M.C.Q

MCQ 11 Mark

If $x - 3$ is a factor of $x^2 - ax - 15, $ then $a =$

✓
$-2$
B
$5$
C
$-5$
D
$3$

Answer

Correct option: A.

$-2$

$x - 3$ is a factor of $x^2 - ax - 15,$
then at $x = 3,$
$x^2 - ax - 15 = 0$
$i.e. (3)^2 - a(3) - 15 = 0$
$9 - 3a - 15 = 0$
$a = -2$

View full question & answer→

MCQ 21 Mark

If $x - a$ is a factor of $x^3 - 3x^2a + 2a^2x + b,$ then the value of $b$ is:

✓
$0$
B
$2$
C
$1$
D
$3$

Answer

Correct option: A.

$0$

Let $p(x) = x^3 - 3x^2a + 2a^2x + b$
$(x - a)$ is a factor of $p(x).$
So,
$p(a) = 0$
$a^3 - 3a^2a + 2a^2a + b = 0$
$a^3 - 3a^3 + 2a^3 + b = 0$
$3a^3 - 3a^3 + b = 0$
$b = 0$

View full question & answer→

MCQ 31 Mark

The value of $k$ for which $x - 1$ is a factor of $4x^3 + 3x^2 - 4x + k,$ is:

A
$3$
B
$1$
C
$-2$
✓
$-3$

Answer

Correct option: D.

$-3$

Let $p(x) = 4x^3 + 3x^2 - 4x + k$
Now,
if $(x - 1)$ is a factor of $p(x),$ then at $x = 1, p(x) = 0$
So, $p(1) = 0$
$\Rightarrow 4(1)^3 + 3(1)^2 - 4(1) + k = 0$
$\Rightarrow 4 + 3 - 4 + k = 0$
$\Rightarrow k = -3$

View full question & answer→

MCQ 41 Mark

If $x - 2$ is a factor of $x^2 + 3ax - 2a,$ then $a =$

A
$2$
B
$-2$
C
$1$
✓
$-1$

Answer

Correct option: D.

$-1$

Let $p(x) = x^2 + 3ax - 2a$ be the given polynomial.
$x - 2$ is a factor of $p(x).$
Thus,
$p(2) = 0$
$(2)^2 + 3a \times 2 - 2a = 0$
$4 + 4a = 0$
$a = -1$

View full question & answer→

MCQ 51 Mark

If $x + 1$ is a factor of the polynomial $2x^2 + kx,$ then $k =$

A
$-2$
B
$-3$
C
$4$
✓
$2$

Answer

Correct option: D.

$2$

$x + 1$ is a factor of $p(x) = 2x^2 + kx$
Then$, p(-1) = 0$
$i.e. 2(-1)^2 + k(-1) = 0$
$2 - k = 0$
$k = 2$

View full question & answer→

MCQ 61 Mark

If both $x - 2$ and $\text{x}-\frac{1}{2}$ are factor of $px^2 + 5x + r,$ then

✓
$p = r$
B
$p + r = 0$
C
$2p + r = 0$
D
$p + 2r = 0$

Answer

Correct option: A.

$p = r$

Let $f(x) = px^2 + 5x + r$
Now,
If $x - 2$ and $\text{x}-\frac{1}{2}$ are factors of f(x).
Then at $x = 2$ and $\text{x}-\frac{1}{2},$ f(x) = 0.
So, $f(2) = 0, \text{f}\Big(\frac{1}{2}\Big)=0$
$\Rightarrow p(2)^2 + 5(2) + r = 0$
And, $\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
$\Rightarrow 4p + r + 10 = 0 ...(1)$
And $4r + p + 10 = 0 ...(2)$
Subtracting equation $(2)$ from $(1),$ we have
$3p - 3r = 0$
$\Rightarrow p = r$

View full question & answer→

Question 71 Mark

Let f(x) be a polynomial such that $\text{f}\Big(-\frac{1}{2}\Big)=0,$ then a factor of f(x) is:

2x - 1
2x + 1
x - 1
x + 1

Answer

2x + 1

Solution:
If f(x) is a polynomial and $\text{f}(\alpha)=0.$ Then $(\text{x}-\alpha)$ is a factor of f(x) or vice versa if $(\text{x}-\alpha)$ is a factor of f(x) then $\text{f}(\alpha)=0.$
Now,
$\text{f}\Big(\frac{-1}{2}\Big)=0$
So, at $\text{x}=\frac{-1}{2},\text{f(x)}=0$
Or at 2x = -1, f(x) = 0
Or at 2x + 1 = 0, f(x) = 0
⇒ (2x + 1) is a factor of f(x).

