Polynomials and Their Properties: MCQ test, Set 6

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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.

$3x^{2} - 4x + 6xy^{3}$ is
A a binomial B a monomial C a polynomial of four terms D a trinomial

Answer and solution

Answer: (D) a trinomial

$3x^{2} - 4x + 6xy^{3}$ has three terms, so it is a trinomial.
The degree of $4y^{4} - 4x^{5}y^{3} - 8x^{4} - 4xy^{4} + 2$ is
A $4$ B $8$ C $9$ D $5$

Answer and solution

Answer: (B) $8$

Degrees of the terms: $4, 8, 4, 5, 0$

The greatest is $8$ .
When $p(x) = 7x^{2} - 6x^{3} - 8$ is written in standard form $a_nx^n + \cdots + a_1x + a_0$ , the value of $a_n$ is
A $-8$ B $-6$ C $3$ D $7$

Answer and solution

Answer: (B) $-6$

Standard form: $-6x^{3} + 7x^{2} - 8$

$n = 3$ and $a_n$ is the coefficient of $x^{3}$ : $a_n = -6$
Which of the following is a bi-quadratic polynomial?
A $8 - 9x^{3}$ B $2 - 4x^{2}$ C $-1 + 8x - x^{2} + x^{4}$ D $1 - 7x$

Answer and solution

Answer: (C) $-1 + 8x - x^{2} + x^{4}$

$-1 + 8x - x^{2} + x^{4}$ has highest power $4$ , so it is a bi-quadratic polynomial.
Pick the expression that is NOT a polynomial.
A $\displaystyle -\frac{5}{2}x + 9x^{-3} - \frac{7}{3}$ B $-7x^{2} + 8x - 2$ C $\displaystyle \frac{2}{3}x - 2x^{2}$ D $-8x + 7 + 2\sqrt{5}x^{4}$

Answer and solution

Answer: (A) $\displaystyle -\frac{5}{2}x + 9x^{-3} - \frac{7}{3}$

In $\displaystyle -\frac{5}{2}x + 9x^{-3} - \frac{7}{3}$ , the power $-3$ of $x$ is negative.

So it is not a polynomial. (A coefficient like $\sqrt{2}$ or $\displaystyle \frac{2}{3}$ is allowed.)
If $(a + 3)y^2 + (b - 3a)y + (c + 3b)$ is a zero polynomial, then
A $a = -3, b = -9, c = -27$ B $a = -3, b = -9, c = 27$ C $a = -3, b = 9, c = -27$ D $a = 3, b = 9, c = -27$

Answer and solution

Answer: (B) $a = -3, b = -9, c = 27$

$a + 3 = 0 \Rightarrow a = -3$

$b - 3a = 0 \Rightarrow b + 9 = 0 \Rightarrow b = -9$

$c + 3b = 0 \Rightarrow c - 27 = 0 \Rightarrow c = 27$
Which one of these is a quadratic equation?
A $-3x + 3 = 0$ B $9x^{3} - 1 = 0$ C $2x^{2} - 4x = 6$ D $\displaystyle x^2 - \frac{2}{x^2} = 3$

Answer and solution

Answer: (C) $2x^{2} - 4x = 6$

$2x^{2} - 4x = 6$ : it has the form $ax^2 + bx + c = 0$ with $a \ne 0$ .

The others have a negative or fractional power of $x$ , or a degree other than $2$ .
If $p(x) = -2x^{2} + 3x - 6$ , then $p(-1) =$
A $1$ B $-7$ C $-5$ D $-11$

Answer and solution

Answer: (D) $-11$

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

$= -2 - 3 - 6$

$= -11$
If $-2$ is a zero of $p(x) = x^{3} + 8x^{2} + ax + 8$ , then $a =$
A $17$ B $-24$ C $-16$ D $16$

Answer and solution

Answer: (D) $16$

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

$= -8 + 32 - 2a + 8$

$= -2a + 32$

$-2a + 32 = 0 \Rightarrow a = 16$
The roots of $x^{2} - 3x - 18 = 0$ are
A $-3, 6$ B $3, 6$ C $-3, -6$ D $3, -6$

