Question paper, Set 6

(i) $\displaystyle -\frac{2}{x^{4}}$: $\displaystyle -\frac{2}{x^{4}} = -2x^{-4}$, the power of $x$ is negative, so it is not a monomial.

(ii) $8x^{3}y^{3}$: the powers of $x$ and $y$ are $3$ and $3$, non-negative integers, so it is a monomial.

(iii) $\displaystyle \frac{5}{x^{3}}$: $\displaystyle \frac{5}{x^{3}} = 5x^{-3}$, the power of $x$ is negative, so it is not a monomial.

(iv) $\displaystyle \frac{8}{x^{3}}$: $\displaystyle \frac{8}{x^{3}} = 8x^{-3}$, the power of $x$ is negative, so it is not a monomial.

Answer: (ii) only is a monomial

The greatest power of $x$ in the polynomial is $5$.

So its degree is $5$.

Answer: $5$

Standard form: terms in descending powers of $x$.

$p(x) = 9x^{3} - x^{2} - 3x + 9$

Answer: $p(x) = 9x^{3} - x^{2} - 3x + 9$

$0 = 0 \times x^0 = 0 \times x^1 = 0 \times x^2 = \cdots$

So no one power of $x$ can be taken as its degree.

Answer: Not defined

In a zero polynomial all the coefficients are zero.

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

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

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

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

The highest power of $x$ in the equation is $1$.

So it is an linear equation.

Answer: Linear equation

$\displaystyle p\left(\frac{1}{2}\right) = -2\left(\frac{1}{2}\right)^{3} + 6\left(\frac{1}{2}\right)^{2} + \left(\frac{1}{2}\right) - 1$

$\displaystyle = -\frac{1}{4} + \frac{3}{2} + \frac{1}{2} - 1$

$\displaystyle = \frac{3}{4}$

Answer: $\displaystyle \frac{3}{4}$

$1 \times (-14) = -14 = (-2) \times 7$ and $(-2) + 7 = 5$

$x^{2} + 5x - 14 = x^{2} - 2x + 7x - 14$

$= x(x - 2) + 7(x - 2)$

$= (x - 2)(x + 7)$

$\Rightarrow x - 2 = 0$ (or) $x + 7 = 0$

$\Rightarrow x = -7, 2$

Answer: $x = -7, 2$

$a = 1, b = 4, c = -12$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 - 4(1)(-12)}}{2(1)}$

$\displaystyle = \frac{-4 \pm \sqrt{64}}{2}$

$\displaystyle = \frac{-4 + 8}{2}, \frac{-4 - 8}{2} = 2, -6$

Answer: $x = 2, -6$

$-2x^{3} \div x = -2x^{2}$; $(-2x^{2}) \times (x - 2) = -2x^{3} + 4x^{2}$; subtract: $-4x^{2} + 11x - 17$

$-4x^{2} \div x = -4x$; $(-4x) \times (x - 2) = -4x^{2} + 8x$; subtract: $3x - 17$

$3x \div x = 3$; $3 \times (x - 2) = 3x - 6$; subtract: $-11$

Quotient $= -2x^{2} - 4x + 3$, Remainder $= -11$

Answer: Quotient $= -2x^{2} - 4x + 3$, Remainder $= -11$

$x - 3 = 0 \Rightarrow x = 3$

By the remainder theorem, remainder $= f(3)$

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

$= 18 - 15 - 1$

$= 2$

Answer: $2$

$x^{2} - 7x + 12 = x^{2} - 3x - 4x + 12$

$= x(x - 3) - 4(x - 3)$

$= (x - 3)(x - 4)$

Let $f(x) = 3x^{4} - 24x^{3} + 52x^{2} - x - 60$.

$f(3) = 3(3)^{4} - 24(3)^{3} + 52(3)^{2} - (3) - 60$

$= 243 - 648 + 468 - 3 - 60$

$= 0$

So $x - 3$ is a factor of $f(x)$.

$f(4) = 3(4)^{4} - 24(4)^{3} + 52(4)^{2} - (4) - 60$

$= 768 - 1536 + 832 - 4 - 60$

$= 0$

So $x - 4$ is a factor of $f(x)$.

$\therefore (x - 3)(x - 4) = x^{2} - 7x + 12$ divides $f(x)$ exactly.

Answer: $x^{2} - 7x + 12 = (x - 3)(x - 4)$ and $f(3) = f(4) = 0$, so it divides $f(x)$

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

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

Remainder $= -2x - 5$

So $-(-2x - 5) = 2x + 5$ must be added.

Answer: $2x + 5$

$x^{2} - 4 = (x + 2)(x - 2)$

Let $f(x) = x^4 + px^3 - 2x^2 - 8x + q$; $x + 2$ and $x - 2$ are factors.

