Polynomials and Their Properties: MCQ test, Set 3

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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.

$x^{3}y^{2} - 7x^{4}$ is
A a binomial B a polynomial of four terms C a monomial D a trinomial

Answer and solution

Answer: (A) a binomial

$x^{3}y^{2} - 7x^{4}$ has two terms, so it is a binomial.
The degree of the monomial $-6x^{5}y^{2}z^{6}$ is
A $60$ B $6$ C $14$ D $13$

Answer and solution

Answer: (D) $13$

Degree = sum of the exponents $= 5 + 2 + 6 = 13$
The coefficient of $x$ in $-9x - 6x^{5} - 2$ is
A $1$ B $9$ C $-9$ D $-2$

Answer and solution

Answer: (C) $-9$

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

Answer and solution

Answer: (C) Cubic polynomial

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

Answer and solution

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

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

Each other option has a negative or fractional power of $x$ .
If $px^{n-1} - qx^{n-2} - rx^{n-3}$ is a bi-quadratic polynomial ( $n \in N$ ), then $n =$
A $7$ B $6$ C $4$ D $5$

Answer and solution

Answer: (D) $5$

The degree of a bi-quadratic polynomial is $4$ .

So the highest power: $n-1 = 4$

$\Rightarrow n = 5$
The equation $9x^{2} + 6 = 0$ is a
A Linear equation B Quadratic equation C Bi-quadratic equation D Cubic equation

Answer and solution

Answer: (B) Quadratic equation

The highest power of $x$ is $2$ , so it is a quadratic equation.
If $p(x) = x^{2} - 7x - 2$ , then $p(-4) =$
A $-14$ B $10$ C $46$ D $42$

Answer and solution

Answer: (D) $42$

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

$= 16 + 28 - 2$

$= 42$
The zero of the polynomial $9x - 11$ is
A $\displaystyle -\frac{11}{9}$ B $\displaystyle -\frac{9}{11}$ C $\displaystyle \frac{11}{9}$ D $\displaystyle \frac{9}{11}$

Answer and solution

Answer: (C) $\displaystyle \frac{11}{9}$

$\displaystyle 9x - 11 = 0 \Rightarrow 9x = 11 \Rightarrow x = \frac{11}{9}$
The roots of $x^{2} + 16x + 63 = 0$ are
A $9, -7$ B $-9, -7$ C $9, 7$ D $-9, 7$

Answer and solution

Answer: (B) $-9, -7$

$x^{2} + 16x + 63 = x^{2} + 7x + 9x + 63$

$= x(x + 7) + 9(x + 7)$

$= (x + 7)(x + 9)$

$\Rightarrow x = -9, -7$
The discriminant of $x^{2} - 4x + 1 = 0$ is
A $15$ B $20$ C $12$ D $-20$

Answer and solution

Answer: (C) $12$

$\Delta = b^2 - 4ac = (-4)^2 - 4(1)(1) = 12$
When $4x^{3} + 9x^{2} + 7x + 9$ is divided by $x + 1$ , the quotient is
A $4x^{2} + 5x + 9$ B $4x^{2} + 5x + 2$ C $4x^{2} + 13x + 20$ D $4x^{2} - 5x + 2$

Answer and solution

Answer: (B) $4x^{2} + 5x + 2$

$4x^{3} \div x = 4x^{2}$ ; $4x^{2} \times (x + 1) = 4x^{3} + 4x^{2}$ ; subtract: $5x^{2} + 7x + 9$

$5x^{2} \div x = 5x$ ; $5x \times (x + 1) = 5x^{2} + 5x$ ; subtract: $2x + 9$

$2x \div x = 2$ ; $2 \times (x + 1) = 2x + 2$ ; subtract: $7$
If $2x^{3} + 2x^{2} + ax + 8$ leaves the remainder $-6$ when divided by $x + 1$ , then $a =$
A $14$ B $2$ C $-18$ D $8$

