Question 13 Marks
Using the remainder theorem, find the remainder, when $p(x)$ is divided by$g(x)$,
where, $p(x)=2 x^3+x^2-15 x-12, g(x)=x+2$.
Answer$p ( x )=2 x ^3+ x ^2-15 x -12$
$g(x)=x+2$ by remainder theorem,
when $p ( x )$ is divided by $( x +2)$,
then the remainder $=p(-2)$.
Putting $x=-2$ in $p(x)$,
we get $p(-2)=(-2)^3+(-2)^2-15 \times(-2)-12=-16+4+30-12=6$
$\therefore$ Remainder $=6$
Thus, the remainder when $p(x)$ is divided by $g(x)$ is 6 .
View full question & answer→Question 23 Marks
Using the remainder theorem, find the remainder, when $p(x)$ is divided by $g(x)$, where,
$p(x)=x^3-6 x^2+9 x+3, g(x)=x-1$.
Answer$p(x)=x^3-6 x^2+9 x+3 g(x)=x-1$ by remainder theorem,
when $p(x)$ is divided by $(x-1)$,
then the remainder $=p(1)$.
Putting $x=1$ in $p(x)$,
we get $p(1)=1^3-6 \times 1^2+9 \times 1+3=1-6+9+3=7$
$\therefore$ Remainder $=7$
Thus, the remainder when $p ( x )$ is divided by $g ( x )$ is 7 .
View full question & answer→Question 33 Marks
What must be added to $2 x^4-5 x^3+2 x^2-x-3$ so that the result is exactly divisible by $(x-2) ?$
AnswerLet the required number to be added be $k$.
Then, $p ( x )=2 x ^4-5 x ^3+2 x ^2- x -3+ k$
$g(x)=x-2$
Thus, we have,
$p(2)=0$
$\Rightarrow 2(2)^4-5(2)^3+2(2)^2-2-3+k=0$
$\Rightarrow 32-40+8-5+k=0$
$\Rightarrow k-5=0$
$\Rightarrow k=5$
Hence, the required number to be added is 5 .
View full question & answer→Question 43 Marks
Using the remainder theorem, find the remainder, when $p ( x )$ is divided by $g ( x )$, where,
$p(x)=x^3-2 x^2-8 x-1, g(x)=x+1$
Answer$p(x)=x^3-2 x^2-8 x-1$
$g(x)=x+1 \text { by remainder theorem, }$
$\text { when } p(x) \text { is divided by }(x+1) \text {, }$
$\text { then the remainder }=p(-1)$
$\text { Putting } x=-1 \text { in } p(x),$
$\text { we get } p(-1)=(-1)^3-2 \times(-1)^2-8 \times(-1)-1$
$=-1-2+8-1$
$=4$
$\therefore \text { Remainder }=4$
Thus, the remainder when $p ( x )$ is divided by $g ( x )$ is 4 .
View full question & answer→Question 53 Marks
Show that $(p-1)$ is a factor of $\left(p^{10}-1\right)$ and also of $\left(p^{11}-1\right)$.
AnswerLet $q(p)=\left(p^{10}-1\right)$ and $f(p)=\left(p^{11}-1\right)$
By the factor theorem, $(p-1)$ will be a factor of $q(p)$ and $f(p)$ if $q(1)$ and $f(1)=0$.
Now, $q(p)=p^{10}-1$
$\Rightarrow q(1)=1^{10}-1=1-1=0$
Hence, $(p-1)$ is a factor of $p^{10}-1$.
And, $f(p)=p^{11}-1$
$\Rightarrow f(1)=1^{11}-1=1-1=0$
Hence, $(p-1)$ is also a factor of $p ^{11}-1$.
View full question & answer→Question 63 Marks
The polynomials$ (2x^3 + x^2 - ax + 2)$ and $(2x^3 - 3x^2 - 3x + a)$ when divided by $(x - 2)$ leave the same remainder. Find the value of a.
AnswerLet $f(x)=2 x^3+x^2-a x+2$
$g(x)=2 x^3-3 x^2-3 x+a .$
By remainder theorem, when $f(x)$ is divided by $(x-2)$, then the remainder $=f(2)$.
Putting $x=2$ in $f(x)$, we get
$f(2)=2 \times 2^3+2^2-a \times 2+2$
$=16+4-2 a+2$
$=-2 a+22$
By remainder theorem, when $g(x)$ is divided by $(x-2)$, then the remainder $=g(2)$.
