MCQ test, Set 8

Answer: (C) $\displaystyle 4x^{\frac{2}{3}}y^{3}$

The power $\displaystyle \frac{2}{3}$ of $x$ is not a whole number.

So $\displaystyle 4x^{\frac{2}{3}}y^{3}$ is not a monomial; the others have only non-negative integer powers.

Answer: (C) $8$

Degrees of the terms: $6, 8, 4, 3, 0$

The greatest is $8$.

Answer: (D) $3$

Standard form: $3x^{5} + 5x + 4$

$n = 5$ and $a_n$ is the coefficient of $x^{5}$: $a_n = 3$

Answer: (A) $8 - 3x$

$8 - 3x$ has highest power $1$, so it is an linear polynomial.

Answer: (A) $\displaystyle 4 + 2x^{\frac{1}{3}}$

In $\displaystyle 4 + 2x^{\frac{1}{3}}$, the power $\displaystyle \frac{1}{3}$ of $x$ is not a whole number.

So it is not a polynomial. (A coefficient like $\sqrt{2}$ or $\displaystyle \frac{2}{3}$ is allowed.)

Answer: (C) $p = 5, q = -5, r = 5$

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

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

$r + q = 0 \Rightarrow r - 5 = 0 \Rightarrow r = 5$

Answer: (B) $4x^{2} - 2 = 0$

$4x^{2} - 2 = 0$: 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: (B) $49$

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

$= 32 + 14 + 3$

$= 49$

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

$\displaystyle -9x + 6 = 0 \Rightarrow -9x = -6 \Rightarrow x = \frac{2}{3}$

Answer: (C) $-8, 7$

$x^{2} + x - 56 = x^{2} - 7x + 8x - 56$

$= x(x - 7) + 8(x - 7)$

$= (x - 7)(x + 8)$

$\Rightarrow x = -8, 7$

Answer: (B) $97$

$\Delta = b^2 - 4ac = 1^2 - 4(2)(-12) = 97$

Answer: (C) $7$

$f(x) = g(x)q(x) + r(x)$: degree of $q(x)$ = degree of $f(x)$ $-$ degree of $g(x)$

$= 9 - 2 = 7$

Answer: (A) $4$

$f(1) = (1)^{3} - 4(1)^{2} + a(1) - 7$

$= 1 - 4 + a - 7$

$= a - 10$

$a - 10 = -6 \Rightarrow a = 4$

Answer: (B) $-6$

$\displaystyle 2x - 1 = 0 \Rightarrow x = \frac{1}{2}$; $\displaystyle f\left(\frac{1}{2}\right) = 0$

$\displaystyle f\left(\frac{1}{2}\right) = 8\left(\frac{1}{2}\right)^{3} - 4\left(\frac{1}{2}\right)^{2} + a\left(\frac{1}{2}\right) + 3$

$\displaystyle = 1 - 1 + \frac{1}{2}a + 3$

$\displaystyle = \frac{1}{2}a + 3$

$\displaystyle \frac{1}{2}a + 3 = 0 \Rightarrow a = -6$

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

$2\sqrt{5}x^{2} - 7x - 12\sqrt{5} = 2\sqrt{5}x^{2} - 15x + 8x - 12\sqrt{5}$

$= \sqrt{5}x(2x - 3\sqrt{5}) + 4(2x - 3\sqrt{5}) = (\sqrt{5}x + 4)(2x - 3\sqrt{5})$

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

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

Remainder $= -5x + 3$, so $-5x + 3$ must be subtracted.

Answer: (C) $5$

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

$f(2) = (2)^{3} - 5(2)^{2} + 4(2) + 9$

$= 8 - 20 + 8 + 9$

$= 5$

The remainder is $5$, so $5$ must be subtracted.

Answer: (D) $a = 1, b = -16$

$f(4) = 0 \Rightarrow 4a + b = -12$ ... (1)

$f(-1) = 0 \Rightarrow a - b = 17$ ... (2)

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

$\Rightarrow a = 1$

Put $a = 1$ in (1): $4 + b = -12 \Rightarrow b = -16$

Answer: (B) $(x - 1 - \sqrt{5})(x - 1 + \sqrt{5})$

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

$\displaystyle = \frac{2 \pm \sqrt{20}}{2}$

$\displaystyle = \frac{2 \pm 2\sqrt{5}}{2}$

$= 1 \pm \sqrt{5}$

$= 1 + \sqrt{5}, 1 - \sqrt{5}$

So $x^{2} - 2x - 4 = (x - \alpha)(x - \beta) = (x - 1 - \sqrt{5})(x - 1 + \sqrt{5})$

Answer: (A) $(x + 4)(x^{2} + 5x - 6)$

$x^{3} \div x = x^{2}$; $x^{2} \times (x + 4) = x^{3} + 4x^{2}$; subtract: $5x^{2} + 14x - 24$

$5x^{2} \div x = 5x$; $5x \times (x + 4) = 5x^{2} + 20x$; subtract: $-6x - 24$

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

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

Answer: (A) $(x + 3)(x + 2)(x + 1)$

$f(-1) = -1 + 6 - 11 + 6 = 0$

Dividing and splitting the quotient gives $f(x) = (x + 3)(x + 2)(x + 1)$.

Answer: (C) $9$

Second row: $0 \quad 2 \quad 8 \quad 8$

Bottom row: $2 \quad 8 \quad 8 \;\big|\; 9$

Remainder $= 9$

Answer: (B) $32$

$n = 155$: $x + a$ is a factor of $x^n + a^n$ (n odd).

So $29 + 3 = 32$ divides it.

Answer: (A) $\displaystyle \frac{9}{4}a^{2} - \frac{3}{4}ax + \frac{1}{16}x^{2}$

$\displaystyle \frac{27}{8}a^{3} \div \frac{3}{2}a = \frac{9}{4}a^{2}$; $\displaystyle \frac{9}{4}a^{2} \times (\frac{3}{2}a - \frac{1}{4}x) = \frac{27}{8}a^{3} - \frac{9}{16}a^{2}x$; subtract: $\displaystyle -\frac{9}{8}a^{2}x + \frac{9}{32}ax^{2} - \frac{1}{64}x^{3}$

$\displaystyle -\frac{9}{8}a^{2}x \div \frac{3}{2}a = -\frac{3}{4}ax$; $\displaystyle (-\frac{3}{4}ax) \times (\frac{3}{2}a - \frac{1}{4}x) = -\frac{9}{8}a^{2}x + \frac{3}{16}ax^{2}$; subtract: $\displaystyle \frac{3}{32}ax^{2} - \frac{1}{64}x^{3}$

$\displaystyle \frac{3}{32}ax^{2} \div \frac{3}{2}a = \frac{1}{16}x^{2}$; $\displaystyle \frac{1}{16}x^{2} \times (\frac{3}{2}a - \frac{1}{4}x) = \frac{3}{32}ax^{2} - \frac{1}{64}x^{3}$; subtract: $0$

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

Check each remainder with the remainder theorem: $f(-1) = 0$, $f(0) = -5$, $f(3) = -32$.

Only $-x^{2} - 6x - 5$ gives all three.