[2 Mark Question Answer]

Question 12 Marks

What number must be added to $2x^3 – 7x^2 + 2x$ so that the resulting polynomial leaves the remainder – 2 when divided by $2x – 3?$

Answer

Let $a$ be added to $2 x^3-7 x 2+2 x$ dividing it by $2 x-3$, then
$(2 x-3) \overline{2 x^3-7 x^2+2 x+a}\left(x^2-2 x-2)\right.$
$2 x^3-3 x^2$
$\frac{-}{-4 \times 2+2 x}$
$-4 x^2+6 x$
$+\quad-$
$-4 x+6$
$\frac{+-}{a-6}$
But remainder is $-2$ , then
$ a-6=-2$
$\Rightarrow a=-2+6$
$\Rightarrow a=4$
Hence $4$ is to be added.

View full question & answer→

Question 22 Marks

What number must be subtracted from $2 x^2-5 x$ so that the resulting polynomial leaves the remainder $2$ , when divided by $2 x+1 ?$

Answer

Let a be subtracted from $2 x^2-5 x$,
Dividing $2 x ^2-5 x$ by $2 x +1$,
$2 x+1) \overline{2 x^2-5 x-a}(x-3$
$
\begin{gathered}
2 x^2+x \\
\hline-\quad- \\
-6 x-a \\
-6 x-3 \\
+\quad+ \\
\hline-a+3 \\
\hline
\end{gathered}
$
Here remainder is $(3-a)$
but we are given that remainder is $2 .$
$ \therefore 3-a=2 $
$ \Rightarrow-a=2-3=-1$
$ \Rightarrow a =1$
Hence 1 is to be subtracted.

View full question & answer→

Question 32 Marks

Using remainder theorem, find the value of a if the division of $x^3 + 5x^2 – ax + 6 by (x – 1)$ leaves the remainder 2a.

Answer

Let $x – 1 = 0$, then $x = 1$
Substituting the value of x in $f(x)$
$f(x) = x^3 + 5x^3 – ax + 6$
$= (1)^3 + 5(1)^2 – a(1) + 6$
$= 1 + 5 – a + 6$
$= 12 – a$
$\because $ Remainder $= 2a$
$\therefore 12 – a = 2a$
$\Rightarrow 12 = a + 2a$
$\Rightarrow 3a – 12$
$\therefore a = 4.$

View full question & answer→

Question 42 Marks

Using remainder theorem, find the value of k if on dividing $2x^3 + 3x^2 - kx + 5$ by $x - 2$, leaves a remainder $7$

Answer

Let $f(x)=2 x^3+3 x^2-k x+5$
Using remainder theorem,
$ f (2)=7$
$\therefore 2(2)^3+3(2)^2-k(2)+5=7$
$\therefore 2(8)+3(4)-k(2)+5=7$
$\therefore 16+12-2 k +5=7$
$\therefore 2 k =16+12+5-7$
$\therefore 2 k =26$
$\therefore k =13$

View full question & answer→

Question 52 Marks

Find the remainder (without division) on dividing $3x^2 + 5x – 9 by (3x + 2)$

Answer

Let $3 x+2=0$,
then $3 x=-2$
$\Rightarrow x=\frac{-2}{3}$
Substituting the value of $x$ in $f(x)$
$ f(x)=3 x^2+5 x-9$
$=3\left(-\frac{2}{3}\right)^2+5\left(-\frac{2}{3}\right)-9$
$=3 \times \frac{4}{9}-5 \times \frac{2}{3}-9$
$=\frac{4}{3}-\frac{10}{3}-9$
$=-\frac{6}{3}-9$
$=-2-9$
$=-11$
$\therefore \text { Remainder }=-11 .$

View full question & answer→

Question 62 Marks

Find the remainder (without division) on dividing $f(x)$ by $(2x + 1)$ where $f(x) = 3x^3 – 7x^2 + 4x + 11$

Answer

Let $2 x+1=0$, then $x=$
Substituting the value of $x$ in $f(x)$ :
$
\begin{aligned}
& f(x)=3 x^3-7 x^2+4 x+11 \\
& =3\left(-\frac{1}{2}\right)^3-7\left(-\frac{1}{2}\right)^2+4\left(-\frac{1}{2}\right)+11 \\
& =3\left(-\frac{1}{8}\right)-7\left(\frac{1}{4}\right)+4\left(-\frac{1}{2}\right)+11 \\
& =-\frac{3}{8}-\frac{7}{4}-2+11 \\
& =\frac{-3-14-16+88}{8} \\
& =\frac{55}{8} \\
& =6 \frac{7}{8} \\
& \therefore \text { Remainder }=6 \frac{7}{8} .
\end{aligned}
$

