Question paper, Set 8

$-7x^{3}y^{3}$ has one term: $-7x^{3}y^{3}$.

An expression with one term is a monomial.

Answer: Monomial

Degree of a monomial = sum of the exponents of its variables

$= 1 + 3 + 2 = 6$

Answer: $6$

In standard form the powers of $x$ go down: $p(x) = 8x^{5} - 3x^{4} - 8x^{2} - 8x - 5$

Highest power $= 5$, so $n = 5$; its coefficient $a_5 = 8$

Constant term $a_0 = -5$

Answer: $p(x) = 8x^{5} - 3x^{4} - 8x^{2} - 8x - 5$; $n = 5$, $a_5 = 8$, $a_0 = -5$

In $4x^{3} + 6x^{-1}$, the power $-1$ of $x$ is negative.

So it is not a polynomial.

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

In a zero polynomial all the coefficients are zero.

$a - 7 = 0 \Rightarrow a = 7$

$b - a = 0 \Rightarrow b - 7 = 0 \Rightarrow b = 7$

$c + 2b = 0 \Rightarrow c + 14 = 0 \Rightarrow c = -14$

Answer: $a = 7, b = 7, c = -14$

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

So it is a cubic equation.

Answer: Cubic equation

$p(-2) = 4(-2)^{3} - 4(-2)^{2} + 7(-2) - 5$

$= -32 - 16 - 14 - 5$

$= -67$

Answer: $-67$

$-6x - 4 = 0$

$\Rightarrow -6x = 4$

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

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

Let $p(x) = x^{2} + x - 72$.

$p(8) = (8)^{2} + (8) - 72$

$= 64 + 8 - 72$

$= 0$

$p(-9) = (-9)^{2} + (-9) - 72$

$= 81 - 9 - 72$

$= 0$

Both $8$ and $-9$ are roots.

Answer: Both $8$ and $-9$ are roots

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

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

$\displaystyle = \frac{3 \pm \sqrt{33}}{6}$

$\displaystyle = \frac{3 + \sqrt{33}}{6}, \frac{3 - \sqrt{33}}{6}$

Answer: $\displaystyle x = \frac{3 + \sqrt{33}}{6}, \frac{3 - \sqrt{33}}{6}$

$x^{3} \div x = x^{2}$; $x^{2} \times (x + 1) = x^{3} + x^{2}$; subtract: $6x^{2} + 12x + 4$

$6x^{2} \div x = 6x$; $6x \times (x + 1) = 6x^{2} + 6x$; subtract: $6x + 4$

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

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

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

$\displaystyle 3x + 4 = 0 \Rightarrow x = -\frac{4}{3}$

Remainder $\displaystyle = f\left(-\frac{4}{3}\right)$

$\displaystyle f\left(-\frac{4}{3}\right) = 81\left(-\frac{4}{3}\right)^{3} + 9\left(-\frac{4}{3}\right)^{2} + 6\left(-\frac{4}{3}\right) + 7$

$= -192 + 16 - 8 + 7$

$= -177$

Answer: $-177$

Let $f(x) = ax^{3} + bx^{2} + 2x - 1$.

$x - 1$ is a factor $\Rightarrow f(1) = 0$

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

$= a + b + 2 - 1$

$= a + b + 1$

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

Remainder on dividing by $x - 2$ is $-17$ $\Rightarrow f(2) = -17$

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

$= 8a + 4b + 4 - 1$

$= 8a + 4b + 3$

$\Rightarrow 2a + b = -5$ ... (2)

$(1) - (2)$: $-a = 4$

$\Rightarrow a = -4$

Put $a = -4$ in (1): $-4 + b = -1 \Rightarrow b = 3$

Answer: $a = -4, b = 3$

$8 \times (-21) = -168 = (-6) \times 28$ and $(-6) + 28 = 22$

$8x^{2} + 22x - 21 = 8x^{2} - 6x + 28x - 21$

$= 2x(4x - 3) + 7(4x - 3)$

$= (4x - 3)(2x + 7)$

Answer: $(4x - 3)(2x + 7)$

$3x^{3} \div x^{2} = 3x$; $3x \times (x^{2} + 3x + 1) = 3x^{3} + 9x^{2} + 3x$; subtract: $-7x^{2} - 30x$

$-7x^{2} \div x^{2} = -7$; $(-7) \times (x^{2} + 3x + 1) = -7x^{2} - 21x - 7$; subtract: $-9x + 7$

Remainder $= -9x + 7$

Subtracting the remainder leaves an exact multiple: $-9x + 7$ must be subtracted.

Answer: $-9x + 7$

$a = 1, b = 2, c = -6$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{4 - 4(1)(-6)}}{2(1)}$

$\displaystyle = \frac{-2 \pm \sqrt{28}}{2}$

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

$= -1 \pm \sqrt{7}$

$= -1 + \sqrt{7}, -1 - \sqrt{7}$

$\therefore x^{2} + 2x - 6 = a(x - \alpha)(x - \beta)$

$= \left[x - \left(-1 + \sqrt{7}\right)\right]\left[x - \left(-1 - \sqrt{7}\right)\right]$

$= (x + 1 - \sqrt{7})(x + 1 + \sqrt{7})$

Answer: $(x + 1 - \sqrt{7})(x + 1 + \sqrt{7})$

$4y^{3} \div 2y = 2y^{2}$; $2y^{2} \times (2y + 9) = 4y^{3} + 18y^{2}$; subtract: $10y^{2} + 49y + 18$

$10y^{2} \div 2y = 5y$; $5y \times (2y + 9) = 10y^{2} + 45y$; subtract: $4y + 18$

$4y \div 2y = 2$; $2 \times (2y + 9) = 4y + 18$; subtract: $0$

$\therefore 4y^{3} + 28y^{2} + 49y + 18 = (2y + 9)(2y^{2} + 5y + 2)$

$2 \times 2 = 4 = 1 \times 4$ and $1 + 4 = 5$

$2y^{2} + 5y + 2 = 2y^{2} + y + 4y + 2$

$= y(2y + 1) + 2(2y + 1)$

$= (2y + 1)(y + 2)$

$\therefore 4y^{3} + 28y^{2} + 49y + 18 = (2y + 9)(2y + 1)(y + 2)$

Answer: $(2y + 9)(2y + 1)(y + 2)$

Let $f(x) = 3x^{3} + x^{2} - 10x - 8$.

$f(1) = 3 + 1 - 10 - 8 \ne 0$

$f(-1) = -3 + 1 + 10 - 8 = 0$

$\therefore x + 1$ is a factor

Write the coefficients and $-1$ on the left: $-1 \;\big|\; 3 \quad 1 \quad -10 \quad -8$

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

Second row: $0 \quad -3 \quad 2 \quad 8$

Bottom row: $3 \quad -2 \quad -8 \;\big|\; 0$

$\therefore f(x) = (x + 1)(3x^{2} - 2x - 8)$

$3x^{2} - 2x - 8 = 3x^{2} + 4x - 6x - 8$

$= x(3x + 4) - 2(3x + 4)$

$= (3x + 4)(x - 2)$

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

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

$x - a$ is a factor of $x^n - a^n$ for every positive integer $n$.

So $30 - 7 = 23$ divides $30^{142} - 7^{142}$.

Remainder $= 0$

Answer: $0$

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

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

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

Remainder $= 0$; Quotient $= a^{2} + 4ax - x^{2}$

Answer: Quotient $= a^{2} + 4ax - x^{2}$