Question paper, Set 1

The power of $x$ is $-1$, a negative number.

So $-2x^{-1}y$ is not a monomial.

Answer: No, the power of $x$ is $-1$, a negative number

Degrees of the terms: $4, 8, 3, 9, 0$

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 $3$, so the degree is $3$.

A polynomial of degree $3$ is a cubic polynomial.

Answer: Cubic polynomial

The degree of a quadratic polynomial is $2$.

So the highest power: $n-4 = 2$

$\Rightarrow n = 6$

Answer: $n = 6$

(i) $\displaystyle \frac{5}{x^2} = 5x^{-2}$ has a negative power, so it is not a quadratic equation.

(ii) Its degree is $3$, so it is a cubic equation.

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

Answer: (iii) only

$p(-2) = 7(-2)^{2} + 6(-2) - 11$

$= 28 - 12 - 11$

$= 5$

Answer: $5$

$\displaystyle p\left(-\frac{1}{2}\right) = 16\left(-\frac{1}{2}\right)^{3} + a\left(-\frac{1}{2}\right)^{2} - 2\left(-\frac{1}{2}\right) + 6$

$\displaystyle = -2 + \frac{1}{4}a + 1 + 6$

$\displaystyle = \frac{1}{4}a + 5$

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

$\displaystyle \frac{1}{4}a + 5 = 0$

$\displaystyle \Rightarrow \frac{1}{4}a = -5$

$\Rightarrow a = -20$

Answer: $a = -20$

$5 \times (-4) = -20 = 4 \times (-5)$ and $4 + (-5) = -1$

$5x^{2} - x - 4 = 5x^{2} + 4x - 5x - 4$

$= x(5x + 4) - 1(5x + 4)$

$= (5x + 4)(x - 1)$

$\Rightarrow 5x + 4 = 0$ (or) $x - 1 = 0$

$\displaystyle \Rightarrow x = -\frac{4}{5}, 1$

Answer: $\displaystyle x = -\frac{4}{5}, 1$

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

$\Delta = b^2 - 4ac = 3^2 - 4(5)(-10)$

$= 9 + 200 = 209$

Answer: $\Delta = 209$

$4x^{4} \div x^{2} = 4x^{2}$; $4x^{2} \times (x^{2} + x - 5) = 4x^{4} + 4x^{3} - 20x^{2}$; subtract: $-x^{3} - x^{2} - 6$

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

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

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

Let $f(x) = ax^{3} - 4x^{2} - x + 9$ and $g(x) = 2x^{3} - 4x + a$.

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

$f(-1) = a(-1)^{3} - 4(-1)^{2} - (-1) + 9$

$= -a - 4 + 1 + 9$

$= -a + 6$

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

$= -2 + 4 + a$

$= a + 2$

$-a + 6 = a + 2$

$\Rightarrow -2a = -4$

$\Rightarrow a = 2$

Answer: $a = 2$

Let $f(x) = x^{3} - 5x^{2} + 4x - 20$; $x - 5 = 0 \Rightarrow x = 5$

$f(5) = (5)^{3} - 5(5)^{2} + 4(5) - 20$

$= 125 - 125 + 20 - 20$

$= 0$

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

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

$15 \times (-24) = -360 = 9 \times (-40)$ and $9 + (-40) = -31$

$15x^{2} - 31x - 24 = 15x^{2} + 9x - 40x - 24$

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

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

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

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

$= x(x - 1) + 3(x - 1)$

$= (x - 1)(x + 3)$

Let $f(x) = x^4 + px^3 - x^2 - 24x + q$; $x - 1$ and $x + 3$ are factors.

$f(1) = 0$:

$f(1) = (1)^{4} + p(1)^{3} - (1)^{2} - 24(1) + q$

$= 1 + p - 1 - 24 + q$

$= p + q - 24$

$\Rightarrow p + q = 24$ ... (1)

$f(-3) = 0$:

$f(-3) = (-3)^{4} + p(-3)^{3} - (-3)^{2} - 24(-3) + q$

$= 81 - 27p - 9 + 72 + q$

$= -27p + q + 144$

$\Rightarrow 27p - q = 144$ ... (2)

$(1) + (2)$: $28p = 168$

$\Rightarrow p = 6$

Put $p = 6$ in (1): $6 + q = 24 \Rightarrow q = 18$

Answer: $p = 6, q = 18$

Divide $x^{3} - 9x^{2} + 8x + 60$ by $x - 6$:

$x^{3} \div x = x^{2}$; $x^{2} \times (x - 6) = x^{3} - 6x^{2}$; subtract: $-3x^{2} + 8x + 60$

$-3x^{2} \div x = -3x$; $(-3x) \times (x - 6) = -3x^{2} + 18x$; subtract: $-10x + 60$

$-10x \div x = -10$; $(-10) \times (x - 6) = -10x + 60$; subtract: $0$

$\therefore x^{3} - 9x^{2} + 8x + 60 = (x - 6)(x^{2} - 3x - 10)$

$x^{2} - 3x - 10 = x^{2} + 2x - 5x - 10$

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

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

$\therefore x^{3} - 9x^{2} + 8x + 60 = (x - 6)(x + 2)(x - 5)$

Answer: $(x - 6)(x + 2)(x - 5)$

Let $f(x) = x^{3} + 2x^{2} - 23x - 60$. By trial and error:

$f(1) = 1 + 2 - 23 - 60 \ne 0$

$f(-1) = -1 + 2 + 23 - 60 \ne 0$

$f(2) = 8 + 8 - 46 - 60 \ne 0$

$f(-2) = -8 + 8 + 46 - 60 \ne 0$

$f(3) = 27 + 18 - 69 - 60 \ne 0$

$f(-3) = -27 + 18 + 69 - 60 = 0$

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

$x^{3} \div x = x^{2}$; $x^{2} \times (x + 3) = x^{3} + 3x^{2}$; subtract: $-x^{2} - 23x - 60$

$-x^{2} \div x = -x$; $(-x) \times (x + 3) = -x^{2} - 3x$; subtract: $-20x - 60$

$-20x \div x = -20$; $(-20) \times (x + 3) = -20x - 60$; subtract: $0$

$\therefore f(x) = (x + 3)(x^{2} - x - 20)$

$x^{2} - x - 20 = x^{2} + 4x - 5x - 20$

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

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

$\therefore f(x) = (x + 3)(x + 4)(x - 5)$

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

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

The missing power of $x$ gets the coefficient $0$.

Write the coefficients and $3$ on the left: $3 \;\big|\; 5 \quad 0 \quad -1 \quad 5$

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

Second row: $0 \quad 15 \quad 45 \quad 132$

Bottom row: $5 \quad 15 \quad 44 \;\big|\; 137$

Quotient $= 5x^{2} + 15x + 44$, Remainder $= 137$

Answer: Quotient $= 5x^{2} + 15x + 44$, Remainder $= 137$

$x^{3} + 27 = x^{3} + 3^{3}$

$f(-3) = (-3)^{3} + 27 = 0$

So $x + 3$ is a factor.

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

Answer: Yes, $f(-3) = 0$

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

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

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

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

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

$(3) - (2)$: $-3a + b = 16$ ... (5)

$(5) - (4)$: $2a = 20 \Rightarrow a = 10$

From (4): $b = 46$; from (1): $c = 45$

$\therefore f(x) = 10x^{2} + 46x + 45$

Answer: $f(x) = 10x^{2} + 46x + 45$