$g(2)=2 \times 2^3-3 \times 2^2-3 \times 2+a$
$=16-12-6+a$
$=-2+a$
It is given that,
$\Rightarrow-2 a+22=-2+a$
$\Rightarrow-3 a=-24$
$\Rightarrow a=8$
Thus, the value of $a$ is 8 .
View full question & answer→Question 73 Marks
Using the remainder theorem, find the remainder, when $p(x)$ is divided by $g(x)$, where,
$p(x) = 6x^3 + 13x^2 + 3, g(x) = 3x + 2.$
Answer$p(x) = 6x^3 + 13x^2 + 3$
$g(x) = 3x + 2$ $=3\Big(\text{x}+\frac{2}{3}\Big)=3\Big[\text{x}-\Big(-\frac{2}{3}\Big)\Big]$ by remainder theorem,
when p(x) is divided by $(3x + 2)$, then the remainder $=\text{p}\Big(-\frac{2}{3}\Big).$
Putting $\text{x}=-\frac{2}{3}$ in
p(x), we get$\text{p}\Big(-\frac{2}{3}\Big)=6\times\Big(-\frac{2}{3}\Big)^3+13\times\Big(-\frac{2}{3}\Big)^2+3$
$=\frac{-16+52+27}{9}$
$=\frac{63}{9}=7$
$\therefore$ Remainder = $7$
Thus, the remainder when $p(x)$ is divided by $g(x)$ is $7$.
View full question & answer→Question 83 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x)$, when $p(x) = 3x^3 + x^2 - 20x + 12, g(x) = 3x - 2$
AnswerBy the factor theorem, $g(x) = 3x - 2$ will be a factor of p(x) if $\text{p}\Big(\frac{2}{3}\Big)=0.$
Now, $\text{p}(\text{x}) = 3\text{x}^3 + \text{x}^2 - 20\text{x} + 12$
$\Rightarrow\text{p}\Big(\frac{2}{3}\Big)=3\Big(\frac{2}{3}\Big)^3+\Big(\frac{2}{3}\Big)^2-20\times\frac{2}{3}+12$
$=3\times\frac{8}{27}+\frac{4}{9}-\frac{40}{3}+12$
$=\frac{8}{9}+\frac{4}{9}-\frac{40}{3}+12$
$=\frac{8+4-120+108}{9}$
$=\frac{0}{9}$
$=0$
Hence, $g(x) = 3x - 2$ is a factor of the given polynomial $p(x).$
View full question & answer→Question 93 Marks
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where,
$p(x) = x^3 - 6x^2 + 2x - 4$, $\text{g}(\text{x})=1-\frac{3}{2}\text{x}.$
Answer$p(x) = x^3 - 6x^2 + 2x - 4$$\text{g}(\text{x})=1-\frac{3}{2}\text{x}=-\frac{3}{2}\Big(\text{x}-\frac{2}{3}\Big)$
By remainder theorem, when p(x) is divided by $\Big(1-\frac{3}{2}\text{x}\Big),$
then the remainder $=\text{p}\Big(\frac{2}{3}\Big).
$Putting $\text{x}=\frac{2}{3}$ in p(x),
we get$\text{p}\Big(\frac{2}{3}\Big)=\Big(\frac{2}{3}\Big)^3-6\times\Big(\frac{2}{3}\Big)^2+2\times\Big(\frac{2}{3}\Big)-4$
$=\frac{8}{27}-\frac{8}{3}+\frac{4}{3}-4$
$=\frac{8-72+36-108}{27}=-\frac{136}{27}$
$\therefore$ Remainder $=-\frac{136}{27}$
Thus, the remainder when p(x) is divided by g(x) is $=-\frac{136}{27}.$
View full question & answer→Question 103 Marks
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where,
$p(x) = 2x^3 - 7x^2 + 9x - 13, g(x) = x - 3.$
Answer$p(x)=2 x^3-7 x^2+9 x-13 g(x)=x-3$ by remainder theorem,
when $p(x)$ is divided by $(x-3)$,then the remainder $=p(3)$.
Putting $x=3$ in $p(x)$, we get $p(3)=2 \times 3^3-7 \times 3^2+9 \times 3-13=54-63+27-13=5$
$\therefore$ Remainder $=5$
Thus, the remainder when $p(x)$ is divided by $q(x)$ is 5 .
