Question 12 Marks
Find the value of ‘m’, if $mx^3 + 2x^2 – 3$ and $x^2 – mx + 4$ leave the same remainder when each is divided by $x – 2$
Answer$\text { Let } f(x)=m x^3+2 x^2-3$
$g(x)=x^2-m x+4$
It is given that $f(x)$ and $g(x)$ leave the same remainder when divided by $(x – 2).$ Therefore, we have:
$f (2) = g (2)$
$m(2)^3+2(2)^2-3=(2)^2-m(2)+4$
$8 m+8-3=4-2 m+4$
$10 m=3$
$m=\frac{3}{10}$
View full question & answer→Question 22 Marks
What should be subtracted from $3x^3 – 8x^2 + 4x – 3$, so that the resulting expression has $x + 2$ as a factor?
AnswerLet the required number be $k.$
Let $f(x) = 3x^3 – 8x^2 + 4x – 3 – k$
According to the given information,
$f (-2) = 0$
$3(-2)^3 – 8(-2)^2 + 4(-2) – 3 – k = 0$
$-24 – 32 – 8 – 3 – k = 0$
$-67 – k = 0$
$k = -67$
Thus, the required number is $-67.$
View full question & answer→Question 32 Marks
Using remainder theorem, find the value of k if on dividing $2x^3 + 3x^2 - kx + 5$ by $x - 2,$ leaves a remainder $7$
AnswerLet $f(x)=2 x^3+3 x^2-k x+5$
Using remainder theorem,
$f(2) = 7$
$\therefore 2(2)^3+3(2)^2-k(2)+5=7$
$\therefore 2(8)+3(4)-k(2)+5=7$
$\therefore 16 + 12 - 2k + 5 = 7$
$\therefore 2k = 16 + 12 + 5 - 7$
$\therefore 2k = 26$
$\therefore k = 13$
View full question & answer→Question 42 Marks
When divided by $x-3$ the polynomials $x^3-p x^2+x+6$ and $2 x^3-x^2-(p+3) x-6$ leave the same remainder. Find the value of 'p'.
AnswerIf $(x – 3)$ divides$ f(x) = x^3 – px^2 + x + 6$, then,
Remainder $= f(3) = 3^3 – p(3)^2 + 3 + 6 = 36 – 9p$
If $(x – 3)$ divides $g(x) = 2x^3 – x^2 – (p + 3) x – 6$, then
Remainder $= g(3) = 2(3)^3 – (3)^2 – (p + 3) (3) – 6 = 30 – 3p$
Now,$ f(3) = g(3)$
$\Rightarrow 36 – 9p = 30 – 3p$
$\Rightarrow -6p = -6$
$\Rightarrow p = 1$
View full question & answer→Question 52 Marks
Find the number which should be added to $x^2 + x + 3$ so that the resulting polynomial is completely divisible by $(x + 3).$
AnswerLet the required number be $k.$
Let $f(x) = x^2 + x + 3 + k$
It is given that $f(x)$ is divisible by $(x + 3).$
$\therefore$ Remainder $= 0$
$f (-3) = 0$
$(-3)^2 + (-3) + 3 + k = 0$
$9 – 3 + 3 + k = 0$
$9 + k = 0$
$k = -9$
Thus, the required number is$ -9.$
View full question & answer→Question 62 Marks
Find the value of ‘a’, if $(x – a)$ is a factor of $x^3 – ax^2 + x + 2$.