Answer and solution

Answer: (A) $14$

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

$= -2 + 2 - a + 8$

$= -a + 8$

$-a + 8 = -6 \Rightarrow a = 14$
The remainder when $3x^{3} - x^{2} + 4x - 5$ is divided by $x + 1$ is
A $-5$ B $1$ C $13$ D $-13$

Answer and solution

Answer: (D) $-13$

Remainder $= f(-1)$

$f(-1) = 3(-1)^{3} - (-1)^{2} + 4(-1) - 5$

$= -3 - 1 - 4 - 5$

$= -13$
Which of the following is a factor of $x^{3} + 4x^{2} + x - 6$ ?
A $x - 4$ B $x - 2$ C $x + 2$ D $x + 1$

Answer and solution

Answer: (C) $x + 2$

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

$= -8 + 16 - 2 - 6$

$= 0$

$f(-2) = 0$ , so $x + 2$ is a factor.
$12x^{2} - 53x + 56 =$
A $(4x - 8)(3x - 7)$ B $(4x + 7)(3x + 8)$ C $(4x - 7)(3x - 8)$ D $(4x + 8)(3x + 7)$

Answer and solution

Answer: (C) $(4x - 7)(3x - 8)$

$12x^{2} - 53x + 56 = 12x^{2} - 21x - 32x + 56$

$= 3x(4x - 7) - 8(4x - 7)$

$= (4x - 7)(3x - 8)$
What must be added to $x^{3} - 6x^{2} - 3x + 1$ so that it is exactly divisible by $x - 2$ ?
A $-21$ B $21$ C $-25$ D $25$

Answer and solution

Answer: (B) $21$

Remainder $= f(2)$ (remainder theorem)

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

$= 8 - 24 - 6 + 1$

$= -21$

The remainder is $-21$ , so $21$ must be added.
If $x + 2$ and $x + 1$ are factors of $x^3 + ax^2 + bx + 8$ , then
A $a = 7, b = 14$ B $a = -7, b = -14$ C $a = 7, b = -14$ D $a = -7, b = 14$

Answer and solution

Answer: (A) $a = 7, b = 14$

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

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

$(1) - (2)$ : $a = 7$

$\Rightarrow a = 7$

Put $a = 7$ in (1): $14 - b = 0 \Rightarrow b = 14$
A cubic polynomial has zeros $4$ , $-4$ and $3$ and leading coefficient $2$ . The polynomial is
A $2x^{3} - 6x^{2} - 32x + 96$ B $x^{3} - 3x^{2} - 16x + 48$ C $2x^{3} + 6x^{2} - 32x - 96$ D $2x^{3} - 22x^{2} + 80x - 96$

Answer and solution

Answer: (A) $2x^{3} - 6x^{2} - 32x + 96$

With zeros $4, -4, 3$ and leading coefficient $2$ : $2(x - 4)(x + 4)(x - 3)$

$= 2x^{3} - 6x^{2} - 32x + 96$
Given that $x - 3$ is a factor, $x^{3} - 2x^{2} - 23x + 60 =$
A $(x - 3)(x^{2} - x - 20)$ B $(x - 3)(x^{2} + x + 20)$ C $(x - 3)(x^{2} - 5x - 8)$ D $(x - 3)(x^{2} + x - 20)$

Answer and solution

Answer: (D) $(x - 3)(x^{2} + x - 20)$

$x^{3} \div x = x^{2}$ ; $x^{2} \times (x - 3) = x^{3} - 3x^{2}$ ; subtract: $x^{2} - 23x + 60$

$x^{2} \div x = x$ ; $x \times (x - 3) = x^{2} - 3x$ ; subtract: $-20x + 60$

$-20x \div x = -20$ ; $(-20) \times (x - 3) = -20x + 60$ ; subtract: $0$

Quotient $= x^{2} + x - 20$
$x^{3} - 4x^{2} - 11x + 30 =$
A $(x - 3)(x + 5)(x + 2)$ B $(x + 3)(x + 5)(x - 2)$ C $(x + 3)(x - 5)(x - 2)$ D $(x - 3)(x - 5)(x - 2)$