Answer and solution

Answer: (A) $-3, 6$

$x^{2} - 3x - 18 = x^{2} + 3x - 6x - 18$

$= x(x + 3) - 6(x + 3)$

$= (x + 3)(x - 6)$

$\Rightarrow x = -3, 6$
The roots of $x^{2} + 3x + 1 = 0$ are
A $-3 \pm \sqrt{5}$ B $\displaystyle \frac{3 \pm \sqrt{5}}{2}$ C $\displaystyle \frac{-3 \pm \sqrt{5}}{2}$ D $\displaystyle \frac{-3 \pm \sqrt{13}}{2}$

Answer and solution

Answer: (C) $\displaystyle \frac{-3 \pm \sqrt{5}}{2}$

$a = 1, b = 3, c = 1$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{9 - 4(1)(1)}}{2(1)}$

$\displaystyle = \frac{-3 + \sqrt{5}}{2}, \frac{-3 - \sqrt{5}}{2}$
The discriminant of $4x^{2} + 9x - 8 = 0$ is
A $209$ B $47$ C $-47$ D $113$

Answer and solution

Answer: (A) $209$

$\Delta = b^2 - 4ac = 9^2 - 4(4)(-8) = 209$
When $x^{3} - x^{2} - 3x - 5$ is divided by $x - 2$ , the quotient is
A $x^{2} + x - 8$ B $x^{2} + x - 1$ C $x^{2} - 3x + 3$ D $x^{2} - x - 1$

Answer and solution

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

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

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

$-x \div x = -1$ ; $(-1) \times (x - 2) = -x + 2$ ; subtract: $-7$
The remainder when $2x^{3} - 6x^{2} + 7x - 2$ is divided by $x + 4$ is
A $254$ B $-2$ C $-254$ D $58$

Answer and solution

Answer: (C) $-254$

Remainder $= f(-4)$

$f(-4) = 2(-4)^{3} - 6(-4)^{2} + 7(-4) - 2$

$= -128 - 96 - 28 - 2$

$= -254$
Which of the following is a factor of $x^{3} + 6x^{2} + 3x - 10$ ?
A $x - 1$ B $x - 5$ C $x - 4$ D $x + 1$

Answer and solution

Answer: (A) $x - 1$

$f(1) = (1)^{3} + 6(1)^{2} + 3(1) - 10$

$= 1 + 6 + 3 - 10$

$= 0$

$f(1) = 0$ , so $x - 1$ is a factor.
$2\sqrt{2}x^{2} + 4x - 3\sqrt{2} =$
A $(\sqrt{2}x + 3)(2x + \sqrt{2})$ B $(\sqrt{2}x - 3)(2x - \sqrt{2})$ C $(\sqrt{2}x - 3)(2x + \sqrt{2})$ D $(\sqrt{2}x + 3)(2x - \sqrt{2})$

Answer and solution

Answer: (D) $(\sqrt{2}x + 3)(2x - \sqrt{2})$

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

$= \sqrt{2}x(2x - \sqrt{2}) + 3(2x - \sqrt{2}) = (\sqrt{2}x + 3)(2x - \sqrt{2})$
What must be added to $-3x^{3} - 2x^{2} - 8x - 10$ so that it is exactly divisible by $x^{2} - x + 4$ ?
A $-3x - 5$ B $x - 10$ C $-x + 10$ D $x + 10$

Answer and solution

Answer: (B) $x - 10$

$-3x^{3} \div x^{2} = -3x$ ; $(-3x) \times (x^{2} - x + 4) = -3x^{3} + 3x^{2} - 12x$ ; subtract: $-5x^{2} + 4x - 10$

$-5x^{2} \div x^{2} = -5$ ; $(-5) \times (x^{2} - x + 4) = -5x^{2} + 5x - 20$ ; subtract: $-x + 10$

Remainder $= -x + 10$ , so $x - 10$ must be added.
If $x - 1$ and $x + 4$ are factors of $x^3 + ax^2 + bx - 12$ , then
A $a = 6, b = 5$ B $a = -6, b = -5$ C $a = 6, b = -5$ D $a = -6, b = 5$

