MCQ 11 Mark
If $x +1$ is a factor of the polynomial $2 x ^2+ kx$, then $k =$
Answer$x+1$ is a factor of $p(x)=2 x^2+k x$
Then, $p(-1)=0$
i.e. $2(-1)^2+k(-1)=0$
$2-k=0$
$k = 2$
View full question & answer→MCQ 21 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
$p + 2r = 0$
AnswerCorrect option: A. $p = r$
Let $f(x)=p x^2+5 x+r$
Now,
If $x-2$ and $x-\frac{1}{2}$ are factors of $f(x)$.
Then at $x=2$ and $x-\frac{1}{2}, f(x)=0$.
So, $f(2)=0, f\left(\frac{1}{2}\right)=0$
$\Rightarrow p (2)^2+5(2)+ r =0$
And, $P \left(\frac{1}{2}\right)^2+5\left(\frac{1}{2}\right)+ r =0$
$\Rightarrow 4 p+r+10=0 \ldots(1)$
And $4 r+p+10=0$
Subtracting equation $(2)$ from $(1),$ we have
$3p - 3r = 0$
$\Rightarrow p = r$
View full question & answer→MCQ 31 Mark
If $x-3$ is a factor of $x^2-a x-15$, then $a=$
Answer$x-3$ is a factor of $x^2-a x-15$,
then at $x=3$,
$x^2-a x-15=0$
i.e. $(3)^2-a(3)-15=0$
$9-3 a-15=0$
$a=-2$
View full question & answer→MCQ 41 Mark
If $x^3+6 x^2+4 x+k$ is exactly divisible by $x+2$, then $k$
AnswerSince, $p(x)=x^3+6 x^2+4 x+k$ 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 51 Mark
If $x - a$ is a factor of $x^3 - 3x^2a + 2a^2x + b$, then the value of $b$ is:
AnswerLet $p(x)=x^3-3 x^2 a+2 a^2 x+b$
$(x-a)$ is a factor of $p(x)$.
So,
$p(a)=0$
$a^3-3 a^2 a+2 a^2 a+b=0$
$a^3-3 a^3+2 a^3+b=0$
$3 a^3-3 a^3+b=0$
$b=0$
View full question & answer→MCQ 61 Mark
The value of $k$ for which $x-1$ is a factor of $4 x^3+3 x^2-4 x+k$, is
AnswerLet $p(x)=4 x^3+3 x^2-4 x+k$
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 71 Mark
If $x - 2$ is a factor of $x^2 + 3ax - 2a$, then $a =$
AnswerLet 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 81 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 $-19$
AnswerCorrect 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 91 Mark
Let $f(x)$ be a polynomial such that $\text{f}\Big(-\frac{1}{2}\Big)=0,$ then a factor of $f(x)$ is:
- A
$2x - 1$
- ✓
$2x + 1$
- C
$x - 1$
- D
$x + 1$
AnswerCorrect option: B. $2x + 1$
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 $2x + 1 = 0, f(x) = 0$
$\Rightarrow (2x + 1)$ is a factor of $f(x)$.
View full question & answer→MCQ 101 Mark
If $x^{51} + 51$ is divided by $x + 1$, the remainder is:
AnswerWhen 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 111 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
${f(x) + g(x)}\ g(x)$
AnswerCorrect 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 $x = 1$
$ (a)\ f(1)g(1) = 0 \times g(1) = 0$
$ (b)\ -f(1) + g(1) = 0 + g(1) = g(1) \neq 0$
$ (c)\ f(1) - g(1) = 0 - g(1) = -g(1) \neq 0$
$ (d)\ {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 121 Mark
$(x + 1)$ is a factor of $x^n + 1$ only if:
AnswerCorrect 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→MCQ 131 Mark
If $x^{140} + 2x^{151}$ + $k$ is divisible by $x + 1,$ then the value of $k$ is:
AnswerLet $p(x) = x^{140} + 2x^{151} + k$
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 141 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
$x + 4$
AnswerCorrect 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 151 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
$b + c + d = a + e$
AnswerCorrect 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 161 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
$-1$
AnswerCorrect option: B. $\frac{2}{5}$
Let $p ( x )=3 x ^3+8 x ^2+8 x +3+5 k$ 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,
$p(x)=3 x^3+3 x^2+3 x+5 x^2+5 x+3+2-2+5 k[\text { Adding }+2,-2 \text { in } p(x)]$
$=3 x\left(x^2+x+1\right)+5\left(x^2+x+1\right)+5 k-2$
$p(x)=\left(x^2+x+1\right)(3 x+5)+5 k-2 \ldots(1)$
From equation $(1),$ we can see if we divide $p(x)$ by $q(x)$, then quotient will be $(3 x+5)$ and remainder will be $(5 k-2)$
But $q(x)$ is a factor of $p(x)$.
So, remainder $=0$
$\Rightarrow 5 k-2=0$
$\Rightarrow k=\frac{2}{5}$
View full question & answer→MCQ 171 Mark
If $(3 x-1)^7=a_7 x^7+a_6 x^6+a_5 x^5+\ldots+a_1 x+a_0$ then $a_7+a_6+a_5+\ldots+a_1+a_0=$
Answer$(3x - 1)^7 = a_7x^7 + a_6x^6 + ... + a_1x + a_{0 ...(1)}$
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 181 Mark
If $x + 2$ is a factor of $x^2 + mx + 14,$ then $m =$
AnswerIf $x + 2$ 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 191 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 $6$
AnswerCorrect 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 201 Mark
If $x + a$ is a factor of $x^4 - a^2x^2 + 3x - 6a$, then $a$ is:
Answer$x + a$ 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→