Question paper, Set 4

In $-2x^{3}y^{3}$, the powers of $x$ and $y$ are $3$ and $3$, non-negative integers.

So $-2x^{3}y^{3}$ is a monomial.

Answer: Yes, the powers of $x$ and $y$ are $3$ and $3$, non-negative integers

Degrees of the terms: $3, 2, 5, 9, 0$

Greatest among them $= 9$

So the degree of the polynomial is $9$.

Answer: $9$

The term with $x^{2}$ in $p(x)$ is $4x^{2}$.

So the coefficient of $x^{2}$ is $4$.

Answer: $4$

The highest power of $x$ is $4$, so the degree is $4$.

A polynomial of degree $4$ is a bi-quadratic polynomial.

Answer: Bi-quadratic polynomial

In a zero polynomial all the coefficients are zero.

$p + 4 = 0 \Rightarrow p = -4$

$q - 3p = 0 \Rightarrow q + 12 = 0 \Rightarrow q = -12$

$r + 3q = 0 \Rightarrow r - 36 = 0 \Rightarrow r = 36$

Answer: $p = -4, q = -12, r = 36$

(i) $\displaystyle \frac{1}{x^2} = 1x^{-2}$ has a negative power, so it is not a quadratic equation.

(ii) It has the form $ax^2 + bx + c = 0$ with $a \ne 0$.

(iii) It has the form $ax^2 + bx + c = 0$ with $a \ne 0$.

Answer: (ii) and (iii)

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

$= -32 - 32 - 2$

$= -66$

Answer: $-66$

$\displaystyle p\left(-\frac{1}{2}\right) = 8\left(-\frac{1}{2}\right)^{3} - 6\left(-\frac{1}{2}\right)^{2} + a\left(-\frac{1}{2}\right) - 2$

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

$\displaystyle = -\frac{1}{2}a - \frac{9}{2}$

Given $\displaystyle p\left(-\frac{1}{2}\right) = 0$:

$\displaystyle -\frac{1}{2}a - \frac{9}{2} = 0$

$\displaystyle \Rightarrow -\frac{1}{2}a = \frac{9}{2}$

$\Rightarrow a = -9$

Answer: $a = -9$

$9 \times (-14) = -126 = 6 \times (-21)$ and $6 + (-21) = -15$

$9x^{2} - 15x - 14 = 9x^{2} + 6x - 21x - 14$

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

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

$\Rightarrow 3x + 2 = 0$ (or) $3x - 7 = 0$

$\displaystyle \Rightarrow x = -\frac{2}{3}, \frac{7}{3}$

Answer: $\displaystyle x = -\frac{2}{3}, \frac{7}{3}$

$a = 2, b = -7, c = -8$

$\Delta = b^2 - 4ac = (-7)^2 - 4(2)(-8)$

$= 49 + 64 = 113$

Answer: $\Delta = 113$

Write the divisor in descending powers: $7 - 2x = -2x + 7$

$-6x^{3} \div (-2x) = 3x^{2}$; $3x^{2} \times (-2x + 7) = -6x^{3} + 21x^{2}$; subtract: $6x^{2} - 35x + 58$

$6x^{2} \div (-2x) = -3x$; $(-3x) \times (-2x + 7) = 6x^{2} - 21x$; subtract: $-14x + 58$

$-14x \div (-2x) = 7$; $7 \times (-2x + 7) = -14x + 49$; subtract: $9$

Quotient $= 3x^{2} - 3x + 7$, Remainder $= 9$

Answer: Quotient $= 3x^{2} - 3x + 7$, Remainder $= 9$

Let $f(x) = ax^{3} + 2x^{2} - x + 7$ and $g(x) = x^{3} - x^{2} - a$.

Given $f(2) = g(2)$

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

$= 8a + 8 - 2 + 7$

$= 8a + 13$

$g(2) = (2)^{3} - (2)^{2} - a$

$= 8 - 4 - a$

$= -a + 4$

$8a + 13 = -a + 4$

$\Rightarrow 9a = -9$

$\Rightarrow a = -1$

Answer: $a = -1$

Let $f(x) = 2x^{3} - 9x^{2} + x - 30$; $x - 5 = 0 \Rightarrow x = 5$

$f(5) = 2(5)^{3} - 9(5)^{2} + (5) - 30$

$= 250 - 225 + 5 - 30$

$= 0$

$\Rightarrow x - 5$ is a factor of $f(x)$.

