Polynomials and Their Properties: MCQ test, Set 9

Download the PDF (with key sheet and solutions, 92.7 KB)

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

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

Answer and solution

Answer: (B) a monomial

$-4x^{3}y$ has one term, so it is a monomial.
The degree of the monomial $8xy^{2}z$ is
A $2$ B $3$ C $5$ D $4$

Answer and solution

Answer: (D) $4$

Degree = sum of the exponents $= 1 + 2 + 1 = 4$
The coefficient of $x^{2}$ in $3 - 6x^{2} + 3x^{3}$ is
A $3$ B $-6$ C $2$ D $6$

Answer and solution

Answer: (B) $-6$

The term with $x^{2}$ is $-6x^{2}$ , so the coefficient is $-6$ .
$5\sqrt{5}$ is a
A Linear polynomial B Quadratic polynomial C Constant polynomial D Zero polynomial

Answer and solution

Answer: (C) Constant polynomial

$5\sqrt{5} = 5\sqrt{5} \times x^0$ has degree $0$ : a constant polynomial.
Identify the polynomial among these.
A $3x^{5} - 2\sqrt{5}x^{3}$ B $\displaystyle -3 - \frac{4}{x}$ C $4\sqrt{2}x - 3\sqrt{x}$ D $\displaystyle 2x^{\frac{3}{2}} - 5x^{4} + \frac{3}{2}x$

Answer and solution

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

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

Each other option has a negative or fractional power of $x$ .
If $px^{n-2} - qx^{n-3} - rx^{n-4}$ is a bi-quadratic polynomial ( $n \in N$ ), then $n =$
A $8$ B $6$ C $7$ D $4$

Answer and solution

Answer: (B) $6$

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

So the highest power: $n-2 = 4$

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

Answer and solution

Answer: (D) Bi-quadratic equation

The highest power of $x$ is $4$ , so it is a bi-quadratic equation.
If $p(x) = 3x^{3} - x^{2} - 7x - 1$ , then $p(-1) =$
A $-6$ B $-12$ C $2$ D $4$

Answer and solution

Answer: (C) $2$

$p(-1) = 3(-1)^{3} - (-1)^{2} - 7(-1) - 1$

$= -3 - 1 + 7 - 1$

$= 2$
If $3$ is a zero of $p(x) = x^{2} + ax + 6$ , then $a =$
A $-5$ B $5$ C $-7$ D $-4$

Answer and solution

Answer: (A) $-5$

$p(3) = (3)^{2} + a(3) + 6$

$= 9 + 3a + 6$

$= 3a + 15$

$3a + 15 = 0 \Rightarrow a = -5$
The roots of $2x^{2} + x - 6 = 0$ are
A $\displaystyle 2, -\frac{3}{2}$ B $-2, 3$ C $\displaystyle -\frac{1}{2}, \frac{2}{3}$ D $\displaystyle -2, \frac{3}{2}$

Answer and solution

Answer: (D) $\displaystyle -2, \frac{3}{2}$

$2x^{2} + x - 6 = 2x^{2} - 3x + 4x - 6$

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

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

$\displaystyle \Rightarrow x = -2, \frac{3}{2}$
The roots of $2x^{2} + 7x - 11 = 0$ are
A $\displaystyle \frac{-7 \pm \sqrt{137}}{2}$ B $\displaystyle \frac{7 \pm \sqrt{137}}{4}$ C $\displaystyle \frac{7 \pm \sqrt{137}}{2}$ D $\displaystyle \frac{-7 \pm \sqrt{137}}{4}$

Answer and solution

Answer: (D) $\displaystyle \frac{-7 \pm \sqrt{137}}{4}$

$a = 2, b = 7, c = -11$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{49 - 4(2)(-11)}}{2(2)}$

$\displaystyle = \frac{-7 + \sqrt{137}}{4}, \frac{-7 - \sqrt{137}}{4}$
When $2x^{3} + 6x^{2} - 7x - 25$ is divided by $x + 3$ , the quotient is
A $2x^{2} - 7$ B $2x^{2} + 12x + 29$ C $2x^{2} - 3$ D $2x^{2} - 11$

Answer and solution

Answer: (A) $2x^{2} - 7$

$2x^{3} \div x = 2x^{2}$ ; $2x^{2} \times (x + 3) = 2x^{3} + 6x^{2}$ ; subtract: $-7x - 25$

