MCQ test, Set 4

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

$-6x^{4}y^{2}$: the powers of $x$ and $y$ are $4$ and $2$, non-negative integers, so it is a monomial.

Each of the others has a negative or fractional power.

Answer: (C) $7$

Degrees of the terms: $3, 7, 2, 0$

The greatest is $7$.

Answer: (B) $1$

Standard form: $x^{4} + 9x - 9$

$n = 4$ and $a_n$ is the coefficient of $x^{4}$: $a_n = 1$

Answer: (D) $x - 8x^{2} + 7$

$x - 8x^{2} + 7$ has highest power $2$, so it is a quadratic polynomial.

Answer: (A) $-4x^{-3} - 9x^{2} + 2$

In $-4x^{-3} - 9x^{2} + 2$, 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.)

Answer: (A) $p = -5, q = 10, r = -20$

$p + 5 = 0 \Rightarrow p = -5$

$q + 2p = 0 \Rightarrow q - 10 = 0 \Rightarrow q = 10$

$r + 2q = 0 \Rightarrow r + 20 = 0 \Rightarrow r = -20$

Answer: (C) $5\sqrt{2}x^2 - x + 4\sqrt{2} = 0$

$5\sqrt{2}x^2 - x + 4\sqrt{2} = 0$: 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 $2$.

Answer: (A) $-11$

$p(-1) = (-1)^{3} + 5(-1)^{2} + 7(-1) - 8$

$= -1 + 5 - 7 - 8$

$= -11$

Answer: (D) $-1$

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

$= 1 + 3 - a - 5$

$= -a - 1$

$-a - 1 = 0 \Rightarrow a = -1$

Answer: (A) $\displaystyle -3, -\frac{4}{5}$

$5x^{2} + 19x + 12 = 5x^{2} + 4x + 15x + 12$

$= x(5x + 4) + 3(5x + 4)$

$= (5x + 4)(x + 3)$

$\displaystyle \Rightarrow x = -3, -\frac{4}{5}$

Answer: (C) $1, 9$

$x^{2} - 10x + 9 = x^{2} - x - 9x + 9$

$= x(x - 1) - 9(x - 1)$

$= (x - 1)(x - 9)$

$\Rightarrow x = 1, 9$

Answer: (B) $5 \pm \sqrt{30}$

$a = 1, b = -10, c = -5$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{100 - 4(1)(-5)}}{2(1)}$

$= 5 + \sqrt{30}, 5 - \sqrt{30}$

Answer: (D) $2x - 1$

$x^{4} \div x^{2} = x^{2}$; $x^{2} \times (x^{2} - x - 1) = x^{4} - x^{3} - x^{2}$; subtract: $5x^{3} + 2x^{2} - 10x - 8$

$5x^{3} \div x^{2} = 5x$; $5x \times (x^{2} - x - 1) = 5x^{3} - 5x^{2} - 5x$; subtract: $7x^{2} - 5x - 8$

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

Remainder $= 2x - 1$

Answer: (C) $-73$

$\displaystyle 2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$; remainder $\displaystyle = f\left(-\frac{3}{2}\right)$

$= -73$

Answer: (A) $8$

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

$= -8 + 20 - 2a + 4$

$= -2a + 16$

$-2a + 16 = 0 \Rightarrow a = 8$

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

$3\sqrt{2}x^{2} + 7x - 3\sqrt{2} = 3\sqrt{2}x^{2} - 2x + 9x - 3\sqrt{2}$

$= \sqrt{2}x(3x - \sqrt{2}) + 3(3x - \sqrt{2}) = (\sqrt{2}x + 3)(3x - \sqrt{2})$

Answer: (B) $5x + 5$

$-3x^{3} \div x^{2} = -3x$; $(-3x) \times (x^{2} - 1) = -3x^{3} + 3x$; subtract: $6x^{2} + 5x - 1$

$6x^{2} \div x^{2} = 6$; $6 \times (x^{2} - 1) = 6x^{2} - 6$; subtract: $5x + 5$

Remainder $= 5x + 5$, so $5x + 5$ must be subtracted.

Answer: (D) $a = -3, b = -6$

$f(4) = 0 \Rightarrow 4a + b = -18$ ... (1)

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

$(1) + (2)$: $6a = -18$

$\Rightarrow a = -3$

Put $a = -3$ in (1): $-12 + b = -18 \Rightarrow b = -6$

Answer: (A) $(x + 2 - \sqrt{11})(x + 2 + \sqrt{11})$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 - 4(1)(-7)}}{2(1)}$

$\displaystyle = \frac{-4 \pm \sqrt{44}}{2}$

$\displaystyle = \frac{-4 \pm 2\sqrt{11}}{2}$

$= -2 \pm \sqrt{11}$

$= -2 + \sqrt{11}, -2 - \sqrt{11}$

So $x^{2} + 4x - 7 = (x - \alpha)(x - \beta) = (x + 2 - \sqrt{11})(x + 2 + \sqrt{11})$

Answer: (C) $(x + 5)(x^{2} - 5x + 6)$

$x^{3} \div x = x^{2}$; $x^{2} \times (x + 5) = x^{3} + 5x^{2}$; subtract: $-5x^{2} - 19x + 30$

$-5x^{2} \div x = -5x$; $(-5x) \times (x + 5) = -5x^{2} - 25x$; subtract: $6x + 30$

$6x \div x = 6$; $6 \times (x + 5) = 6x + 30$; subtract: $0$

Quotient $= x^{2} - 5x + 6$

Answer: (B) $(x - 4)(x + 3)(x - 2)$

$f(2) = 8 - 12 - 20 + 24 = 0$

Dividing and splitting the quotient gives $f(x) = (x - 4)(x + 3)(x - 2)$.

Answer: (B) $9$

Second row: $0 \quad 1 \quad 4 \quad 4$

Bottom row: $1 \quad 4 \quad 4 \;\big|\; 9$

Remainder $= 9$

Answer: (D) $x^{3} + 3x^{2} + 9x + 27$

$x^{4} - 81 = (x - 3)(x^{3} + 3x^{2} + 9x + 27)$

Signs are all $+$; the powers of $3$ go up as the powers of $x$ go down.

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

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

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

$\displaystyle \frac{1}{2}ax^{2} \div 2a = \frac{1}{4}x^{2}$; $\displaystyle \frac{1}{4}x^{2} \times (2a + \frac{1}{2}x) = \frac{1}{2}ax^{2} + \frac{1}{8}x^{3}$; subtract: $0$

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

Check each remainder with the remainder theorem: $f(1) = -7$, $f(2) = -10$, $f(3) = -11$.

Only $x^{2} - 6x - 2$ gives all three.