View full question & answer→Question 113 Marks
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where,
p(x) = $x^3 - ax^2 + 6x - a, g(x) x - a.$
Answerp(x) = $x^3 - ax^2 + 6x - a,$
g(x) $x - a$
By remainder theorem,
when p(x) is divided by $(x - a)$, then the remainder = p(a).
Putting $x = a$ in p(x),
we get p(a) =$ a^3 - a \times a^2 + 6 \times a - a$
$= a^3 - a^3 + 6a - a = 5a$
$\therefore$ Remainder = $5a$
Thus, the remainder when p(x) is divided by g(x) is 5a.
View full question & answer→Question 123 Marks
Find the value of m for which $(2x - 1)$ is a factor of $(8x^4 + 4x^3 - 16x^2 + 10x + m).$
AnswerLet $p(x) = 8x^4 + 4x^3 - 16x^2 + 10x + m$
It is given that $(2x - 1)$ is a factor of $p(x).$
$\Rightarrow\text{p}\Big(\frac{1}{2}\Big)=0$
$\Rightarrow8\Big(\frac{1}{2}\Big)^4+4\Big(\frac{1}{2}\Big)^3-16\Big(\frac{1}{2}\Big)^2+10\times\frac{1}{2}+\text{m}=0$
$\Rightarrow8\times\frac{1}{16}+4\times\frac{1}{8}-16\times\frac{1}{4}+5+\text{m}=0$
$\Rightarrow\frac{1}{2}+\frac{1}{2}-4+5+\text{m}=0$
$\Rightarrow1+1+\text{m}=0$
$\Rightarrow2+\text{m}=0$
$\Rightarrow\text{m}=-2$
View full question & answer→Question 133 Marks
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where,
$p(x) = 2x^3 - 9x^2 + x + 15, g(x) = 2x - 3.$
Answer$p(x) = 2x^3 - 9x^2 + x + 15$
$g(x) = 2x - 3$ by remainder theorem,
when p(x) is divided by (2x - 3), then the remainder $=\text{p}\Big(\frac{3}{2}\Big).$ Putting $\text{x}=\frac{3}{2}$ in p(x), we get$\text{p}\Big(\frac{3}{2}\Big)=2\times\Big(\frac{3}{2}\Big)^2+\frac{3}{2}+15$
$=\frac{27}{4}-\frac{81}{4}+\frac{3}{2}+15$
$=\frac{27-81+6+60}{4}$
$=\frac{12}{4}=3$
$\therefore$ Remainder = $3$
Thus, the remainder when $p(x)$ is divided by $g(x)$ is $3.$
View full question & answer→Question 143 Marks
Using the remainder theorem, find the remainder, when $p(x)$ is divided by $g(x)$, where,
$p(x)=3 x^4-6 x^2-8 x-2, g(x)=x-2$.
Answer$p(x)=3 x^4-6 x^2-8 x-2 g(x)=x-2$ by remainder theorem,
when $p(x)$ is divided by $(x-2)$,
then the remainder $=p(2)$.
Putting $x=2$ in $p(x)$,
we get $p(2)=3 \times 2^4-6 \times 2^2-8 \times 2-2=48-24-16-2=6$
$\therefore$ Remainder $=6$
Thus, the remainder when $p ( x )$ is divided by $g ( x )$ is 6 .
View full question & answer→Question 153 Marks
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where,
p(x) = 2x^3 + 3x^2 - 11x - 3$, $\text{g}(\text{x})=\Big(\text{x}+\frac{1}{2}\Big).$
Answer$p(x) =2x^3 + 3x^2 - 11x - 3$
$\text{g}(\text{x})=\Big(\text{x}+\frac{1}{2}\Big)=\Big[\text{x}-\Big(-\frac{1}{2}\Big)\Big]$
By remainder theorem, when p(x) is divided by $\Big(\text{x}+\frac{1}{2}\Big),$ then the remainder $=\text{p}\Big(-\frac{1}{2}\Big).$
Putting $\text{x}=-\frac{1}{2}$ in p(x), we get
$\text{p}\Big(-\frac{1}{2}\Big)=2\times\Big(-\frac{1}{2}\Big)^3\\+3\times\Big(-\frac{1}{2}\Big)^2-11\times\Big(-\frac{1}{2}\Big)-3$
$=-\frac{1}{4}+\frac{3}{4}+\frac{11}{2}-3$
$=\frac{-1+3+22-12}{4}$
$=\frac{12}{4}=3$
$\therefore$ Remainder = $3$
Thus, the remainder when $p(x)$ is divided by $g(x)$ is $3.$
View full question & answer→