Question paper, Set 3

(i) $-8\sqrt{x}y^{2}$: $\displaystyle \sqrt{x} = x^{\frac{1}{2}}$, the power $\displaystyle \frac{1}{2}$ is not a whole number, so it is not a monomial.

(ii) $\displaystyle \frac{7}{3}x^{4}$: the powers of $x$ are $4$, non-negative integers, so it is a monomial.

(iii) $\displaystyle 2x^{\frac{1}{3}}y$: the power $\displaystyle \frac{1}{3}$ of $x$ is not a whole number, so it is not a monomial.

(iv) $-4x^{2}y^{4}$: the powers of $x$ and $y$ are $2$ and $4$, non-negative integers, so it is a monomial.

Answer: (ii) and (iv) are monomials

The greatest power of $x$ in the polynomial is $5$.

So its degree is $5$.

Answer: $5$

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

$p(x) = 5x^{3} - 9x^{2} + 2x + 7$

Answer: $p(x) = 5x^{3} - 9x^{2} + 2x + 7$

$\displaystyle 4\sqrt{2}x - 6x^{\frac{2}{3}} + 2$ is an algebraic expression with three terms, so it is a multinomial.

The power $\displaystyle \frac{2}{3}$ of $x$ is not a whole number, so it is not a polynomial.

Answer: It is a multinomial but not a polynomial

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

So the highest power: $n-3 = 4$

$\Rightarrow n = 7$

Answer: $n = 7$

The highest power of $x$ in the equation is $3$.

So it is a cubic equation.

Answer: Cubic equation

$p(-3) = 5(-3)^{3} + 2(-3)^{2} + (-3) + 8$

$= -135 + 18 - 3 + 8$

$= -112$

Answer: $-112$

$1 \times 24 = 24 = (-4) \times (-6)$ and $(-4) + (-6) = -10$

$x^{2} - 10x + 24 = x^{2} - 4x - 6x + 24$

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

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

$\Rightarrow x - 4 = 0$ (or) $x - 6 = 0$

$\Rightarrow x = 4, 6$

Answer: $x = 4, 6$

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

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{9 - 4(2)(1)}}{2(2)}$

$\displaystyle = \frac{3 \pm \sqrt{1}}{4}$

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

Answer: $\displaystyle x = 1, \frac{1}{2}$

$x^{3} \div x = x^{2}$; $x^{2} \times (x + 2) = x^{3} + 2x^{2}$; subtract: $-x^{2} - 6x - 11$

$-x^{2} \div x = -x$; $(-x) \times (x + 2) = -x^{2} - 2x$; subtract: $-4x - 11$

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

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

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

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

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

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

$= 36 - 12 + 4$

$= 28$

Answer: $28$

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

Divisible means remainder $= f(1) = 0$

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

$= 1 - 6 + 6 - a + 3$

$= -a + 4$

$-a + 4 = 0$

$\Rightarrow -a = -4$

$\Rightarrow a = 4$

Answer: $a = 4$

$-4x^{4} \div x^{2} = -4x^{2}$; $(-4x^{2}) \times (x^{2} - 3x - 2) = -4x^{4} + 12x^{3} + 8x^{2}$; subtract: $-6x^{3} + 17x^{2} + 21x + 5$

$-6x^{3} \div x^{2} = -6x$; $(-6x) \times (x^{2} - 3x - 2) = -6x^{3} + 18x^{2} + 12x$; subtract: $-x^{2} + 9x + 5$

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

Remainder $= 6x + 3$

Subtracting the remainder leaves an exact multiple: $6x + 3$ must be subtracted.

Answer: $6x + 3$

$x^{2} - 4 = (x + 2)(x - 2)$

Let $f(x) = x^4 + px^3 - 8x^2 + 12x + q$; $x + 2$ and $x - 2$ are factors.

