Question paper, Set 10

In $9xy^{3}$, the powers of $x$ and $y$ are $1$ and $3$, non-negative integers.

So $9xy^{3}$ is a monomial.

Answer: Yes, the powers of $x$ and $y$ are $1$ and $3$, non-negative integers

Degrees of the terms: $6, 7, 9, 6, 0$

Greatest among them $= 9$

So the degree of the polynomial is $9$.

Answer: $9$

The term with $x^{3}$ in $p(x)$ is $-4x^{3}$.

So the coefficient of $x^{3}$ is $-4$.

Answer: $-4$

The highest power of $x$ is $2$, so the degree is $2$.

A polynomial of degree $2$ is a quadratic polynomial.

Answer: Quadratic polynomial

In a zero polynomial all the coefficients are zero.

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

$q - 2p = 0 \Rightarrow q - 8 = 0 \Rightarrow q = 8$

$r - q = 0 \Rightarrow r - 8 = 0 \Rightarrow r = 8$

Answer: $p = 4, q = 8, r = 8$

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

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

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

Answer: (iii) only

$p(-1) = 2(-1)^{2} + 6(-1) - 12$

$= 2 - 6 - 12$

$= -16$

Answer: $-16$

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

$= -16 - 16 - 2a + 2$

$= -2a - 30$

Given $p(-2) = 0$:

$-2a - 30 = 0$

$\Rightarrow -2a = 30$

$\Rightarrow a = -15$

Answer: $a = -15$

$12 \times 21 = 252 = (-9) \times (-28)$ and $(-9) + (-28) = -37$

$12x^{2} - 37x + 21 = 12x^{2} - 9x - 28x + 21$

$= 3x(4x - 3) - 7(4x - 3)$

$= (4x - 3)(3x - 7)$

$\Rightarrow 4x - 3 = 0$ (or) $3x - 7 = 0$

$\displaystyle \Rightarrow x = \frac{3}{4}, \frac{7}{3}$

Answer: $\displaystyle x = \frac{3}{4}, \frac{7}{3}$

$a = 1, b = -2, c = -2$

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

$= 4 + 8 = 12$

Answer: $\Delta = 12$

Write the divisor in descending powers: $7 - 3x = -3x + 7$

$-9x^{3} \div (-3x) = 3x^{2}$; $3x^{2} \times (-3x + 7) = -9x^{3} + 21x^{2}$; subtract: $-3x^{2} + 19x - 23$

$-3x^{2} \div (-3x) = x$; $x \times (-3x + 7) = -3x^{2} + 7x$; subtract: $12x - 23$

$12x \div (-3x) = -4$; $(-4) \times (-3x + 7) = 12x - 28$; subtract: $5$

Quotient $= 3x^{2} + x - 4$, Remainder $= 5$

Answer: Quotient $= 3x^{2} + x - 4$, Remainder $= 5$

Let $f(x) = ax^{3} + 4x^{2} - x - 6$ and $g(x) = 2x^{3} + x + a$.

Given $f(-1) = g(-1)$

$f(-1) = a(-1)^{3} + 4(-1)^{2} - (-1) - 6$

$= -a + 4 + 1 - 6$

$= -a - 1$

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

$= -2 - 1 + a$

$= a - 3$

$-a - 1 = a - 3$

$\Rightarrow -2a = -2$

$\Rightarrow a = 1$

Answer: $a = 1$

Let $f(x) = x^{3} - x^{2} - 13x + 21$; $x - 3 = 0 \Rightarrow x = 3$

$f(3) = (3)^{3} - (3)^{2} - 13(3) + 21$

$= 27 - 9 - 39 + 21$

$= 0$

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

Answer: $f(3) = 0$, so $x - 3$ is a factor

$5 \times 12 = 60 = 3 \times 20$ and $3 + 20 = 23$

$5x^{2} + 23x + 12 = 5x^{2} + 3x + 20x + 12$

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

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

Answer: $(5x + 3)(x + 4)$

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

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

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

Let $f(x) = x^4 + px^3 + 26x^2 - 43x + q$; $x - 3$ and $x - 2$ are factors.