$-7x \div x = -7$ ; $(-7) \times (x + 3) = -7x - 21$ ; subtract: $-4$
The remainder when $x^{3} + 4x^{2} - 5x - 6$ is divided by $x + 1$ is
A $-2$ B $-6$ C $0$ D $2$

Answer and solution

Answer: (D) $2$

Remainder $= f(-1)$

$f(-1) = (-1)^{3} + 4(-1)^{2} - 5(-1) - 6$

$= -1 + 4 + 5 - 6$

$= 2$
Which of the following is a factor of $x^{3} - 7x^{2} + 7x + 15$ ?
A $x + 1$ B $x + 5$ C $x - 1$ D $x - 6$

Answer and solution

Answer: (A) $x + 1$

$f(-1) = (-1)^{3} - 7(-1)^{2} + 7(-1) + 15$

$= -1 - 7 - 7 + 15$

$= 0$

$f(-1) = 0$ , so $x + 1$ is a factor.
$5x^{2} + 22x - 15 =$
A $(x + 3)(5x - 5)$ B $(5x - 3)(x + 5)$ C $(x - 5)(5x + 3)$ D $(x - 3)(5x + 5)$

Answer and solution

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

$5x^{2} + 22x - 15 = 5x^{2} - 3x + 25x - 15$

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

$= (5x - 3)(x + 5)$
What must be added to $-4x^{3} + 12x^{2} - 8x + 15$ so that it is exactly divisible by $x^{2} - 2x + 2$ ?
A $-8x + 7$ B $-8x - 7$ C $-4x + 4$ D $8x + 7$

Answer and solution

Answer: (B) $-8x - 7$

$-4x^{3} \div x^{2} = -4x$ ; $(-4x) \times (x^{2} - 2x + 2) = -4x^{3} + 8x^{2} - 8x$ ; subtract: $4x^{2} + 15$

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

Remainder $= 8x + 7$ , so $-8x - 7$ must be added.
If $x + 2$ and $x - 1$ are factors of $x^3 + ax^2 + bx + 6$ , then
A $a = 2, b = 5$ B $a = -2, b = 5$ C $a = 2, b = -5$ D $a = -2, b = -5$

Answer and solution

Answer: (D) $a = -2, b = -5$

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

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

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

$\Rightarrow a = -2$

Put $a = -2$ in (1): $-4 - b = 1 \Rightarrow b = -5$
A quadratic polynomial has zeros $-1$ and $\displaystyle \frac{5}{2}$ and leading coefficient $4$ . The polynomial is
A $4x^{2} - 6x - 10$ B $4x^{2} + 6x - 10$ C $4x^{2} + 14x + 10$ D $\displaystyle x^{2} - \frac{3}{2}x - \frac{5}{2}$

Answer and solution

Answer: (A) $4x^{2} - 6x - 10$

With zeros $\displaystyle -1, \frac{5}{2}$ and leading coefficient $4$ : $\displaystyle 4(x + 1)\left(x - \frac{5}{2}\right)$

$= 4x^{2} - 6x - 10$
Given that $x + 3$ is a factor, $x^{3} + 11x^{2} + 36x + 36 =$
A $(x + 3)(x^{2} + 14x + 78)$ B $(x + 3)(x^{2} - 8x + 12)$ C $(x + 3)(x^{2} + 8x - 12)$ D $(x + 3)(x^{2} + 8x + 12)$

Answer and solution

Answer: (D) $(x + 3)(x^{2} + 8x + 12)$

$x^{3} \div x = x^{2}$ ; $x^{2} \times (x + 3) = x^{3} + 3x^{2}$ ; subtract: $8x^{2} + 36x + 36$

$8x^{2} \div x = 8x$ ; $8x \times (x + 3) = 8x^{2} + 24x$ ; subtract: $12x + 36$

$12x \div x = 12$ ; $12 \times (x + 3) = 12x + 36$ ; subtract: $0$

Quotient $= x^{2} + 8x + 12$
$x^{3} - x^{2} - 17x - 15 =$
A $(x - 3)(x + 1)(x - 5)$ B $(x + 3)(x - 1)(x - 5)$ C $(x + 3)(x + 1)(x - 5)$ D $(x - 3)(x - 1)(x + 5)$

