) \end{aligned}$

(iv) Let f (x) =  $2 x^{3}+9 x^{2}+7 x-6$

Factor of 6 =$\pm 1, \pm 2, \pm 3, \pm 6$

Then by trial and error method, 

Let x =-2

$\begin{aligned} \therefore f(-2) &=2 x(-2)^{3}+9 x(-2)^{2}+7 x(-2)-6 . \\ &=2 x(-8)+9 \times 4-14-6 \\ &=-16+36-20 \\ &=-16+16 \\ &=0 \end{aligned}$

$\therefore x+2$ is the factor of $f(x)$

Then, dividing $f(x)$ by $x+2$

(TO BE ADDED)

$\begin{aligned} f(x) &=(x+2)\left(2 x^{2}+5 x-3\right) . \\ &=(x+2)\left\{2 x^{2}+6 x-x-3\right\} . \\ &=(x+2)\{2 x(x+3)-1(x+3)\} . \\ &=(x+2)(x+3)(2 x-1) \end{aligned}$

Question 16

Ans: Let p (x) = $x^{3}-2 x^{2}+p x+6$

$\begin{aligned} q(x) &=x^{2}-5 x+q . \\ H \cdot C \cdot F &=x-3 . \end{aligned}$

∴ H.C.F is a factor of p(x) and q (x)

Let x-3=0

x=3

So, $\begin{aligned} p(3) &=(3)^{3}-2 \times(3)^{2}+p \times 3+6 \\  \end{aligned}$

$=27-18+3 p+6$

$=9+3 p+6$

$=15+3 p$

∴ Remainder =0

So, 15 +3p =0

3p=-15

$p=\frac{-15}{3}$

$p=-5$

and $\begin{aligned} q(3) &=(3)^{2}-5 \times 3+q . \\ &=9-15+q \\ &=-6+q . \end{aligned}$

Remainder =0 

So, -6+q = 0 

q=6

Hence , p =-5 and q=6

Then 6p +5q = $6 (-5)+5 \times 6$

$=-30+30$

$=0$

Question 17

Ans (i) Let $f(x)=x^{3}-6 x^{2}-15 x+80$

and let $p$ be subtracted from $f(x)$ so that it may he exactly divisible by $x+4$.

$\therefore P(x)=x^{3}-6 x^{2}-15 x+80-p$.

and let $x+4=0$,

$x=-4$.

$\begin{aligned} P(-4) &=(-4)^{3}-6 \times(4)^{2}-15 \times(-4)+8 0-p \\ &=-64-6 \times 16+60+8 0-p . \\ &=-64-96+140-p \\ &=-160+14 0-p \\ &=-20-p . \end{aligned}$

$\therefore p(x)$ is divisible by $x+4$.

$\therefore$ Remainder $=0$

So, $-20-p=0$

$\begin{aligned}-p &=20 \\ p &=-20 \end{aligned}$

(ii) Let q( x) = $x^{3}-3 x^{2}-12 x+19$

and let p be added of q (x) 

So (x) = $x^{3}-3 x^{2}-12 x+19+p$

But it is divisible by x-2

x-2 =0 

x=2

$\begin{aligned} q(2) &=(2)^{3}-3 \times(2)^{2}-12 \times 2+19+p \\ &=8-12-24+19+p \\ &=-4-5+p \\ &=-9+p \end{aligned}$

$\therefore Q(x)$ is divivible by $x-2$.

$\therefore$ Remainder 0

$-9+p=0$

$p=9$