View full question & answer→

Question 72 Marks

Find the remainder (without division) on dividing $f(x)$ by $(2x + 1)$ where $f(x) = 4x^2 + 5x + 3$

Answer

Let $2 x+1=0$, then $x=-\frac{1}{2}$
Substituting the value of $x$ in $f(x)$ :
$
\begin{aligned}
& f(x)=4 x^2+5 x+3 \\
& =4\left(-\frac{1}{2}\right)^2+5 \times\left(-\frac{1}{2}\right)+3 \\
& =4 \times \frac{1}{4}-\frac{5}{2}+3 \\
& =1-\frac{5}{2}+3 \\
& =4-\frac{5}{2} \\
& =\frac{3}{2} \\
& \therefore \text { Remainder }=\frac{3}{2} .
\end{aligned}
$

View full question & answer→

Question 82 Marks

Find the value of $‘K’$ for which $x = 3$ is a solution of the quadratic equation, $(K + 2)x^2 – Kx + 6 = 0$. Also, find the other root of the equation.

Answer

$(K + 2)x^2 – Kx + 6 = 0 …(1)$
Substitute x = 3 in equation $(1)$
$(–4 + 2)x^2 –(–4)x + 6 = 0$
$\Rightarrow –2x^2+ 4x + 6 = 0$
$\Rightarrow x^2– 2x – 3 = 0 ...$(Dividing by $2)$
$\Rightarrow x^2 – 3x + x – 3 = 0$
$\Rightarrow x(x – 3) + 1(x – 3) = 0$
$(x + 1)(x – 3) = 0$
So, the roots are $x = –1$ and $x = 3$
Thus, the other root of the equation is $x = –1.$

View full question & answer→

Question 92 Marks

When $2x^3 – 9x^2 + 10x – p$ is divided by $(x + 1)$, the remainder is $– 24$.Find the value of $p.$

Answer

Let $x + 1 = 0 $then $x = -1$
Substituting the value of x in $f(x)$
$f(x) = 2x^3 – 9x^2 + 10x – p$
$f(–1) = 2(–1)^3 – 9(–1)^2 + 10(–1) – p$
$= –2 – 9 – 10 – p$
$= –21 – p$
$\because $ the remainder $= –24$
$\therefore –21 – p = –24$
$\Rightarrow p = –24 + 21$
$= –3$
$\therefore p = 3.$

View full question & answer→

Question 102 Marks

Find the remainder when $2x^3 – 3x^2 + 4x + 7$ is divided by $2x + 1$

Answer

Let $2 x+1=0$,
then $2 x-1$
$
\Rightarrow x =-\frac{1}{2}
$
Now substituting the value of $x$ in $f(x)$
$
\begin{aligned}
& f\left(-\frac{1}{2}\right)=2\left(-\frac{1}{2}\right)^2-3\left(-\frac{1}{2}\right)^2+4\left(-\frac{1}{2}\right)+7 \\
& =2\left(-\frac{1}{8}\right)-3\left(\frac{1}{4}\right)+4\left(-\frac{1}{2}\right)+7 \\
& =-\frac{1}{4}-\frac{3}{4}-2+7 \\
& =-1+-2+7 \\
& =4 \\
& \therefore \text { Remainder }=4 .
\end{aligned}
$

View full question & answer→

Question 112 Marks

Find the remainder when $2x^3 – 3x^2 + 4x + 7$ is divided by $x + 3$

Answer

$f(x) = 2x^3 – 3x^2 + 4x + 7$
Let $x + 3 = 0,$ then $x = – 3$
Substituting the value of x in $f(x)$
$f(–3) = 2(–3)^3 – 3(–3)^2 + 4(–3) + 7$
$= 2 x (–27) – 3(9) + 4(–3) + 7$
$= –54 – 27 – 12 + 7$
$= – 93 + 7$
$= – 86$
$\therefore $ Remainder $= – 86.$

View full question & answer→

Question 122 Marks

Find the remainder when$ 2x^3 – 3x^2 + 4x + 7$ is divided by $x – 2$

Answer

$f(x) = 2x^3 – 3x^2 + 4x + 7$
Let $x – 2 = 0,$
then $x = 2$
Substituting value of x in $f(x)$
$f(2) = 2 (2)^3 – 3 (2)^2 + 4 (2) + 7$
$= 2 \times 8 – 3 \times 4 + 4 \times 2 + 7$
$= 16 – 12 + 8 + 7$
$= 19$
Remainder $= 19$

View full question & answer→

Mathematics [2 Mark Question Answer]

Test yourself on this topic