Answer$\operatorname{left} f(x)=x^3-a x^2+x+2$
it is given that (x-a) is a factor of f (x)
Remainder $=f(a)=0$
$a^3-a^3+a+2=0$
$a+2=0$
$a=-2$
View full question & answer→Question 72 Marks
Find the number that must be subtracted from the polynomial $3y^3 + y^2 – 22y + 15$, so that the resulting polynomial is completely divisible by $y + 3.$
AnswerLet the number to be subtracted from the given polynomial be $k$
let $f(y)=3 y^3+y^2-22 y+15-k$
it is given that $f ( y )$ is divisible by $(y+3)$
Remainder $=f(-3)=0$
$3(-3)^3+(-3)^2-22(-3)+15-k=0$
$-81+9+66+15-k=0$
$9-k=0$
$k=9$
View full question & answer→Question 82 Marks
When $x^3+2 x^2-k x+4$ is divided by $x – 2$, the remainder is $k.$ Find the value of constant $k.$
AnswerLeft $f ( x )=x^3+2 x^2-k x+4$
$x-2=0$
$\Rightarrow x=2$
on dividing f (x) by x-2,, it leaves a remainder $K.$
$\therefore f(2)=k$
$(2)^3+2(2)^2-k(2)+4=k$
$8+8-2 k+4=k$
$20=3 k$
$k=\frac{20}{3}=6 \frac{2}{3}$
View full question & answer→Question 92 Marks
Find the value of a, if $x – 2$ is a factor of $2x^5 – 6x^4 – 2ax^3 + 6ax^2 + 4ax + 8.$
Answer$f(x) = 2x^5 – 6x^4 – 2ax^3 + 6ax^2 + 4ax + 8$
$x – 2 = 0$
$\Rightarrow x = 2$
Since, x – 2 is a factor of f(x), remainder $= 0.$
$2(2)^5 – 6(2)^4 – 2a(2)^3 + 6a(2)^2 + 4a(2) + 8 = 0$
$64 – 96 – 16a + 24a + 8a + 8 = 0$
$-24 + 16a = 0$
$16a = 24$
$a = 1.5$
View full question & answer→Question 102 Marks
Find the value of k, if $3x – 4$ is a factor of expression
$3 x^2+2 x-k$
AnswerFind the value of k, if $3x – 4$ is a factor of expression $3 x^2+2 x - k$.
$\therefore f\left(\frac{4}{3}\right)=0$
$\Rightarrow 3\left(\frac{4}{3}\right)^2+2\left(\frac{4}{3}\right)-k=0$
$\Rightarrow \frac{16}{3}+\frac{8}{3}-k=0 $
$ \Rightarrow \frac{24}{3}=k$
$ \Rightarrow k =8$
View full question & answer→Question 112 Marks
If $2x + 1$ is a factor of $2x^2 + ax – 3$, find the value of a.
Answer$2 x+1$ is a factor of $f ( x )=2 x^2+a x-3$
$\therefore f\left(\frac{-1}{2}\right)=0$
$\Rightarrow 2\left(\frac{-1}{2}\right)^2+a\left(\frac{-1}{2}\right)-3=0$
$\Rightarrow \frac{1}{2}-\frac{a}{2}=3$
$\Rightarrow 1-a=6$
$\Rightarrow a=-5$
View full question & answer→Question 122 Marks
Use the Remainder Theorem to find which of the following is a factor of $2x^3 + 3x^2 – 5x – 6.x + 2$
AnswerBy remainder theorem we know that when a polynomial $f (x)$ is divided by $x – a$, then the remainder is $f(a).$
Let $f(x) = 2x^3 + 3x^2 – 5x – 6$
$f(-2)=2(-2)^3(-2)^2-5(-2)-6$
$=-16+12+10-6=0`$
Thus, $(x + 2)$ is a factor of the polynomial $f(x).$
View full question & answer→Question 132 Marks
Use the Remainder Theorem to find which of the following is a factor of $2x^3 + 3x^2 – 5x – 6.2x – 1$
AnswerBy remainder theorem we know that when a polynomial f (x) is divided by $x – a,$ then the remainder is f(a).
Let $f(x)=2 x^3+3 x^2-5 x-6$
$f\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)^3+3\left(\frac{1}{2}\right)^2-5\left(\frac{1}{2}\right)-6$
$\frac{1}{4}+\frac{3}{4}-\frac{5}{2}-6$
$ -\frac{5}{2}-5=-\frac{15}{2} \neq 0$
Thus, $(2x – 1)$ is not a factor of the polynomial f(x)
View full question & answer→Question 142 Marks
Use the Remainder Theorem to find which of the following is a factor of $2x^3 + 3x^2 – 5x – 6.x + 1$
AnswerBy remainder theorem we know that when a polynomial f (x) is divided by x – a, then the remainder is f(a)
let $f(x)=2 x^3+3 x^2-5 x-6$
$f(-1)=2(-1)^3+3(-1)^2-5(-1)-6=-2+3+5=0$
Thus, (x + 1) is a factor of the polynomial f(x).
