Polynomials and Their Properties: MCQ test, Set 1

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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 of the following is a monomial?
A $2x^{4}$ B $\displaystyle -9x^{\frac{1}{3}}y^{2}$ C $\displaystyle -\frac{3}{x^{3}}$ D $\displaystyle -6x^{\frac{2}{3}}y^{3}$

Answer and solution

Answer: (A) $2x^{4}$

$2x^{4}$ : the powers of $x$ are $4$ , non-negative integers, so it is a monomial.

Each of the others has a negative or fractional power.
The degree of the monomial $2x^{6}y^{6}z^{5}$ is
A $6$ B $180$ C $18$ D $17$

Answer and solution

Answer: (D) $17$

Degree = sum of the exponents $= 6 + 6 + 5 = 17$
The coefficient of $x^{2}$ in $-4x + 4 - 2x^{3} + 8x^{5} + 5x^{2}$ is
A $2$ B $-2$ C $5$ D $-5$

Answer and solution

Answer: (C) $5$

The term with $x^{2}$ is $5x^{2}$ , so the coefficient is $5$ .
$3 - 2x^{2}$ is a
A Linear polynomial B Bi-quadratic polynomial C Cubic polynomial D Quadratic polynomial

Answer and solution

Answer: (D) Quadratic polynomial

The highest power of $x$ is $2$ , so it is a quadratic polynomial.
Among the following, the polynomial is
A $-6x^{4} - 2\sqrt{x}$ B $\displaystyle \frac{5}{3}x^{3} + 2\sqrt{5}x^{4} - 9$ C $-8x - 4\sqrt{x}$ D $-3\sqrt{5}x^{4} + 4x^{3} - 7x^{-3}$

Answer and solution

Answer: (B) $\displaystyle \frac{5}{3}x^{3} + 2\sqrt{5}x^{4} - 9$

In $\displaystyle \frac{5}{3}x^{3} + 2\sqrt{5}x^{4} - 9$ , all powers of $x$ are whole numbers, so it is a polynomial.

Each other option has a negative or fractional power of $x$ .
If $px^{n-10} - qx^{n-11} - rx^{n-12}$ is a cubic polynomial ( $n \in N$ ), then $n =$
A $3$ B $13$ C $15$ D $14$

Answer and solution

Answer: (B) $13$

The degree of a cubic polynomial is $3$ .

So the highest power: $n-10 = 3$

$\Rightarrow n = 13$
The equation $-5x^{3} - 7x^{2} + 7x = -5$ is a
A Quadratic equation B Linear equation C Cubic equation D Bi-quadratic equation

Answer and solution

Answer: (C) Cubic equation

The highest power of $x$ is $3$ , so it is a cubic equation.
If $p(x) = 3x^{2} + 3x + 8$ , then $p(-4) =$
A $68$ B $-52$ C $44$ D $28$

Answer and solution

Answer: (C) $44$

$p(-4) = 3(-4)^{2} + 3(-4) + 8$

$= 48 - 12 + 8$

$= 44$
The zero of the polynomial $-8x - 6$ is
A $\displaystyle -\frac{3}{4}$ B $\displaystyle -\frac{4}{3}$ C $\displaystyle \frac{4}{3}$ D $\displaystyle \frac{3}{4}$

Answer and solution

Answer: (A) $\displaystyle -\frac{3}{4}$

$\displaystyle -8x - 6 = 0 \Rightarrow -8x = 6 \Rightarrow x = -\frac{3}{4}$
The roots of $x^{2} + 12x + 35 = 0$ are
A $7, 5$ B $7, -5$ C $-7, -5$ D $-7, 5$

Answer and solution

Answer: (C) $-7, -5$

$x^{2} + 12x + 35 = x^{2} + 5x + 7x + 35$

$= x(x + 5) + 7(x + 5)$

$= (x + 5)(x + 7)$

$\Rightarrow x = -7, -5$
The discriminant of $x^{2} - 10x - 8 = 0$ is
A $132$ B $-68$ C $68$ D $108$

Answer and solution

Answer: (A) $132$

$\Delta = b^2 - 4ac = (-10)^2 - 4(1)(-8) = 132$
When $x^{3} - 6x^{2} - 6x + 7$ is divided by $x + 1$ , the quotient is
A $x^{2} + 7x + 1$ B $x^{2} - 5x - 11$ C $x^{2} - 7x + 1$ D $x^{2} - 7x + 7$

