Polynomials and Their Properties: MCQ test, Set 5

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

Which one of these is NOT a monomial?
A $\displaystyle -\frac{7}{5}x^{5}y^{3}$ B $\displaystyle 6x^{\frac{1}{3}}y^{3}$ C $2x^{5}y^{4}$ D $-8x^{2}y^{2}$

Answer and solution

Answer: (B) $\displaystyle 6x^{\frac{1}{3}}y^{3}$

The power $\displaystyle \frac{1}{3}$ of $x$ is not a whole number.

So $\displaystyle 6x^{\frac{1}{3}}y^{3}$ is not a monomial; the others have only non-negative integer powers.
The degree of the monomial $-2x^{2}y^{3}z^{3}$ is
A $8$ B $3$ C $18$ D $9$

Answer and solution

Answer: (A) $8$

Degree = sum of the exponents $= 2 + 3 + 3 = 8$
The coefficient of $x^{2}$ in $9x^{2} + 5x^{3} + 2 - 9x^{4}$ is
A $2$ B $-9$ C $9$ D $5$

Answer and solution

Answer: (C) $9$

The term with $x^{2}$ is $9x^{2}$ , so the coefficient is $9$ .
$-7x + 6 + 9x^{4} - x^{3} - 7x^{2}$ is a
A Quadratic polynomial B Bi-quadratic polynomial C Linear polynomial D Cubic polynomial

Answer and solution

Answer: (B) Bi-quadratic polynomial

The highest power of $x$ is $4$ , so it is a bi-quadratic polynomial.
Which one of these expressions is a polynomial?
A $\displaystyle 8x^{\frac{1}{3}} - 2x^{3} - \frac{5}{3}x$ B $\displaystyle -2\sqrt{3}x^{2} + \frac{5}{3}x + 7\sqrt{x}$ C $2\sqrt{x} + 6$ D $-3\sqrt{5}x^{5} + 5$

Answer and solution

Answer: (D) $-3\sqrt{5}x^{5} + 5$

In $-3\sqrt{5}x^{5} + 5$ , all powers of $x$ are whole numbers, so it is a polynomial.

Each other option has a negative or fractional power of $x$ .
If $5x^{n-9} + 5x^{n-10} - 6x^{n-11}$ is a quadratic polynomial ( $n \in N$ ), then $n =$
A $11$ B $12$ C $2$ D $13$

Answer and solution

Answer: (A) $11$

The degree of a quadratic polynomial is $2$ .

So the highest power: $n-9 = 2$

$\Rightarrow n = 11$
The equation $-4x^{4} + 7x^{2} + 6x + 7 = 0$ is a
A Cubic equation B Bi-quadratic equation C Quadratic equation D Linear equation

Answer and solution

Answer: (B) Bi-quadratic equation

The highest power of $x$ is $4$ , so it is a bi-quadratic equation.
If $p(x) = 4x^{2} + 7x + 10$ , then $p(-3) =$
A $-47$ B $67$ C $25$ D $5$

Answer and solution

Answer: (C) $25$

$p(-3) = 4(-3)^{2} + 7(-3) + 10$

$= 36 - 21 + 10$

$= 25$
If $p(x) = -2x^{3} - 5x^{2} - 4x - 4$ , then $p(-3) =$
A $17$ B $-115$ C $-7$ D $107$

Answer and solution

Answer: (A) $17$

$p(-3) = -2(-3)^{3} - 5(-3)^{2} - 4(-3) - 4$

$= 54 - 45 + 12 - 4$

$= 17$
The zero of the polynomial $4x + 11$ is
A $\displaystyle -\frac{11}{4}$ B $\displaystyle \frac{4}{11}$ C $\displaystyle -\frac{4}{11}$ D $\displaystyle \frac{11}{4}$

Answer and solution

Answer: (A) $\displaystyle -\frac{11}{4}$

$\displaystyle 4x + 11 = 0 \Rightarrow 4x = -11 \Rightarrow x = -\frac{11}{4}$
The roots of $15x^{2} + 13x + 2 = 0$ are
A $-2, -1$ B $\displaystyle \frac{2}{3}, \frac{1}{5}$ C $\displaystyle -\frac{2}{3}, -\frac{1}{5}$ D $\displaystyle -\frac{3}{2}, -5$

Answer and solution

Answer: (C) $\displaystyle -\frac{2}{3}, -\frac{1}{5}$

$15x^{2} + 13x + 2 = 15x^{2} + 3x + 10x + 2$

$= 3x(5x + 1) + 2(5x + 1)$

$= (5x + 1)(3x + 2)$

$\displaystyle \Rightarrow x = -\frac{2}{3}, -\frac{1}{5}$
The discriminant of $3x^{2} + 10x - 11 = 0$ is
A $-32$ B $32$ C $133$ D $232$