View full question & answer→Question 152 Marks
show that
$3 x+2$ is a factor of $3 x^2-x-2$
Answer(x – a) is a factor of a polynomial f(x) if the remainder, when f(x) is divided by (x – a), is 0, i.e., if f(a) = 0.
$f(x)=3 x^2-x-2$
$f\left(\frac{-2}{3}\right)=3\left(\frac{-2}{3}\right)^2-\left(\frac{-2}{3}\right)-2=\frac{4}{3}+\frac{2}{3}-2=2-2=0$
Hence, $3 x +2$ is a factor of $3 x^2-x-2$
View full question & answer→Question 162 Marks
show that
$x-2$ is a factor of $5 x^2+15 x-50$
Answer(x – a) is a factor of a polynomial f(x) if the remainder, when f(x) is divided by (x – a), is 0, i.e., if f(a) = 0.
$f(x)=5 x^2+15 x-50$
$f(2)=5(2)^2+15(2)-50=20+30-50=0$
Hence, $x-2$ is a factor of $5 x^2+15 x-50$
View full question & answer→Question 172 Marks
What number should be subtracted from $x^3 + 3x^2 – 8x + 14$ so that on dividing it with $x – 2$, the remainder is $10.$
AnswerLet the number to be subtracted be $k$ and the resulting polynomial be $f(x).$
So, $f(x)=x^3+3 x^2-8 x+14-k$
it is given that when $f(x)$ is divided by $(x-2),$ the remainder is $10.$
$f(2)=10$
$(2)^3+3(2)^2-8(2)+14-k=10$
$8+12-16+14-k=10$
$18-k=10$
$k=8$
Thus, the requied number is $8.$
View full question & answer→Question 182 Marks
What number should be added to $3x^3 – 5x^2 + 6x$ so that when resulting polynomial is divided by $x – 3,$ the remainder is $8?$
AnswerLet the numbw-er k be added and resulting polnomic be $f(x).$
So, $f(x)=3 x^3+5 x^2+6 x+k$
it is given that when $f(x)$ is dividing by $(x-3),$ the remainder is $8.$
$\therefore f(3)=8$
$3(3)^3-5(3)^2+6(3)+k=8$
$81-45+18+k=8$
$54+k=8$
$k=-46$
Thus, the required number is $-46.$
View full question & answer→Question 192 Marks
Find, in given case, the remainder when:
$x^3+3 x^2-12 x+4$ is divided by $x -2$
Answer$f(x)=x^4-3 x^2-12 x+4$
$\begin{aligned} \text { Remainded } & =f(2)=(2)^3+3(2)^2-12(2)+4 \\ & =8+12-24+4 \\ & =0\end{aligned}$
View full question & answer→Question 202 Marks
Find , in given case, the remainder when :
$x^4-3 x^2+2 x+1$ is dividend by $x-1$
AnswerBy remainder theorem we know that when a polynomial f (x) is divided by x – a, then the remainder is f(a).
$f(x)=x^4-3 x^2+2 x+1$
Remainder $=f(1)=(1)^4-3(1)^2+2(1)+1=1-3+2+1=1$
View full question & answer→Question 212 Marks
Find the value of a, if the division of $ax^3 + 9x^2 + 4x – 10$ by $x + 3$ leaves a remainder $5.$
AnswerLeft $f(x) = ax^3 + 9x^2 + 4x - 10x + 3 = 0$
$\Rightarrow x = -3$
On dividing f(x) by x+3, it leaves a remainder 5.
$\therefore f(-3) = 5$
$a(-3)^3 + 9(-3)^2 + 4(-3) - 10 = 5$
$-27a + 81 - 12 - 10 = 5$
$54 = 27a$
$a=2$
View full question & answer→