$f(-2) = 0$:

$f(-2) = (-2)^{4} + p(-2)^{3} - 8(-2)^{2} + 12(-2) + q$

$= 16 - 8p - 32 - 24 + q$

$= -8p + q - 40$

$\Rightarrow 8p - q = -40$ ... (1)

$f(2) = 0$:

$f(2) = (2)^{4} + p(2)^{3} - 8(2)^{2} + 12(2) + q$

$= 16 + 8p - 32 + 24 + q$

$= 8p + q + 8$

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

$(1) + (2)$: $16p = -48$

$\Rightarrow p = -3$

Put $p = -3$ in (1): $-24 - q = -40 \Rightarrow q = 16$

Answer: $p = -3, q = 16$

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

$a = 15, b = 49, c = 40$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-49 \pm \sqrt{2401 - 4(15)(40)}}{2(15)}$

$\displaystyle = \frac{-49 \pm \sqrt{1}}{30}$

$\displaystyle = \frac{-49 + 1}{30}, \frac{-49 - 1}{30} = -\frac{8}{5}, -\frac{5}{3}$

$\displaystyle \therefore 15x^{2} + 49x + 40 = 15\left(x + \frac{8}{5}\right)\left(x + \frac{5}{3}\right)$

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

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

Let $f(x) = 2x^{3} + 7x^{2} - 3x - 18$. By trial and error:

$f(1) = 2 + 7 - 3 - 18 \ne 0$

$f(-1) = -2 + 7 + 3 - 18 \ne 0$

$f(2) = 16 + 28 - 6 - 18 \ne 0$

$f(-2) = -16 + 28 + 6 - 18 = 0$

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

$2x^{3} \div x = 2x^{2}$; $2x^{2} \times (x + 2) = 2x^{3} + 4x^{2}$; subtract: $3x^{2} - 3x - 18$

$3x^{2} \div x = 3x$; $3x \times (x + 2) = 3x^{2} + 6x$; subtract: $-9x - 18$

$-9x \div x = -9$; $(-9) \times (x + 2) = -9x - 18$; subtract: $0$

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

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

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

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

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

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

$x^{3} - 64 = x^{3} - 4^{3}$

$f(4) = (4)^{3} - 64 = 0$

So $x - 4$ is a factor.

Answer: Yes, $f(4) = 0$

$x - a$ is a factor of $x^n - a^n$ for every positive integer $n$: so $27 - 11 = 16$ divides it.

$x + a$ is also a factor of $x^n - a^n$ when $n$ is even.

Here $n = 120$ is even, so $27 + 11 = 38$ also divides it.

Answer: Divisible by $16$ and by $38$

$\displaystyle \frac{8}{27}a^{3} \div \frac{2}{3}a = \frac{4}{9}a^{2}$; $\displaystyle \frac{4}{9}a^{2} \times (\frac{2}{3}a + 2x) = \frac{8}{27}a^{3} + \frac{8}{9}a^{2}x$; subtract: $\displaystyle \frac{16}{9}a^{2}x + 8ax^{2} + 8x^{3}$

$\displaystyle \frac{16}{9}a^{2}x \div \frac{2}{3}a = \frac{8}{3}ax$; $\displaystyle \frac{8}{3}ax \times (\frac{2}{3}a + 2x) = \frac{16}{9}a^{2}x + \frac{16}{3}ax^{2}$; subtract: $\displaystyle \frac{8}{3}ax^{2} + 8x^{3}$

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

Remainder $= 0$; Quotient $\displaystyle = \frac{4}{9}a^{2} + \frac{8}{3}ax + 4x^{2}$

Answer: Quotient $\displaystyle = \frac{4}{9}a^{2} + \frac{8}{3}ax + 4x^{2}$

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

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

$f(2) = 10 \Rightarrow 4a + 2b + c = 10$ ... (2)

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

$(2) - (1)$, divide by $3$: $a + b = 4$ ... (4)

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

$(5) - (4)$: $4a = -12 \Rightarrow a = -3$

From (4): $b = 7$; from (1): $c = 8$

$\therefore f(x) = -3x^{2} + 7x + 8$

Answer: $f(x) = -3x^{2} + 7x + 8$