Answer and solution

Answer: (D) $232$

$\Delta = b^2 - 4ac = 10^2 - 4(3)(-11) = 232$
A polynomial of degree $6$ is divided by a polynomial of degree $4$ . The degree of the quotient is
A $4$ B $24$ C $10$ D $2$

Answer and solution

Answer: (D) $2$

$f(x) = g(x)q(x) + r(x)$ : degree of $q(x)$ = degree of $f(x)$ $-$ degree of $g(x)$

$= 6 - 4 = 2$
If $x^{3} + ax^{2} - 5x - 6$ leaves the remainder $-2$ when divided by $x - 1$ , then $a =$
A $0$ B $8$ C $12$ D $10$

Answer and solution

Answer: (B) $8$

$f(1) = (1)^{3} + a(1)^{2} - 5(1) - 6$

$= 1 + a - 5 - 6$

$= a - 10$

$a - 10 = -2 \Rightarrow a = 8$
If $16x^{4} + 4x^{3} - ax^{2} + 3x - 4$ is divisible by $1 - 2x$ , then $a =$
A $19$ B $4$ C $-4$ D $-20$

Answer and solution

Answer: (C) $-4$

$\displaystyle 1 - 2x = 0 \Rightarrow x = \frac{1}{2}$ ; $\displaystyle f\left(\frac{1}{2}\right) = 0$

$\displaystyle f\left(\frac{1}{2}\right) = 16\left(\frac{1}{2}\right)^{4} + 4\left(\frac{1}{2}\right)^{3} - a\left(\frac{1}{2}\right)^{2} + 3\left(\frac{1}{2}\right) - 4$

$\displaystyle = 1 + \frac{1}{2} - \frac{1}{4}a + \frac{3}{2} - 4$

$\displaystyle = -\frac{1}{4}a - 1$

$\displaystyle -\frac{1}{4}a - 1 = 0 \Rightarrow a = -4$
$5x^{2} + 37x - 24 =$
A $(x + 3)(5x - 8)$ B $(5x - 3)(x + 8)$ C $(x - 3)(5x + 8)$ D $(x - 8)(5x + 3)$

Answer and solution

Answer: (B) $(5x - 3)(x + 8)$

$5x^{2} + 37x - 24 = 5x^{2} - 3x + 40x - 24$

$= x(5x - 3) + 8(5x - 3)$

$= (5x - 3)(x + 8)$
What must be subtracted from $x^{3} - 2x^{2} + x - 5$ so that it is exactly divisible by $x - 4$ ?
A $31$ B $-105$ C $105$ D $-31$

Answer and solution

Answer: (A) $31$

Remainder $= f(4)$ (remainder theorem)

$f(4) = (4)^{3} - 2(4)^{2} + (4) - 5$

$= 64 - 32 + 4 - 5$

$= 31$

The remainder is $31$ , so $31$ must be subtracted.
If $x - 2$ and $x - 3$ are factors of $x^3 + ax^2 + bx + 6$ , then
A $a = 4, b = 1$ B $a = 4, b = -1$ C $a = -4, b = -1$ D $a = -4, b = 1$

Answer and solution

Answer: (D) $a = -4, b = 1$

$f(2) = 0 \Rightarrow 2a + b = -7$ ... (1)

$f(3) = 0 \Rightarrow 3a + b = -11$ ... (2)

$(1) - (2)$ : $-a = 4$

$\Rightarrow a = -4$

Put $a = -4$ in (1): $-8 + b = -7 \Rightarrow b = 1$
A quadratic polynomial has zeros $3$ and $\displaystyle \frac{5}{3}$ and leading coefficient $3$ . The polynomial is
A $3x^{2} + 14x + 15$ B $3x^{2} - 14x + 15$ C $3x^{2} - 4x - 15$ D $\displaystyle x^{2} - \frac{14}{3}x + 5$

Answer and solution

Answer: (B) $3x^{2} - 14x + 15$

With zeros $\displaystyle 3, \frac{5}{3}$ and leading coefficient $3$ : $\displaystyle 3(x - 3)\left(x - \frac{5}{3}\right)$

$= 3x^{2} - 14x + 15$
Given that $x - 4$ is a factor, $x^{3} - 5x^{2} + 2x + 8 =$
A $(x - 4)(x^{2} - 9x + 38)$ B $(x - 4)(x^{2} - x + 2)$ C $(x - 4)(x^{2} + x - 2)$ D $(x - 4)(x^{2} - x - 2)$