$f(3) = 0$:

$f(3) = (3)^{4} + p(3)^{3} + 26(3)^{2} - 43(3) + q$

$= 81 + 27p + 234 - 129 + q$

$= 27p + q + 186$

$\Rightarrow 27p + q = -186$ ... (1)

$f(2) = 0$:

$f(2) = (2)^{4} + p(2)^{3} + 26(2)^{2} - 43(2) + q$

$= 16 + 8p + 104 - 86 + q$

$= 8p + q + 34$

$\Rightarrow 8p + q = -34$ ... (2)

$(1) - (2)$: $19p = -152$

$\Rightarrow p = -8$

Put $p = -8$ in (1): $-216 + q = -186 \Rightarrow q = 30$

Answer: $p = -8, q = 30$

Divide $x^{3} - 18x^{2} + 107x - 210$ by $x - 5$:

$x^{3} \div x = x^{2}$; $x^{2} \times (x - 5) = x^{3} - 5x^{2}$; subtract: $-13x^{2} + 107x - 210$

$-13x^{2} \div x = -13x$; $(-13x) \times (x - 5) = -13x^{2} + 65x$; subtract: $42x - 210$

$42x \div x = 42$; $42 \times (x - 5) = 42x - 210$; subtract: $0$

$\therefore x^{3} - 18x^{2} + 107x - 210 = (x - 5)(x^{2} - 13x + 42)$

$x^{2} - 13x + 42 = x^{2} - 6x - 7x + 42$

$= x(x - 6) - 7(x - 6)$

$= (x - 6)(x - 7)$

$\therefore x^{3} - 18x^{2} + 107x - 210 = (x - 5)(x - 6)(x - 7)$

Answer: $(x - 5)(x - 6)(x - 7)$

Let $f(x) = x^{3} + 5x^{2} + 2x - 8$. By trial and error:

$f(1) = 1 + 5 + 2 - 8 = 0$

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

$x^{3} \div x = x^{2}$; $x^{2} \times (x - 1) = x^{3} - x^{2}$; subtract: $6x^{2} + 2x - 8$

$6x^{2} \div x = 6x$; $6x \times (x - 1) = 6x^{2} - 6x$; subtract: $8x - 8$

$8x \div x = 8$; $8 \times (x - 1) = 8x - 8$; subtract: $0$

$\therefore f(x) = (x - 1)(x^{2} + 6x + 8)$

$x^{2} + 6x + 8 = x^{2} + 2x + 4x + 8$

$= x(x + 2) + 4(x + 2)$

$= (x + 2)(x + 4)$

$\therefore f(x) = (x - 1)(x + 2)(x + 4)$

Answer: $(x - 1)(x + 2)(x + 4)$

$x + 1 = 0 \Rightarrow x = -1$

Write the coefficients and $-1$ on the left: $-1 \;\big|\; 1 \quad 4 \quad -2 \quad -2$

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

Second row: $0 \quad -1 \quad -3 \quad 5$

Bottom row: $1 \quad 3 \quad -5 \;\big|\; 3$

Quotient $= x^{2} + 3x - 5$, Remainder $= 3$

Answer: Quotient $= x^{2} + 3x - 5$, Remainder $= 3$

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

Here $n = 99$ is odd, so $23 + 18 = 41$ divides $23^{99} + 18^{99}$.

Remainder $= 0$

Answer: $0$

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

$f(0) = 5 \Rightarrow c = 5$ ... (1)

$f(1) = -3 \Rightarrow a + b + c = -3$ ... (2)

$f(2) = -7 \Rightarrow 4a + 2b + c = -7$ ... (3)

$(2) - (1)$: $a + b = -8$ ... (4)

$(3) - (2)$: $3a + b = -4$ ... (5)

$(5) - (4)$: $2a = 4 \Rightarrow a = 2$

From (4): $b = -10$; from (1): $c = 5$

$\therefore f(x) = 2x^{2} - 10x + 5$

Answer: $f(x) = 2x^{2} - 10x + 5$