Answer and solution

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

$f(2) = 8 - 16 - 22 + 30 = 0$

Dividing and splitting the quotient gives $f(x) = (x + 3)(x - 5)(x - 2)$ .
Using synthetic division, the quotient when $3x^{3} + 6x^{2} + 4$ is divided by $x + 4$ is
A $3x^{2} - 6x + 24$ B $3x - 6$ C $3x^{2} + 18x + 72$ D $3x^{3} - 6x^{2} + 24x - 92$

Answer and solution

Answer: (A) $3x^{2} - 6x + 24$

Second row: $0 \quad -12 \quad 24 \quad -96$

Bottom row: $3 \quad -6 \quad 24 \;\big|\; -92$

Quotient $= 3x^{2} - 6x + 24$
A factor of $x^{3} + 64$ is
A $4x + 1$ B $x + 64$ C $x - 4$ D $x + 4$

Answer and solution

Answer: (D) $x + 4$

$x^{3} + 64 = x^{3} + 4^{3}$ and $3$ is odd, so $x + 4$ is a factor.
When $\displaystyle \frac{27}{8}a^{3} - \frac{81}{8}a^{2}x + \frac{81}{8}ax^{2} - \frac{27}{8}x^{3}$ is divided by $\displaystyle \frac{3}{2}a - \frac{3}{2}x$ , the quotient is
A $\displaystyle \frac{9}{4}a^{2} + \frac{9}{2}ax + \frac{9}{4}x^{2}$ B $\displaystyle \frac{9}{4}a^{2} - \frac{9}{2}ax + \frac{9}{4}x^{2}$ C $\displaystyle \frac{9}{4}a^{2} + \frac{9}{4}x^{2}$ D $\displaystyle \frac{9}{4}a^{2} - \frac{9}{2}ax - \frac{9}{4}x^{2}$

Answer and solution

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

$\displaystyle \frac{27}{8}a^{3} \div \frac{3}{2}a = \frac{9}{4}a^{2}$ ; $\displaystyle \frac{9}{4}a^{2} \times (\frac{3}{2}a - \frac{3}{2}x) = \frac{27}{8}a^{3} - \frac{27}{8}a^{2}x$ ; subtract: $\displaystyle -\frac{27}{4}a^{2}x + \frac{81}{8}ax^{2} - \frac{27}{8}x^{3}$

$\displaystyle -\frac{27}{4}a^{2}x \div \frac{3}{2}a = -\frac{9}{2}ax$ ; $\displaystyle (-\frac{9}{2}ax) \times (\frac{3}{2}a - \frac{3}{2}x) = -\frac{27}{4}a^{2}x + \frac{27}{4}ax^{2}$ ; subtract: $\displaystyle \frac{27}{8}ax^{2} - \frac{27}{8}x^{3}$

$\displaystyle \frac{27}{8}ax^{2} \div \frac{3}{2}a = \frac{9}{4}x^{2}$ ; $\displaystyle \frac{9}{4}x^{2} \times (\frac{3}{2}a - \frac{3}{2}x) = \frac{27}{8}ax^{2} - \frac{27}{8}x^{3}$ ; subtract: $0$
$x^2 + bx + c$ leaves the remainders $4$ and $19$ when divided by $x - 3$ and $x + 2$ respectively. Then
A $b = -4, c = 7$ B $b = 4, c = -7$ C $b = 4, c = 7$ D $b = -4, c = -7$

Answer and solution

Answer: (A) $b = -4, c = 7$

$f(3) = 9 + 3b + c = 4 \Rightarrow 3b + c = -5$ ... (1)

$f(-2) = 4 - 2b + c = 19 \Rightarrow 2b - c = -15$ ... (2)

$(1) + (2)$ : $5b = -20$

$\Rightarrow b = -4$

Put $b = -4$ in (1): $-12 + c = -5 \Rightarrow c = 7$

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MCQ test, Set 3