Question paper, Set 2

$-6x^{3}y^{3} + 8x^{2}y^{3}$ has two terms: $-6x^{3}y^{3}$, $8x^{2}y^{3}$.

An expression with two terms is a binomial.

Answer: Binomial

Degree of a monomial = sum of the exponents of its variables

$= 4 + 4 + 3 = 11$

Answer: $11$

In standard form the powers of $x$ go down: $p(x) = -5x^{3} + 8x^{2} + 4x + 8$

Highest power $= 3$, so $n = 3$; its coefficient $a_3 = -5$

Constant term $a_0 = 8$

Answer: $p(x) = -5x^{3} + 8x^{2} + 4x + 8$; $n = 3$, $a_3 = -5$, $a_0 = 8$

In $-8x^{2} + 8x^{-3}$, the power $-3$ of $x$ is negative.

So it is not a polynomial.

Answer: No, the power $-3$ of $x$ is negative

In a zero polynomial all the coefficients are zero.

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

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

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

Answer: $p = -2, q = -2, r = 2$

The highest power of $x$ in the equation is $2$.

So it is a quadratic equation.

Answer: Quadratic equation

$\displaystyle p\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{3} + 4\left(\frac{1}{2}\right)^{2} + 9\left(\frac{1}{2}\right) + 1$

$\displaystyle = \frac{1}{8} + 1 + \frac{9}{2} + 1$

$\displaystyle = \frac{53}{8}$

Answer: $\displaystyle \frac{53}{8}$

$7x + 3 = 0$

$\Rightarrow 7x = -3$

$\displaystyle \Rightarrow x = -\frac{3}{7}$

Answer: $\displaystyle x = -\frac{3}{7}$

Let $p(x) = x^{2} + 9x + 8$.

$p(-1) = (-1)^{2} + 9(-1) + 8$

$= 1 - 9 + 8$

$= 0$

$p(-8) = (-8)^{2} + 9(-8) + 8$

$= 64 - 72 + 8$

$= 0$

Both $-1$ and $-8$ are roots.

Answer: Both $-1$ and $-8$ are roots

$a = 1, b = 4, c = -1$

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

$\displaystyle = \frac{-4 \pm \sqrt{20}}{2}$

$\displaystyle = \frac{-4 \pm 2\sqrt{5}}{2}$

$= -2 \pm \sqrt{5}$

$= -2 + \sqrt{5}, -2 - \sqrt{5}$

Answer: $x = -2 + \sqrt{5}, -2 - \sqrt{5}$

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

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

$-4x \div x = -4$; $(-4) \times (x + 1) = -4x - 4$; subtract: $-6$

Quotient $= 2x^{2} + 5x - 4$, Remainder $= -6$

Answer: Quotient $= 2x^{2} + 5x - 4$, Remainder $= -6$

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

Remainder $\displaystyle = f\left(\frac{2}{3}\right)$

$\displaystyle f\left(\frac{2}{3}\right) = 54\left(\frac{2}{3}\right)^{3} + 27\left(\frac{2}{3}\right)^{2} - 9\left(\frac{2}{3}\right) - 9$

$= 16 + 12 - 6 - 9$

$= 13$

Answer: $13$

Let $f(x) = ax^{3} + bx^{2} + 3x - 9$.

$x - 1$ is a factor $\Rightarrow f(1) = 0$

$f(1) = a(1)^{3} + b(1)^{2} + 3(1) - 9$

$= a + b + 3 - 9$

$= a + b - 6$

$\Rightarrow a + b = 6$ ... (1)

Remainder on dividing by $x + 1$ is $-8$ $\Rightarrow f(-1) = -8$

$f(-1) = a(-1)^{3} + b(-1)^{2} + 3(-1) - 9$

$= -a + b - 3 - 9$

$= -a + b - 12$

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

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

$\Rightarrow a = 1$

Put $a = 1$ in (1): $1 + b = 6 \Rightarrow b = 5$

Answer: $a = 1, b = 5$

$3\sqrt{3} \times (-2\sqrt{3}) = -18$; split $3x$ as $-3x$ and $6x$

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

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

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

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

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

$2x^{2} \div x^{2} = 2$; $2 \times (x^{2} - 2x - 1) = 2x^{2} - 4x - 2$; subtract: $3x - 6$

Remainder $= 3x - 6$

Subtracting the remainder leaves an exact multiple: $3x - 6$ must be subtracted.