Answer and solution

Answer: (C) $x^{2} - 7x + 1$

$x^{3} \div x = x^{2}$ ; $x^{2} \times (x + 1) = x^{3} + x^{2}$ ; subtract: $-7x^{2} - 6x + 7$

$-7x^{2} \div x = -7x$ ; $(-7x) \times (x + 1) = -7x^{2} - 7x$ ; subtract: $x + 7$

$x \div x = 1$ ; $1 \times (x + 1) = x + 1$ ; subtract: $6$
The remainder when $x^{4} + 2x^{3} - 11x^{2} - 14x + 24$ is divided by $x^{2} - 4$ is
A $6x + 4$ B $6x - 4$ C $-6x + 4$ D $-6x - 4$

Answer and solution

Answer: (D) $-6x - 4$

$x^{4} \div x^{2} = x^{2}$ ; $x^{2} \times (x^{2} - 4) = x^{4} - 4x^{2}$ ; subtract: $2x^{3} - 7x^{2} - 14x + 24$

$2x^{3} \div x^{2} = 2x$ ; $2x \times (x^{2} - 4) = 2x^{3} - 8x$ ; subtract: $-7x^{2} - 6x + 24$

$-7x^{2} \div x^{2} = -7$ ; $(-7) \times (x^{2} - 4) = -7x^{2} + 28$ ; subtract: $-6x - 4$

Remainder $= -6x - 4$
The remainder when $3x^{3} - x^{2} - 4x - 6$ is divided by $x + 3$ is
A $-6$ B $84$ C $-84$ D $54$

Answer and solution

Answer: (C) $-84$

Remainder $= f(-3)$

$f(-3) = 3(-3)^{3} - (-3)^{2} - 4(-3) - 6$

$= -81 - 9 + 12 - 6$

$= -84$
Which of the following is a factor of $x^{3} + 3x^{2} - 10x - 24$ ?
A $x - 4$ B $x - 1$ C $x + 1$ D $x + 4$

Answer and solution

Answer: (D) $x + 4$

$f(-4) = (-4)^{3} + 3(-4)^{2} - 10(-4) - 24$

$= -64 + 48 + 40 - 24$

$= 0$

$f(-4) = 0$ , so $x + 4$ is a factor.
$6x^{2} - 17x - 45 =$
A $(2x - 5)(3x + 9)$ B $(3x + 5)(2x - 9)$ C $(2x + 9)(3x - 5)$ D $(2x + 5)(3x - 9)$

Answer and solution

Answer: (B) $(3x + 5)(2x - 9)$

$6x^{2} - 17x - 45 = 6x^{2} + 10x - 27x - 45$

$= 2x(3x + 5) - 9(3x + 5)$

$= (3x + 5)(2x - 9)$
What must be subtracted from $x^{3} + 2x^{2} + 8x + 1$ so that it is exactly divisible by $x - 3$ ?
A $70$ B $-70$ C $-32$ D $32$

Answer and solution

Answer: (A) $70$

Remainder $= f(3)$ (remainder theorem)

$f(3) = (3)^{3} + 2(3)^{2} + 8(3) + 1$

$= 27 + 18 + 24 + 1$

$= 70$

The remainder is $70$ , so $70$ must be subtracted.
If $x - 3$ and $x - 1$ are factors of $x^3 + ax^2 + bx - 12$ , then
A $a = 8, b = 19$ B $a = -8, b = 19$ C $a = 8, b = -19$ D $a = -8, b = -19$

Answer and solution

Answer: (B) $a = -8, b = 19$

$f(3) = 0 \Rightarrow 3a + b = -5$ ... (1)

$f(1) = 0 \Rightarrow a + b = 11$ ... (2)

$(1) - (2)$ : $2a = -16$

$\Rightarrow a = -8$

Put $a = -8$ in (1): $-24 + b = -5 \Rightarrow b = 19$
A quadratic polynomial has zeros $-2$ and $\displaystyle \frac{5}{2}$ and leading coefficient $2$ . The polynomial is
A $2x^{2} + 9x + 10$ B $2x^{2} + x - 10$ C $2x^{2} - x - 10$ D $\displaystyle x^{2} - \frac{1}{2}x - 5$