View full question & answer→

MCQ 81 Mark

If $x^{51} + 51$ is divided by $x + 1, $the remainder is:

A
$0$
B
$1$
C
$49$
✓
$50$

Answer

Correct option: D.

$50$

When the polynomial $p(x)$ is divided by $q(x) \ i. e.\ (\text{x}\pm\alpha)$ then $\text{p}(\mp\alpha)$ is the remainder.
If $\text{x}\pm\alpha$ is the factor of polynomial, then remainder is $'0\ '.$
So,
If $x^{51} + 51$ is divided $x + 1.$
Remainder $= (-1)^{51} + 51 $
$= -1 + 51 $
$= 50.$

View full question & answer→

MCQ 91 Mark

If $(x − 1)$ is a factor of polynomial $f(x)$ but not of $g(x) ,$ then it must be a factor of:

✓
$f(x)g(x)$
B
$-f(x) + g(x)$
C
$f(x) - g(x)$
D
${f(x) + g(x)}g(x)$

Answer

Correct option: A.

$f(x)g(x)$

If $x - 1$ is a factor of $f(x)$ then definitely $f(1) = 0$
And,
$x - 1$ is not a factor of $g(x)$, then $\text{g(1)}\neq0.$
So, at $x = 1$
$f(1)g(1) = 0 \times g(1) = 0$
$-f(1) + g(1) = 0 + g(1) = g(1) \neq 0$
$f(1) - g(1) = 0 - g(1) = -g(1) \neq 0$
${f(1) + g(1)}g(1) = {0 + g(1)}g(1) = {g(1)}^2 \neq 0$ So, at $x = 1$ only, $f(x)g(x) = 0$
Thus, $(x - 1)$ is factor of $f(x)g(x)$ too.

View full question & answer→

MCQ 101 Mark

One factor of $x^4 + x^2 - 20$ is $x^2 + 5.$ The other factor is:

✓
$x^2 - 4$
B
$x - 4$
C
$x^{2 }- 5$
D
$x + 4$

Answer

Correct option: A.

$x^2 - 4$

$x^4 + x^2 - 20$
$= x^4 + 5x^2 - 4x^2 - 20$
$=x^2(x^2 + 5) - 4(x^2 + 5)$
$= (x^2 + 5)(x^2 - 4)$
So, other factor is $x^2 - 4.$

View full question & answer→

MCQ 111 Mark

If $x^3 + 6x^2 + 4x + k$ is exactly divisible by $x + 2$, then $k$

A
$-6$
B
$-7$
✓
$-8$
D
$-10$

Answer

Correct option: C.

$-8$

Since, $p(x) = x^3 + 6x^2 + 4x + k$ is exactly divisible by $x + 2,$
$(x + 2)$ is a factor of $p(x).$
So, $p(-2) = 0$
$i.e. (-2)^3 + 6(-2)^2 + 4(-2) + k = 0$
$-8 + 24 - 8 + k = 0$
$24 - 16 + k = 0$
$8 + k = 0$
$k = -8$

View full question & answer→

MCQ 121 Mark

If $x + 2$ and $x - 1$ are the factors of $x^3 + 10x^2 + mx + n,$ then the values of m and $n$ are respectively

A
$5$ and $-3$
B
$17$ and $-8$
✓
$7$ and $-18$
D
$23$ and $-19$

Answer

Correct option: C.