Answer and solution

Answer: (A) $a = 6, b = 5$

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

$f(-4) = 0 \Rightarrow 4a - b = 19$ ... (2)

$(1) + (2)$ : $5a = 30$

$\Rightarrow a = 6$

Put $a = 6$ in (1): $6 + b = 11 \Rightarrow b = 5$
$x^{2} + 8x + 1 =$
A $(x + 4 - \sqrt{17})(x + 4 + \sqrt{17})$ B $(x - 4 - \sqrt{15})(x - 4 + \sqrt{15})$ C $(x + 4 - \sqrt{15})(x + 4 + \sqrt{15})$ D $(x + 8 - \sqrt{15})(x + 8 + \sqrt{15})$

Answer and solution

Answer: (C) $(x + 4 - \sqrt{15})(x + 4 + \sqrt{15})$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{64 - 4(1)(1)}}{2(1)}$

$\displaystyle = \frac{-8 \pm \sqrt{60}}{2}$

$\displaystyle = \frac{-8 \pm 2\sqrt{15}}{2}$

$= -4 \pm \sqrt{15}$

$= -4 + \sqrt{15}, -4 - \sqrt{15}$

So $x^{2} + 8x + 1 = (x - \alpha)(x - \beta) = (x + 4 - \sqrt{15})(x + 4 + \sqrt{15})$
Given that $x - 4$ is a factor, $x^{3} - 11x^{2} + 34x - 24 =$
A $(x - 4)(x^{2} + 7x + 6)$ B $(x - 4)(x^{2} - 7x + 6)$ C $(x - 4)(x^{2} - 7x - 6)$ D $(x - 4)(x^{2} - 15x + 94)$

Answer and solution

Answer: (B) $(x - 4)(x^{2} - 7x + 6)$

$x^{3} \div x = x^{2}$ ; $x^{2} \times (x - 4) = x^{3} - 4x^{2}$ ; subtract: $-7x^{2} + 34x - 24$

$-7x^{2} \div x = -7x$ ; $(-7x) \times (x - 4) = -7x^{2} + 28x$ ; subtract: $6x - 24$

$6x \div x = 6$ ; $6 \times (x - 4) = 6x - 24$ ; subtract: $0$

Quotient $= x^{2} - 7x + 6$
$x^{3} - 12x^{2} + 47x - 60 =$
A $(x + 5)(x - 3)(x - 4)$ B $(x - 5)(x + 3)(x - 4)$ C $(x + 5)(x + 3)(x + 4)$ D $(x - 5)(x - 3)(x - 4)$

Answer and solution

Answer: (D) $(x - 5)(x - 3)(x - 4)$

$f(3) = 27 - 108 + 141 - 60 = 0$

Dividing and splitting the quotient gives $f(x) = (x - 5)(x - 3)(x - 4)$ .
Using synthetic division, the remainder when $2x^{3} - 4x^{2} - 6$ is divided by $x - 1$ is
A $8$ B $-8$ C $-12$ D $-2$

Answer and solution

Answer: (B) $-8$

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

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

Remainder $= -8$
A factor of $x^{5} + 243$ is
A $3x + 1$ B $x - 3$ C $x + 243$ D $x + 3$

Answer and solution

Answer: (D) $x + 3$

$x^{5} + 243 = x^{5} + 3^{5}$ and $5$ is odd, so $x + 3$ is a factor.
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 $\displaystyle \frac{1}{4}a^{2} + 9x^{2}$ B $\displaystyle \frac{1}{4}a^{2} + 3ax - 9x^{2}$ 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: $0$
The quadratic polynomial which leaves the remainders $-12$ , $-8$ and $-47$ when divided by $x + 2$ , $x$ and $x - 3$ respectively is
A $-3x^{2} - 4x - 8$ B $-3x^{2} + 4x - 8$ C $3x^{2} - 4x - 8$ D $-3x^{2} - 4x + 8$

Answer and solution

Answer: (A) $-3x^{2} - 4x - 8$

Check each remainder with the remainder theorem: $f(-2) = -12$ , $f(0) = -8$ , $f(3) = -47$ .

Only $-3x^{2} - 4x - 8$ gives all three.

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MCQ test, Set 6