Question paper, Set 9

(i) $-x^{4}y^{4}$: the powers of $x$ and $y$ are $4$ and $4$, non-negative integers, so it is a monomial.

(ii) $5\sqrt{x}y^{2}$: $\displaystyle \sqrt{x} = x^{\frac{1}{2}}$, the power $\displaystyle \frac{1}{2}$ is not a whole number, so it is not a monomial.

(iii) $\displaystyle \frac{5}{3}x^{2}y$: the powers of $x$ and $y$ are $2$ and $1$, non-negative integers, so it is a monomial.

(iv) $8x^{-4}y^{3}$: the power of $x$ is $-4$, a negative number, so it is not a monomial.

Answer: (i) and (iii) are monomials

The greatest power of $x$ in the polynomial is $8$.

So its degree is $8$.

Answer: $8$

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

$p(x) = 3x^{5} + 9x^{4} + 4x - 1$

Answer: $p(x) = 3x^{5} + 9x^{4} + 4x - 1$

$7x^{2} - 4x^{-2}$ is an algebraic expression with two terms, so it is a multinomial.

The power $-2$ of $x$ is negative, so it is not a polynomial.

Answer: It is a multinomial but not a polynomial

The degree of a quadratic polynomial is $2$.

So the highest power: $n-8 = 2$

$\Rightarrow n = 10$

Answer: $n = 10$

The highest power of $x$ in the equation is $4$.

So it is a bi-quadratic equation.

Answer: Bi-quadratic equation

$p(-2) = 3(-2)^{3} + 5(-2)^{2} - 7(-2) + 1$

$= -24 + 20 + 14 + 1$

$= 11$

Answer: $11$

$1 \times (-21) = -21 = (-3) \times 7$ and $(-3) + 7 = 4$

$x^{2} + 4x - 21 = x^{2} - 3x + 7x - 21$

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

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

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

$\Rightarrow x = -7, 3$

Answer: $x = -7, 3$

$a = 1, b = 7, c = 12$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{49 - 4(1)(12)}}{2(1)}$

$\displaystyle = \frac{-7 \pm \sqrt{1}}{2}$

$\displaystyle = \frac{-7 + 1}{2}, \frac{-7 - 1}{2} = -3, -4$

Answer: $x = -3, -4$

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

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

$3x \div x = 3$; $3 \times (x - 1) = 3x - 3$; subtract: $1$

Quotient $= -2x^{2} + x + 3$, Remainder $= 1$

Answer: Quotient $= -2x^{2} + x + 3$, Remainder $= 1$

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

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

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

$= 54 - 18 + 9 - 6$

$= 39$

Answer: $39$

$\displaystyle 1 - 2x = 0 \Rightarrow x = \frac{1}{2}$

Divisible means remainder $\displaystyle = f\left(\frac{1}{2}\right) = 0$

$\displaystyle f\left(\frac{1}{2}\right) = 8\left(\frac{1}{2}\right)^{3} - a\left(\frac{1}{2}\right)^{2} + 5\left(\frac{1}{2}\right) - 5$

$\displaystyle = 1 - \frac{1}{4}a + \frac{5}{2} - 5$

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

$\displaystyle -\frac{1}{4}a - \frac{3}{2} = 0$

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

$\Rightarrow a = -6$

Answer: $a = -6$

$-3x^{4} \div x^{2} = -3x^{2}$; $(-3x^{2}) \times (x^{2} + 1) = -3x^{4} - 3x^{2}$; subtract: $3x^{3} - 3x^{2} + 8x + 3$

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

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

Remainder $= 5x + 6$

Subtracting the remainder leaves an exact multiple: $5x + 6$ must be subtracted.

Answer: $5x + 6$

$x^{2} - 1 = (x + 1)(x - 1)$

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

$f(-1) = 0$:

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

$= 1 - p + 4 + q$

$= -p + q + 5$

$\Rightarrow p - q = 5$ ... (1)

$f(1) = 0$:

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

$= 1 + p - 4 + q$

$= p + q - 3$

$\Rightarrow p + q = 3$ ... (2)

$(1) + (2)$: $2p = 8$

$\Rightarrow p = 4$

Put $p = 4$ in (1): $4 - q = 5 \Rightarrow q = -1$

Answer: $p = 4, q = -1$

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

$a = 6, b = 13, c = -5$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-13 \pm \sqrt{169 - 4(6)(-5)}}{2(6)}$