$7$ and $-18$

If $(x + 2)$ and $(x - 1)$ are factors of polynomial $x^3 + 10x^2 + mx + n,$
then $x = -2, x = +1$ will satisfy the polynomial.
Let $p(x) = x^3 + 10x^2 + mx + n$
Then, $p(-2) = 0$
$(-2)^3 + 10(-2)^2 + m(-2) + n = 0$
$-8 + 40 - 2m + n = 0$
$32 - 2m + n = 0 ...(1)$
And,$ p(1) = 0$
$(1)^3 + 10(1)^2 + m(1) + n = 0$
$1 + 10 + m + n = 0$
$11 + m + n = 0 ...(2)$
Substracting equation $(1)$ from equation $(2),$ we get
$-21 + 3m = 0$
$3m = 21$
$m = 7$
Substituting $m = 7$ in equation $(2),$
$11 + 7 + n = 0$
$18 + n = 0$
$n = -18$

View full question & answer→

MCQ 131 Mark

If $x^2 - 1$ is a factor of $ax^4+ bx^3 + cx^2 + dx + e,$ then

✓
$a + c + e = b + d$
B
$a + b + e = c + d$
C
$a + b + c = d + e$
D
$b + c + d = a + e$

Answer

Correct option: A.

$a + c + e = b + d$

If $x^2 - 1$ is factor of $p(x) = ax^4+ bx^3 + cx^2 + dx + e.$
Then $(x - 1)$ and $(x + 1)$ will also be factors of $p(x).$
Because $x^2 - 1 = (x - 1)(x + 1)$
Then, at $x = 1$ and $x = -1, p(x) = 0$
$\Rightarrow p(1) = 0$ and $p(-1) = 0$
$\Rightarrow a + b + c + d + e = 0 ...(1)$
And
$\Rightarrow a - b + c - d + e = 0 ...(2)$
Adding equations $(1)$ and $(2).$
$2a + 2c + 2e = 0$
$\Rightarrow a + c + e = 0 ...(3)$
Substracting equation $(2)$ from $(1)$
$2b + 2d = 0$
$\Rightarrow b + d = 0 ...(4)$
From equations $(3) $and $(4),$ we get
$a + c + e = b + d$

View full question & answer→

MCQ 141 Mark

If $x^2 + x + 1$ is a factor of the polynomial $3x^3 + 8x^2{ }+ 8x + 3 + 5k,$ then the value of $k$ is:

A
$0$
✓
$\frac{2}{5}$
C
$\frac{5}{2}$
D
$-1$

Answer

Correct option: B.

$\frac{2}{5}$

Let $p(x) = 3x^3 + 8x^2 + 8x + 3 + 5k$ and $q(x) = x^2 + x + 1$
Now,
If $q(x)$ is a factor of $p(x),$ then arranging $p(x)$ in order to have $q(x)$ in common,
$p(x) = 3x^3 + 3x^2 + 3x + 5x^2 + 5x + 3 + 2 - 2 + 5k [ $Adding $+2, -2 in p(x)]$
$= 3x(x^2 + x + 1) + 5(x^2 + x + 1) + 5k - 2$
$p(x) = (x^2+ x + 1)(3x + 5) + 5k - 2 ...(1)$
From equation $(1), $ we can see if we divide $p(x)$ by $q(x),$ then quotient will be $(3x + 5)$ and remainder will be $(5k - 2)$
But $q(x)$ is a factor of $p(x).$
So, remainder $= 0$
$\Rightarrow 5k - 2 = 0$
$\Rightarrow\text{k}=\frac{2}{5}$

View full question & answer→

MCQ 151 Mark

If $(3x - 1)^7 = a_7x^7 + a_6x^6 + a_5x^5 + ... + a_1x + a_0,$ then $a_7 + a_6 + a_5 + ... + a_1 + a_0 =$

A
$0$
B
$1$
✓
$128$
D
$64$

Answer

Correct option: C.

$128$

$(3x - 1)^7 = a_7x^7 + a_6x^6 + ... + a_1x + a_{0 ...(1)}$
Putting $x = 1$ in equation $(1),$ we have
$[3(1) - 1]^7 = a_7 + a_6 + ..... + a_1 + a_0$
So,
$a_7 + a_6 + a_5 + ..... + a_1 + a_0 = 2^{7 }= 128$

View full question & answer→

MCQ 161 Mark

If $x + 2$ is a factor of $x^2 + mx + 14,$ then $m =$

A
$7$
B
$2$
✓
$9$
D
$14$

Answer

Correct option: C.

