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, $p(x) = x^3 - 6x^2 + 9x + 3, g(x) = x - 1.$
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 $2x^4 - 5x^3 + 2x^2 - x - 3$ so that the result is exactly divisible by $(x - 2)?$
AnswerLet the required number to be added be $k$.
Then, $p(x) = 2x^4 - 5x^3 + 2x^2 - x - 3 + k g(x) = x - 2$
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, $p(x) = x^3 - 2x^2 - 8x - 1, g(x) = x + 1.$
Answer$p(x)=x^3-2 x^2-8 x-1 g(x)=x+1$ by remainder theorem,
when $p(x)$ is divided by $(x+1)$, then the remainder $=p(-1)$. Putting $x=-1$ in $p(x)$,
we get $p(-1)=(-1)^3-2 \times(-1)^2-8 \times(-1)-1=-1-2+8-1=4$
$\therefore$ 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 $\left(2 x^3+x^2-a x+2\right)$ and $\left(2 x^3-3 x^2-3 x+a\right)$ when divided by $(x-2)$ leave the same remainder. Find the value of $a$.
AnswerLet $f(x) = 2x^3 + x^2 - ax + 2 g(x) = 2x^3 - 3x^2 - 3x + a$. By remainder theorem,
when $f(x)$ is divided by $(x - 2)$, then the remainder $= f(2)$. Putting $x = 2$ in $f(x),$
we get $f(2) = 2 \times 2^3 + 2^2 - a \times 2 + 2 = 16 + 4 - 2a + 2 = -2a + 22$ 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,
$\Rightarrow -2a + 22 = -2 + a$
$\Rightarrow -3a = -24$
$\Rightarrow a = 8$
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$p(x) = 2x^3 - 7x^2 + 9x - 13 g(x) = x - 3$ 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 $g(x)$ 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, $p(x) = x^3 - ax^2 + 6x - a, g(x) x - a.$
Answer$p(x) = x^3 - ax^2 + 6x - a, g(x) x - a$ 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 $(2 x-1)$ is a factor of $\left(8 x^4+4 x^3-16 x^2+10 x+m\right)$.
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, $p(x) = 3x^4 - 6x^2 - 8x - 2, g(x) = x - 2.$
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, $p(x)=2 x^3+3 x^2-11 x-3, g(x)=\left(x+\frac{1}{2}\right)$.
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→