Answer and solution

Answer: (D) $(x - 4)(x^{2} - x - 2)$

$x^{3} \div x = x^{2}$ ; $x^{2} \times (x - 4) = x^{3} - 4x^{2}$ ; subtract: $-x^{2} + 2x + 8$

$-x^{2} \div x = -x$ ; $(-x) \times (x - 4) = -x^{2} + 4x$ ; subtract: $-2x + 8$

$-2x \div x = -2$ ; $(-2) \times (x - 4) = -2x + 8$ ; subtract: $0$

Quotient $= x^{2} - x - 2$
$x^{3} + 8x^{2} + 19x + 12 =$
A $(x + 3)(x + 4)(x + 1)$ B $(x + 3)(x - 4)(x + 1)$ C $(x - 3)(x - 4)(x - 1)$ D $(x - 3)(x + 4)(x + 1)$

Answer and solution

Answer: (A) $(x + 3)(x + 4)(x + 1)$

$f(-1) = -1 + 8 - 19 + 12 = 0$

Dividing and splitting the quotient gives $f(x) = (x + 3)(x + 4)(x + 1)$ .
Using synthetic division, the quotient when $x^{3} - 2x + 4$ is divided by $x - 4$ is
A $x^{3} + 4x^{2} + 14x + 60$ B $x + 2$ C $x^{2} - 4x + 14$ D $x^{2} + 4x + 14$

Answer and solution

Answer: (D) $x^{2} + 4x + 14$

Second row: $0 \quad 4 \quad 16 \quad 56$

Bottom row: $1 \quad 4 \quad 14 \;\big|\; 60$

Quotient $= x^{2} + 4x + 14$
$22^{93} - 13^{93}$ is exactly divisible by
A $13$ B $9$ C $22$ D $35$

Answer and solution

Answer: (B) $9$

$n = 93$ : $x - a$ is a factor of $x^n - a^n$ (n odd).

So $22 - 13 = 9$ divides it.
When $\displaystyle \frac{1}{8}a^{3} - \frac{9}{4}a^{2}x + \frac{27}{2}ax^{2} - 27x^{3}$ is divided by $\displaystyle \frac{1}{2}a - 3x$ , the quotient is
A $\displaystyle \frac{1}{4}a^{2} + 3ax + 9x^{2}$ B $\displaystyle \frac{1}{4}a^{2} + 9x^{2}$ C $\displaystyle \frac{1}{4}a^{2} - 3ax + 9x^{2}$ D $\displaystyle \frac{1}{4}a^{2} - 3ax - 9x^{2}$

Answer and solution

Answer: (C) $\displaystyle \frac{1}{4}a^{2} - 3ax + 9x^{2}$

$\displaystyle \frac{1}{8}a^{3} \div \frac{1}{2}a = \frac{1}{4}a^{2}$ ; $\displaystyle \frac{1}{4}a^{2} \times (\frac{1}{2}a - 3x) = \frac{1}{8}a^{3} - \frac{3}{4}a^{2}x$ ; subtract: $\displaystyle -\frac{3}{2}a^{2}x + \frac{27}{2}ax^{2} - 27x^{3}$

$\displaystyle -\frac{3}{2}a^{2}x \div \frac{1}{2}a = -3ax$ ; $\displaystyle (-3ax) \times (\frac{1}{2}a - 3x) = -\frac{3}{2}a^{2}x + 9ax^{2}$ ; subtract: $\displaystyle \frac{9}{2}ax^{2} - 27x^{3}$

$\displaystyle \frac{9}{2}ax^{2} \div \frac{1}{2}a = 9x^{2}$ ; $\displaystyle 9x^{2} \times (\frac{1}{2}a - 3x) = \frac{9}{2}ax^{2} - 27x^{3}$ ; subtract: $0$
$x^2 + bx + c$ leaves the remainders $-4$ and $5$ when divided by $x - 1$ and $x + 2$ respectively. Then
A $b = 2, c = -3$ B $b = 2, c = 3$ C $b = -2, c = -3$ D $b = -2, c = 3$

Answer and solution

Answer: (C) $b = -2, c = -3$

$f(1) = 1 + b + c = -4 \Rightarrow b + c = -5$ ... (1)

$f(-2) = 4 - 2b + c = 5 \Rightarrow 2b - c = -1$ ... (2)

$(1) + (2)$ : $3b = -6$

$\Rightarrow b = -2$

Put $b = -2$ in (1): $-2 + c = -5 \Rightarrow c = -3$

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MCQ test, Set 5