Answer: $3x - 6$

$a = 3, b = -10, c = 5$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{100 - 4(3)(5)}}{2(3)}$

$\displaystyle = \frac{10 \pm \sqrt{40}}{6}$

$\displaystyle = \frac{10 \pm 2\sqrt{10}}{6}$

$\displaystyle = \frac{5 \pm \sqrt{10}}{3}$

$\displaystyle = \frac{5 + \sqrt{10}}{3}, \frac{5 - \sqrt{10}}{3}$

$\therefore 3x^{2} - 10x + 5 = a(x - \alpha)(x - \beta)$

$\displaystyle = 3\left[x - \left(\frac{5 + \sqrt{10}}{3}\right)\right]\left[x - \left(\frac{5 - \sqrt{10}}{3}\right)\right]$

Answer: $\displaystyle 3\left[x - \left(\frac{5 + \sqrt{10}}{3}\right)\right]\left[x - \left(\frac{5 - \sqrt{10}}{3}\right)\right]$

$18t^{3} \div 3t = 6t^{2}$; $6t^{2} \times (3t + 8) = 18t^{3} + 48t^{2}$; subtract: $-51t^{2} - 178t - 112$

$-51t^{2} \div 3t = -17t$; $(-17t) \times (3t + 8) = -51t^{2} - 136t$; subtract: $-42t - 112$

$-42t \div 3t = -14$; $(-14) \times (3t + 8) = -42t - 112$; subtract: $0$

$\therefore 18t^{3} - 3t^{2} - 178t - 112 = (3t + 8)(6t^{2} - 17t - 14)$

$6 \times (-14) = -84 = 4 \times (-21)$ and $4 + (-21) = -17$

$6t^{2} - 17t - 14 = 6t^{2} + 4t - 21t - 14$

$= 2t(3t + 2) - 7(3t + 2)$

$= (3t + 2)(2t - 7)$

$\therefore 18t^{3} - 3t^{2} - 178t - 112 = (3t + 8)(3t + 2)(2t - 7)$

Answer: $(3t + 8)(3t + 2)(2t - 7)$

Let $f(x) = 2x^{3} + 9x^{2} - 6x - 5$.

$f(1) = 2 + 9 - 6 - 5 = 0$

$\therefore x - 1$ is a factor

Write the coefficients and $1$ on the left: $1 \;\big|\; 2 \quad 9 \quad -6 \quad -5$

Bring down the first number; multiply by $1$ and add, each time.

Second row: $0 \quad 2 \quad 11 \quad 5$

Bottom row: $2 \quad 11 \quad 5 \;\big|\; 0$

$\therefore f(x) = (x - 1)(2x^{2} + 11x + 5)$

$2x^{2} + 11x + 5 = 2x^{2} + x + 10x + 5$

$= x(2x + 1) + 5(2x + 1)$

$= (2x + 1)(x + 5)$

$\therefore f(x) = (x - 1)(2x + 1)(x + 5)$

Answer: $(x - 1)(2x + 1)(x + 5)$

$x + a$ is a factor of $x^n + a^n$ when $n$ is odd.

Here $n = 141$ is odd, so $26 + 5 = 31$ divides $26^{141} + 5^{141}$.

Remainder $= 0$

Answer: $0$

$\displaystyle \frac{2}{3}a^{3} \div \frac{1}{3}a = 2a^{2}$; $\displaystyle 2a^{2} \times (\frac{1}{3}a + \frac{1}{4}x) = \frac{2}{3}a^{3} + \frac{1}{2}a^{2}x$; subtract: $\displaystyle -\frac{4}{3}a^{2}x - \frac{7}{3}ax^{2} - x^{3}$

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

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

Remainder $= 0$; Quotient $= 2a^{2} - 4ax - 4x^{2}$

Answer: Quotient $= 2a^{2} - 4ax - 4x^{2}$