Answer and solution

Answer: (C) $2x^{2} - x - 10$

With zeros $\displaystyle -2, \frac{5}{2}$ and leading coefficient $2$ : $\displaystyle 2(x + 2)\left(x - \frac{5}{2}\right)$

$= 2x^{2} - x - 10$
Given that $x - 5$ is a factor, $x^{3} - 8x^{2} + 11x + 20 =$
A $(x - 5)(x^{2} - 3x - 4)$ B $(x - 5)(x^{2} - 3x + 4)$ C $(x - 5)(x^{2} + 3x - 4)$ D $(x - 5)(x^{2} - 13x + 76)$

Answer and solution

Answer: (A) $(x - 5)(x^{2} - 3x - 4)$

$x^{3} \div x = x^{2}$ ; $x^{2} \times (x - 5) = x^{3} - 5x^{2}$ ; subtract: $-3x^{2} + 11x + 20$

$-3x^{2} \div x = -3x$ ; $(-3x) \times (x - 5) = -3x^{2} + 15x$ ; subtract: $-4x + 20$

$-4x \div x = -4$ ; $(-4) \times (x - 5) = -4x + 20$ ; subtract: $0$

Quotient $= x^{2} - 3x - 4$
$x^{3} - 6x^{2} - x + 30 =$
A $(x + 3)(x - 2)(x + 5)$ B $(x - 3)(x + 2)(x - 5)$ C $(x + 3)(x + 2)(x - 5)$ D $(x - 3)(x - 2)(x - 5)$

Answer and solution

Answer: (B) $(x - 3)(x + 2)(x - 5)$

$f(-2) = -8 - 24 + 2 + 30 = 0$

Dividing and splitting the quotient gives $f(x) = (x - 3)(x + 2)(x - 5)$ .
Using synthetic division, the quotient when $x^{3} - x - 6$ is divided by $x - 1$ is
A $x^{2} - x$ B $x^{2} + x$ C $x^{3} + x^{2} - 6$ D $x$

Answer and solution

Answer: (B) $x^{2} + x$

Second row: $0 \quad 1 \quad 1 \quad 0$

Bottom row: $1 \quad 1 \quad 0 \;\big|\; -6$

Quotient $= x^{2} + x$
$27y^{3} + 8 = (3y + 2) \times$
A $9y^{2} - 6y + 2$ B $y^{2} - 2y + 4$ C $9y^{2} + 6y + 4$ D $9y^{2} - 6y + 4$

Answer and solution

Answer: (D) $9y^{2} - 6y + 4$

$27y^{3} + 8 = (3y + 2)(9y^{2} - 6y + 4)$

Signs alternate; the powers of $2$ go up as the powers of $3y$ go down.
When $8a^{3} + 24a^{2}x + 24ax^{2} + 8x^{3}$ is divided by $2a + 2x$ , the quotient is
A $4a^{2} + 8ax + 4x^{2}$ B $4a^{2} + 4x^{2}$ C $4a^{2} + 8ax - 4x^{2}$ D $4a^{2} - 8ax + 4x^{2}$

Answer and solution

Answer: (A) $4a^{2} + 8ax + 4x^{2}$

$8a^{3} \div 2a = 4a^{2}$ ; $4a^{2} \times (2a + 2x) = 8a^{3} + 8a^{2}x$ ; subtract: $16a^{2}x + 24ax^{2} + 8x^{3}$

$16a^{2}x \div 2a = 8ax$ ; $8ax \times (2a + 2x) = 16a^{2}x + 16ax^{2}$ ; subtract: $8ax^{2} + 8x^{3}$

$8ax^{2} \div 2a = 4x^{2}$ ; $4x^{2} \times (2a + 2x) = 8ax^{2} + 8x^{3}$ ; subtract: $0$
$x^2 + bx + c$ leaves the remainders $-9$ and $-10$ when divided by $x + 1$ and $x + 2$ respectively. Then
A $b = -4, c = -6$ B $b = 4, c = 6$ C $b = -4, c = 6$ D $b = 4, c = -6$

Answer and solution

Answer: (D) $b = 4, c = -6$

$f(-1) = 1 - b + c = -9 \Rightarrow b - c = 10$ ... (1)

$f(-2) = 4 - 2b + c = -10 \Rightarrow 2b - c = 14$ ... (2)

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

$\Rightarrow b = 4$

Put $b = 4$ in (1): $4 - c = 10 \Rightarrow c = -6$

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MCQ test, Set 1