$9$

If $x + 2$ is a factor of $x^2 + mx + 14,$
then at $x = -2,$
$x^2 + mx + 14 = 0$
$i.e. (-2)^2 + m(-2) + 14 = 0$
$4 - 2m + 14 = 0$
$2m = 18$
$m = 9$

View full question & answer→

MCQ 171 Mark

If $x^{140} + 2x^{151} + k$ is divisible by $x + 1, $then the value of $k$ is:

✓
$1$
B
$-3$
C
$2$
D
$-2$

Answer

Correct option: A.

$1$

Let $p(x) = x^{140} + 2x^{151} + k$
Since $p(x)$ is divisible by $(x + 1),$
$(x + 1)$ is a factor of $p(x).$
So,
$p(-1) = 0$
$(-1)^{140} + 2(-1)^{151} + k = 0$
$1 + 2(-1) + k = 0$
$1 - 2 + k = 0$
$k - 1 = 0$
$k = 1$

View full question & answer→

MCQ 181 Mark

When $x^3 - 2x^2 + ax - b$ is divided by $x^2 - 2x - 3,$ the remainder is $x - 6$. The values of $a$ and $b$ are respectively

A
$-2, -6$
B
$2$ and $-6$
✓
$-2$ and $6$
D
$2$ and $6$

Answer

Correct option: C.

$-2$ and $6$

Let $p(x) = x^3 - 2x^2 + ax - b, r(x) = x - 6$ and $q(x) = x^2 - 2x - 3$
Then $q(x)$ is a factor of$ [p(x) - r(x)] [$because if $p(x)$ is divided by $q(x),$ remainder is $r(x)].$
So, $[p(x) - r(x)]$ will be exactly divided by $q(x)]$
Now,
$q(x) = x^2 - 2x - 3 = (x - 3)(x + 1)$
If $q(x)$ is a factor of $[p(x) - r(x)]$
Then $(x - 3)$ and $(x + 1)$ are also factors of $[p(x) - r(x)]$
So, at $x = 3$ and $x = -1, p(x) - r(x)$ will be zero.
Now
$p(3) - r(3) = 0$
$i.e. (3)^3 - 2(3)^2 + a(3) - b - (3 - 6) = 0$
$i.e. 27 - 18 + 3a - b + 3 = 0$
$i.e. 3a - b + 12 = 0 ...(1)$
And,
$p(-1) - r(-1) = 0$
$i.e. (-1)^3 - 2(-1)^2 + a(-1) - b - (-1 - 6) = 0$
$i.e. -1 - 2 - a - b + 7 = 0$
$i.e -a - b + 4 = 0 ...(2)$
Subtracting equation $(2)$ from equation $(1),$ we get
$4a + 8 = 0$
$a = -2$
From $(2), -(-2) - b + 4 = 0$
$b = 6$

View full question & answer→

MCQ 191 Mark

Ifn $x + a$ is a factor of $x^4 - a^2x^2 + 3x - 6a,$ then $a$ is:

✓
$0$
B
$-1$
C
$1$
D
$2$

Answer

Correct option: A.

$0$

$x + a$ is a factor of polynomial $f(x) = x^4 - a^2x^2 + 3x - 6a,$
Then at $x = -a, p(x) = 0$
$\Rightarrow (-a)^4 - a^2(-a)^2 + 3(-a) - 6a = 0$
$\Rightarrow a^4 - a^4 - 3a - 6a = 0$
$\Rightarrow -9a = 0$
$\Rightarrow a = 0$

View full question & answer→

MCQ 201 Mark

$(x + 1)$ is a factor of $x^n + 1 $only if:

✓
$n$ is an odd integer
B
$n$ is an even integer
C
$n$ is a negative integer
D
$n$ is a positive integer

Answer

Correct option: A.

$n$ is an odd integer

If $x + 1$ is a factor of $x^n + $1,
then, at $x = -1, x^n + 1 = 0$
$(-1)^n + 1 = 0$
$(-1)^n = -1$
$(-1)^n$ will be equal to $-1$ if and only if $n$ is an odd integer.
If n is even, then $(-1)^n = 1$
So, $n$ should be an odd integer.

View full question & answer→

MATHS M.C.Q

Test yourself on this topic