$f(-2) = 0$:

$f(-2) = (-2)^{4} + p(-2)^{3} - 2(-2)^{2} - 8(-2) + q$

$= 16 - 8p - 8 + 16 + q$

$= -8p + q + 24$

$\Rightarrow 8p - q = 24$ ... (1)

$f(2) = 0$:

$f(2) = (2)^{4} + p(2)^{3} - 2(2)^{2} - 8(2) + q$

$= 16 + 8p - 8 - 16 + q$

$= 8p + q - 8$

$\Rightarrow 8p + q = 8$ ... (2)

$(1) + (2)$: $16p = 32$

$\Rightarrow p = 2$

Put $p = 2$ in (1): $16 - q = 24 \Rightarrow q = -8$

Answer: $p = 2, q = -8$

Zeros of $ax^2 + bx + c$ are $\displaystyle \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

$a = 15, b = -8, c = -63$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{64 - 4(15)(-63)}}{2(15)}$

$\displaystyle = \frac{8 \pm \sqrt{3844}}{30}$

$\displaystyle = \frac{8 + 62}{30}, \frac{8 - 62}{30} = \frac{7}{3}, -\frac{9}{5}$

$\displaystyle \therefore 15x^{2} - 8x - 63 = 15\left(x - \frac{7}{3}\right)\left(x + \frac{9}{5}\right)$

$= (3x - 7)(5x + 9)$

Answer: $(3x - 7)(5x + 9)$

Let $f(x) = x^{3} + 9x^{2} + 26x + 24$. By trial and error:

$f(1) = 1 + 9 + 26 + 24 \ne 0$

$f(-1) = -1 + 9 - 26 + 24 \ne 0$

$f(2) = 8 + 36 + 52 + 24 \ne 0$

$f(-2) = -8 + 36 - 52 + 24 = 0$

$\Rightarrow x + 2$ is a factor of $f(x)$

$x^{3} \div x = x^{2}$; $x^{2} \times (x + 2) = x^{3} + 2x^{2}$; subtract: $7x^{2} + 26x + 24$

$7x^{2} \div x = 7x$; $7x \times (x + 2) = 7x^{2} + 14x$; subtract: $12x + 24$

$12x \div x = 12$; $12 \times (x + 2) = 12x + 24$; subtract: $0$

$\therefore f(x) = (x + 2)(x^{2} + 7x + 12)$

$x^{2} + 7x + 12 = x^{2} + 3x + 4x + 12$

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

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

$\therefore f(x) = (x + 2)(x + 3)(x + 4)$

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

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

Here $n = 131$ is odd, so $27 + 16 = 43$ divides $27^{131} + 16^{131}$.

Remainder $= 0$

Answer: $0$

$y^{4} - 81 = (y^{2})^2 - (9)^2$

$= (y^{2} + 9)(y^{2} - 9)$

$= (y^{2} + 9)(y + 3)(y - 3)$

Answer: $(y^{2} + 9)(y + 3)(y - 3)$

$\displaystyle \frac{1}{8}a^{3} \div \frac{1}{2}a = \frac{1}{4}a^{2}$; $\displaystyle \frac{1}{4}a^{2} \times (\frac{1}{2}a - \frac{1}{3}x) = \frac{1}{8}a^{3} - \frac{1}{12}a^{2}x$; subtract: $\displaystyle -\frac{1}{6}a^{2}x + \frac{1}{6}ax^{2} - \frac{1}{27}x^{3}$

$\displaystyle -\frac{1}{6}a^{2}x \div \frac{1}{2}a = -\frac{1}{3}ax$; $\displaystyle (-\frac{1}{3}ax) \times (\frac{1}{2}a - \frac{1}{3}x) = -\frac{1}{6}a^{2}x + \frac{1}{9}ax^{2}$; subtract: $\displaystyle \frac{1}{18}ax^{2} - \frac{1}{27}x^{3}$

$\displaystyle \frac{1}{18}ax^{2} \div \frac{1}{2}a = \frac{1}{9}x^{2}$; $\displaystyle \frac{1}{9}x^{2} \times (\frac{1}{2}a - \frac{1}{3}x) = \frac{1}{18}ax^{2} - \frac{1}{27}x^{3}$; subtract: $0$

Remainder $= 0$; Quotient $\displaystyle = \frac{1}{4}a^{2} - \frac{1}{3}ax + \frac{1}{9}x^{2}$

Answer: Quotient $\displaystyle = \frac{1}{4}a^{2} - \frac{1}{3}ax + \frac{1}{9}x^{2}$

Let $f(x) = ax^2 + bx + c$. The remainder on dividing by $x - k$ is $f(k)$.

$f(1) = 5 \Rightarrow a + b + c = 5$ ... (1)

$f(2) = 3 \Rightarrow 4a + 2b + c = 3$ ... (2)

$f(4) = 5 \Rightarrow 16a + 4b + c = 5$ ... (3)

$(2) - (1)$: $3a + b = -2$ ... (4)

$(3) - (2)$, divide by $2$: $6a + b = 1$ ... (5)

$(5) - (4)$: $3a = 3 \Rightarrow a = 1$

From (4): $b = -5$; from (1): $c = 9$

$\therefore f(x) = x^{2} - 5x + 9$

Answer: $f(x) = x^{2} - 5x + 9$