Question paper, Set 7

$\displaystyle \sqrt{x} = x^{\frac{1}{2}}$, the power $\displaystyle \frac{1}{2}$ is not a whole number.

So $2\sqrt{x}y^{3}$ is not a monomial.

Answer: No, $\displaystyle \sqrt{x} = x^{\frac{1}{2}}$, the power $\displaystyle \frac{1}{2}$ is not a whole number

Degrees of the terms: $6, 2, 8, 0$

Greatest among them $= 8$

So the degree of the polynomial is $8$.

Answer: $8$

The term with $x$ in $p(x)$ is $-2x$.

So the coefficient of $x$ is $-2$.

Answer: $-2$

A cubic polynomial has degree $3$.

General form: $ay^3 + by^2 + cy + d$, where $a \ne 0$

Answer: $ay^3 + by^2 + cy + d$, where $a \ne 0$

The degree of a quadratic polynomial is $2$.

So the highest power: $n-7 = 2$

$\Rightarrow n = 9$

Answer: $n = 9$

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

(ii) $\displaystyle \frac{5}{x^2} = 5x^{-2}$ 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: (i) and (iii)

$p(-1) = 5(-1)^{2} - 8(-1) + 11$

$= 5 + 8 + 11$

$= 24$

Answer: $24$

$p(-2) = 2(-2)^{3} + a(-2)^{2} - 5(-2) - 6$

$= -16 + 4a + 10 - 6$

$= 4a - 12$

Given $p(-2) = 0$:

$4a - 12 = 0$

$\Rightarrow 4a = 12$

$\Rightarrow a = 3$

Answer: $a = 3$

$3 \times (-6) = -18 = (-2) \times 9$ and $(-2) + 9 = 7$

$3x^{2} + 7x - 6 = 3x^{2} - 2x + 9x - 6$

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

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

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

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

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

$a = 3, b = -1, c = -2$

$\Delta = b^2 - 4ac = (-1)^2 - 4(3)(-2)$

$= 1 + 24 = 25$

Answer: $\Delta = 25$

$-2x^{4} \div x^{2} = -2x^{2}$; $(-2x^{2}) \times (x^{2} - 1) = -2x^{4} + 2x^{2}$; subtract: $5x^{3} + 4x^{2} - 10x - 7$

$5x^{3} \div x^{2} = 5x$; $5x \times (x^{2} - 1) = 5x^{3} - 5x$; subtract: $4x^{2} - 5x - 7$

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

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

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

Let $f(x) = ax^{3} + 4x^{2} - 3x + 7$ and $g(x) = 2x^{3} + 4x + a$.

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

$f(2) = a(2)^{3} + 4(2)^{2} - 3(2) + 7$

$= 8a + 16 - 6 + 7$

$= 8a + 17$

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

$= 16 + 8 + a$

$= a + 24$

$8a + 17 = a + 24$

$\Rightarrow 7a = 7$

$\Rightarrow a = 1$

Answer: $a = 1$

Let $f(x) = x^{3} - 2x^{2} + 7x - 14$; $x - 2 = 0 \Rightarrow x = 2$

$f(2) = (2)^{3} - 2(2)^{2} + 7(2) - 14$

$= 8 - 8 + 14 - 14$

$= 0$

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

Answer: $f(2) = 0$, so $x - 2$ is a factor

$8\sqrt{3} \times (-5\sqrt{3}) = -120$; split $37x$ as $-3x$ and $40x$

$8\sqrt{3}x^{2} + 37x - 5\sqrt{3} = 8\sqrt{3}x^{2} - 3x + 40x - 5\sqrt{3}$

$= \sqrt{3}x(8x - \sqrt{3}) + 5(8x - \sqrt{3})$

$= (\sqrt{3}x + 5)(8x - \sqrt{3})$

Answer: $(\sqrt{3}x + 5)(8x - \sqrt{3})$

Let $f(x) = x^4 + px^3 - 17x^2 + 22x + q$; $x - 1$ and $x - 2$ are factors.