Answer and solution

Answer: (C) $(x + 3)(x + 1)(x - 5)$

$f(-1) = -1 - 1 + 17 - 15 = 0$

Dividing and splitting the quotient gives $f(x) = (x + 3)(x + 1)(x - 5)$ .
Using synthetic division, the quotient when $5x^{3} - x^{2} + 5$ is divided by $x + 4$ is
A $5x^{2} + 19x + 76$ B $5x^{3} - 21x^{2} + 84x - 331$ C $5x^{2} - 21x + 84$ D $5x - 21$

Answer and solution

Answer: (C) $5x^{2} - 21x + 84$

Second row: $0 \quad -20 \quad 84 \quad -336$

Bottom row: $5 \quad -21 \quad 84 \;\big|\; -331$

Quotient $= 5x^{2} - 21x + 84$
Using synthetic division, the remainder when $2x^{3} + 3x^{2} + 7$ is divided by $x + 1$ is
A $8$ B $6$ C $-8$ D $12$

Answer and solution

Answer: (A) $8$

Second row: $0 \quad -2 \quad -1 \quad 1$

Bottom row: $2 \quad 1 \quad -1 \;\big|\; 8$

Remainder $= 8$
A factor of $8x^{3} + 27$ is
A $x + 3$ B $3x + 2$ C $2x + 3$ D $2x - 3$

Answer and solution

Answer: (C) $2x + 3$

$8x^{3} + 27 = (2x)^{3} + 3^{3}$ and $3$ is odd, so $2x + 3$ is a factor.
When $\displaystyle \frac{1}{8}a^{3} + \frac{3}{16}a^{2}x + \frac{3}{32}ax^{2} + \frac{1}{64}x^{3}$ is divided by $\displaystyle \frac{1}{2}a + \frac{1}{4}x$ , the quotient is
A $\displaystyle \frac{1}{4}a^{2} - \frac{1}{4}ax + \frac{1}{16}x^{2}$ B $\displaystyle \frac{1}{4}a^{2} + \frac{1}{4}ax + \frac{1}{16}x^{2}$ C $\displaystyle \frac{1}{4}a^{2} + \frac{1}{4}ax - \frac{1}{16}x^{2}$ D $\displaystyle \frac{1}{4}a^{2} + \frac{1}{16}x^{2}$

Answer and solution

Answer: (B) $\displaystyle \frac{1}{4}a^{2} + \frac{1}{4}ax + \frac{1}{16}x^{2}$

$\displaystyle \frac{1}{8}a^{3} \div \frac{1}{2}a = \frac{1}{4}a^{2}$ ; $\displaystyle \frac{1}{4}a^{2} \times (\frac{1}{2}a + \frac{1}{4}x) = \frac{1}{8}a^{3} + \frac{1}{16}a^{2}x$ ; subtract: $\displaystyle \frac{1}{8}a^{2}x + \frac{3}{32}ax^{2} + \frac{1}{64}x^{3}$

$\displaystyle \frac{1}{8}a^{2}x \div \frac{1}{2}a = \frac{1}{4}ax$ ; $\displaystyle \frac{1}{4}ax \times (\frac{1}{2}a + \frac{1}{4}x) = \frac{1}{8}a^{2}x + \frac{1}{16}ax^{2}$ ; subtract: $\displaystyle \frac{1}{32}ax^{2} + \frac{1}{64}x^{3}$

$\displaystyle \frac{1}{32}ax^{2} \div \frac{1}{2}a = \frac{1}{16}x^{2}$ ; $\displaystyle \frac{1}{16}x^{2} \times (\frac{1}{2}a + \frac{1}{4}x) = \frac{1}{32}ax^{2} + \frac{1}{64}x^{3}$ ; subtract: $0$
$x^2 + bx + c$ leaves the remainders $-3$ and $25$ when divided by $x - 2$ and $x + 2$ respectively. Then
A $b = -7, c = -7$ B $b = 7, c = -7$ C $b = -7, c = 7$ D $b = 7, c = 7$

Answer and solution

Answer: (C) $b = -7, c = 7$

$f(2) = 4 + 2b + c = -3 \Rightarrow 2b + c = -7$ ... (1)

$f(-2) = 4 - 2b + c = 25 \Rightarrow 2b - c = -21$ ... (2)

$(1) + (2)$ : $4b = -28$

$\Rightarrow b = -7$

Put $b = -7$ in (1): $-14 + c = -7 \Rightarrow c = 7$

Menu

MCQ test, Set 9