Answer: $f(5) = 0$, so $x - 5$ is a factor

$15 \times 8 = 120 = 5 \times 24$ and $5 + 24 = 29$

$15x^{2} + 29x + 8 = 15x^{2} + 5x + 24x + 8$

$= 5x(3x + 1) + 8(3x + 1)$

$= (3x + 1)(5x + 8)$

Answer: $(3x + 1)(5x + 8)$

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

Let $f(x) = x^4 + px^3 - 10x^2 + 9x + q$; $x + 3$ and $x - 3$ are factors.

$f(-3) = 0$:

$f(-3) = (-3)^{4} + p(-3)^{3} - 10(-3)^{2} + 9(-3) + q$

$= 81 - 27p - 90 - 27 + q$

$= -27p + q - 36$

$\Rightarrow 27p - q = -36$ ... (1)

$f(3) = 0$:

$f(3) = (3)^{4} + p(3)^{3} - 10(3)^{2} + 9(3) + q$

$= 81 + 27p - 90 + 27 + q$

$= 27p + q + 18$

$\Rightarrow 27p + q = -18$ ... (2)

$(1) + (2)$: $54p = -54$

$\Rightarrow p = -1$

Put $p = -1$ in (1): $-27 - q = -36 \Rightarrow q = 9$

Answer: $p = -1, q = 9$

Divide $x^{3} + 4x^{2} - 20x - 48$ by $x + 2$:

$x^{3} \div x = x^{2}$; $x^{2} \times (x + 2) = x^{3} + 2x^{2}$; subtract: $2x^{2} - 20x - 48$

$2x^{2} \div x = 2x$; $2x \times (x + 2) = 2x^{2} + 4x$; subtract: $-24x - 48$

$-24x \div x = -24$; $(-24) \times (x + 2) = -24x - 48$; subtract: $0$

$\therefore x^{3} + 4x^{2} - 20x - 48 = (x + 2)(x^{2} + 2x - 24)$

$x^{2} + 2x - 24 = x^{2} - 4x + 6x - 24$

$= x(x - 4) + 6(x - 4)$

$= (x - 4)(x + 6)$

$\therefore x^{3} + 4x^{2} - 20x - 48 = (x + 2)(x - 4)(x + 6)$

Answer: $(x + 2)(x - 4)(x + 6)$

Let $f(x) = x^{4} - 5x^{3} + 5x^{2} + 5x - 6$. By trial and error:

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

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

$\Rightarrow x - 1$ and $x + 1$ are factors, so $(x - 1)(x + 1) = x^{2} - 1$ is a factor

$x^{4} \div x^{2} = x^{2}$; $x^{2} \times (x^{2} - 1) = x^{4} - x^{2}$; subtract: $-5x^{3} + 6x^{2} + 5x - 6$

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

$6x^{2} \div x^{2} = 6$; $6 \times (x^{2} - 1) = 6x^{2} - 6$; subtract: $0$

$\therefore f(x) = (x^{2} - 1)(x^{2} - 5x + 6)$

$x^{2} - 5x + 6 = x^{2} - 2x - 3x + 6$

$= x(x - 2) - 3(x - 2)$

$= (x - 2)(x - 3)$

$\therefore f(x) = (x - 1)(x + 1)(x - 2)(x - 3)$

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

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

Write the coefficients and $3$ on the left: $3 \;\big|\; 1 \quad 2 \quad -2 \quad -11$

Bring down the first number; multiply by $3$ and add, each time.

Second row: $0 \quad 3 \quad 15 \quad 39$

Bottom row: $1 \quad 5 \quad 13 \;\big|\; 28$

Quotient $= x^{2} + 5x + 13$, Remainder $= 28$

Answer: Quotient $= x^{2} + 5x + 13$, Remainder $= 28$

$x - a$ is a factor of $x^n - a^n$ for every positive integer $n$.

So $24 - 17 = 7$ divides $24^{42} - 17^{42}$.

Remainder $= 0$

Answer: $0$

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

$f(-3) = 1 \Rightarrow 9a - 3b + c = 1$ ... (1)

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

$f(-1) = 4 \Rightarrow a - b + c = 4$ ... (3)

$(2) - (1)$: $-5a + b = -5$ ... (4)

$(3) - (2)$: $-3a + b = 8$ ... (5)

$(5) - (4)$: $\displaystyle 2a = 13 \Rightarrow a = \frac{13}{2}$

From (4): $\displaystyle b = \frac{55}{2}$; from (1): $c = 25$

$\displaystyle \therefore f(x) = \frac{13}{2}x^{2} + \frac{55}{2}x + 25$

Answer: $\displaystyle f(x) = \frac{13}{2}x^{2} + \frac{55}{2}x + 25$