$f(1) = 0$:

$f(1) = (1)^{4} + p(1)^{3} - 17(1)^{2} + 22(1) + q$

$= 1 + p - 17 + 22 + q$

$= p + q + 6$

$\Rightarrow p + q = -6$ ... (1)

$f(2) = 0$:

$f(2) = (2)^{4} + p(2)^{3} - 17(2)^{2} + 22(2) + q$

$= 16 + 8p - 68 + 44 + q$

$= 8p + q - 8$

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

$(1) - (2)$: $-7p = -14$

$\Rightarrow p = 2$

Put $p = 2$ in (1): $2 + q = -6 \Rightarrow q = -8$

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

Divide $x^{3} - 3x^{2} - 16x - 12$ by $x + 1$:

$x^{3} \div x = x^{2}$; $x^{2} \times (x + 1) = x^{3} + x^{2}$; subtract: $-4x^{2} - 16x - 12$

$-4x^{2} \div x = -4x$; $(-4x) \times (x + 1) = -4x^{2} - 4x$; subtract: $-12x - 12$

$-12x \div x = -12$; $(-12) \times (x + 1) = -12x - 12$; subtract: $0$

$\therefore x^{3} - 3x^{2} - 16x - 12 = (x + 1)(x^{2} - 4x - 12)$

$x^{2} - 4x - 12 = x^{2} + 2x - 6x - 12$

$= x(x + 2) - 6(x + 2)$

$= (x + 2)(x - 6)$

$\therefore x^{3} - 3x^{2} - 16x - 12 = (x + 1)(x + 2)(x - 6)$

Answer: $(x + 1)(x + 2)(x - 6)$

Let $f(x) = 3x^{3} - 10x^{2} - x + 12$. By trial and error:

$f(1) = 3 - 10 - 1 + 12 \ne 0$

$f(-1) = -3 - 10 + 1 + 12 = 0$

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

$3x^{3} \div x = 3x^{2}$; $3x^{2} \times (x + 1) = 3x^{3} + 3x^{2}$; subtract: $-13x^{2} - x + 12$

$-13x^{2} \div x = -13x$; $(-13x) \times (x + 1) = -13x^{2} - 13x$; subtract: $12x + 12$

$12x \div x = 12$; $12 \times (x + 1) = 12x + 12$; subtract: $0$

$\therefore f(x) = (x + 1)(3x^{2} - 13x + 12)$

$3x^{2} - 13x + 12 = 3x^{2} - 4x - 9x + 12$

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

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

$\therefore f(x) = (x + 1)(3x - 4)(x - 3)$

Answer: $(x + 1)(3x - 4)(x - 3)$

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

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

Write the coefficients and $-1$ on the left: $-1 \;\big|\; 1 \quad 7 \quad 0 \quad -4$

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

Second row: $0 \quad -1 \quad -6 \quad 6$

Bottom row: $1 \quad 6 \quad -6 \;\big|\; 2$

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

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

$x^{4} + 16 = x^{4} + 2^{4}$

$f(-2) = (-2)^{4} + 16 = 32$

So $x + 2$ is not a factor.

Answer: No, $f(-2) = 32$

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

$f(-3) = -8 \Rightarrow 9a - 3b + c = -8$ ... (1)

$f(2) = -8 \Rightarrow 4a + 2b + c = -8$ ... (2)

$f(3) = -2 \Rightarrow 9a + 3b + c = -2$ ... (3)

$(2) - (1)$, divide by $5$: $-a + b = 0$ ... (4)

$(3) - (2)$: $5a + b = 6$ ... (5)

$(5) - (4)$: $6a = 6 \Rightarrow a = 1$

From (4): $b = 1$; from (1): $c = -14$

$\therefore f(x) = x^{2} + x - 14$

Answer: $f(x) = x^{2} + x - 14$