$\displaystyle = \frac{-13 \pm \sqrt{289}}{12}$

$\displaystyle = \frac{-13 + 17}{12}, \frac{-13 - 17}{12} = \frac{1}{3}, -\frac{5}{2}$

$\displaystyle \therefore 6x^{2} + 13x - 5 = 6\left(x - \frac{1}{3}\right)\left(x + \frac{5}{2}\right)$

$= (2x + 5)(3x - 1)$

Answer: $(2x + 5)(3x - 1)$

Let $f(x) = 2x^{4} - 9x^{3} - 12x^{2} + 79x - 60$. By trial and error:

$f(1) = 2 - 9 - 12 + 79 - 60 = 0$

$f(-1) = 2 + 9 - 12 - 79 - 60 \ne 0$

$f(2) = 32 - 72 - 48 + 158 - 60 \ne 0$

$f(-2) = 32 + 72 - 48 - 158 - 60 \ne 0$

$f(3) = 162 - 243 - 108 + 237 - 60 \ne 0$

$f(-3) = 162 + 243 - 108 - 237 - 60 = 0$

$\Rightarrow x - 1$ and $x + 3$ are factors, so $(x - 1)(x + 3) = x^{2} + 2x - 3$ is a factor

$2x^{4} \div x^{2} = 2x^{2}$; $2x^{2} \times (x^{2} + 2x - 3) = 2x^{4} + 4x^{3} - 6x^{2}$; subtract: $-13x^{3} - 6x^{2} + 79x - 60$

$-13x^{3} \div x^{2} = -13x$; $(-13x) \times (x^{2} + 2x - 3) = -13x^{3} - 26x^{2} + 39x$; subtract: $20x^{2} + 40x - 60$

$20x^{2} \div x^{2} = 20$; $20 \times (x^{2} + 2x - 3) = 20x^{2} + 40x - 60$; subtract: $0$

$\therefore f(x) = (x^{2} + 2x - 3)(2x^{2} - 13x + 20)$

$2x^{2} - 13x + 20 = 2x^{2} - 5x - 8x + 20$

$= x(2x - 5) - 4(2x - 5)$

$= (2x - 5)(x - 4)$

$\therefore f(x) = (x - 1)(x + 3)(2x - 5)(x - 4)$

Answer: $(x - 1)(x + 3)(2x - 5)(x - 4)$

$x^{5} - 243 = x^{5} - 3^{5}$

$f(-3) = (-3)^{5} - 243 = -486$

So $x + 3$ is not a factor.

Answer: No, $f(-3) = -486$

Use $x^n + a^n = (x + a)(x^{n-1} - x^{n-2}a + \cdots + a^{n-1})$.

$243y^{5} + 1 = (3y)^{5} + 1^{5}$

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

$= (3y + 1)(81y^{4} - 27y^{3} + 9y^{2} - 3y + 1)$

Answer: $(3y + 1)(81y^{4} - 27y^{3} + 9y^{2} - 3y + 1)$

$\displaystyle \frac{1}{6}a^{3} \div \frac{2}{3}a = \frac{1}{4}a^{2}$; $\displaystyle \frac{1}{4}a^{2} \times (\frac{2}{3}a - \frac{1}{2}x) = \frac{1}{6}a^{3} - \frac{1}{8}a^{2}x$; subtract: $\displaystyle 2a^{2}x + \frac{7}{6}ax^{2} - 2x^{3}$

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

$\displaystyle \frac{8}{3}ax^{2} \div \frac{2}{3}a = 4x^{2}$; $\displaystyle 4x^{2} \times (\frac{2}{3}a - \frac{1}{2}x) = \frac{8}{3}ax^{2} - 2x^{3}$; subtract: $0$

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

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

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

$f(-3) = -2 \Rightarrow 9a - 3b + c = -2$ ... (1)

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

$f(0) = 1 \Rightarrow c = 1$ ... (3)

$(2) - (1)$, divide by $2$: $\displaystyle -4a + b = \frac{3}{2}$ ... (4)

$(3) - (2)$: $-a + b = 0$ ... (5)

$(5) - (4)$: $\displaystyle 3a = -\frac{3}{2} \Rightarrow a = -\frac{1}{2}$

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

$\displaystyle \therefore f(x) = -\frac{1}{2}x^{2} - \frac{1}{2}x + 1$

Answer: $\displaystyle f(x) = -\frac{1}{2}x^{2} - \frac{1}{2}x + 1$