Question paper, Set 5

$7x^{4}y^{3} + 3xy^{2} + x^{3}y^{2}$ has three terms: $7x^{4}y^{3}$, $3xy^{2}$, $x^{3}y^{2}$.

An expression with three terms is a trinomial.

Answer: Trinomial

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

$= 3 + 1 + 3 = 7$

Answer: $7$

In standard form the powers of $x$ go down: $p(x) = -x^{4} - 9x^{3} - 4$

Highest power $= 4$, so $n = 4$; its coefficient $a_4 = -1$

Constant term $a_0 = -4$

Answer: $p(x) = -x^{4} - 9x^{3} - 4$; $n = 4$, $a_4 = -1$, $a_0 = -4$

In $\displaystyle -\frac{7}{3}x^{2} + 6x^{4} + \frac{7}{4}$, all powers of $x$ are whole numbers.

So it is a polynomial.

Answer: Yes, all powers of $x$ are whole numbers

The degree of a quadratic polynomial is $2$.

So the highest power: $n-5 = 2$

$\Rightarrow n = 7$

Answer: $n = 7$

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

So it is a bi-quadratic equation.

Answer: Bi-quadratic equation

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

$= -54 - 9 - 3 + 6$

$= -60$

Answer: $-60$

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

$= 1 + 3 - 4$

$= 0$

So $1$ is a zero of $p(x)$.

Answer: Yes, $p(1) = 0$

Let $p(x) = x^{2} + 9x + 18$.

$p(-6) = (-6)^{2} + 9(-6) + 18$

$= 36 - 54 + 18$

$= 0$

$p(6) = (6)^{2} + 9(6) + 18$

$= 36 + 54 + 18$

$= 108$

$-6$ is a root; $6$ is not a root ($p(6) = 108 \ne 0$).

Answer: $-6$ is a root; $6$ is not a root ($p(6) = 108 \ne 0$)

$a = 1, b = 6, c = 7$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{36 - 4(1)(7)}}{2(1)}$

$\displaystyle = \frac{-6 \pm \sqrt{8}}{2}$

$\displaystyle = \frac{-6 \pm 2\sqrt{2}}{2}$

$= -3 \pm \sqrt{2}$

$= -3 + \sqrt{2}, -3 - \sqrt{2}$

Answer: $x = -3 + \sqrt{2}, -3 - \sqrt{2}$

$4x^{3} \div x = 4x^{2}$; $4x^{2} \times (x - 4) = 4x^{3} - 16x^{2}$; subtract: $3x^{2} - 13x + 2$

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

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

Quotient $= 4x^{2} + 3x - 1$, Remainder $= -2$

Answer: Quotient $= 4x^{2} + 3x - 1$, Remainder $= -2$

$\displaystyle 2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$

Remainder $\displaystyle = f\left(-\frac{3}{2}\right)$

$\displaystyle f\left(-\frac{3}{2}\right) = 8\left(-\frac{3}{2}\right)^{3} - 12\left(-\frac{3}{2}\right)^{2} + 8\left(-\frac{3}{2}\right) - 6$

$= -27 - 27 - 12 - 6$

$= -72$

Answer: $-72$

Let $f(x) = ax^{3} + bx^{2} - 6x + 36$.

$x + 3$ is a factor $\Rightarrow f(-3) = 0$

$f(-3) = a(-3)^{3} + b(-3)^{2} - 6(-3) + 36$

$= -27a + 9b + 18 + 36$

$= -27a + 9b + 54$

$\Rightarrow 3a - b = 6$ ... (1)

Remainder on dividing by $x - 3$ is $18$ $\Rightarrow f(3) = 18$

$f(3) = a(3)^{3} + b(3)^{2} - 6(3) + 36$

$= 27a + 9b - 18 + 36$

$= 27a + 9b + 18$

$\Rightarrow 3a + b = 0$ ... (2)

$(1) + (2)$: $6a = 6$

$\Rightarrow a = 1$

Put $a = 1$ in (1): $3 - b = 6 \Rightarrow b = -3$

Answer: $a = 1, b = -3$

$12 \times 15 = 180 = 9 \times 20$ and $9 + 20 = 29$

$12x^{2} + 29x + 15 = 12x^{2} + 9x + 20x + 15$

$= 3x(4x + 3) + 5(4x + 3)$

$= (4x + 3)(3x + 5)$

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

$-3x^{3} \div x^{2} = -3x$; $(-3x) \times (x^{2} + 4x + 3) = -3x^{3} - 12x^{2} - 9x$; subtract: $-7x^{2} - 29x - 31$

$-7x^{2} \div x^{2} = -7$; $(-7) \times (x^{2} + 4x + 3) = -7x^{2} - 28x - 21$; subtract: $-x - 10$

Remainder $= -x - 10$

So $-(-x - 10) = x + 10$ must be added.

Answer: $x + 10$

$a = 3, b = 11, c = 1$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-11 \pm \sqrt{121 - 4(3)(1)}}{2(3)}$

$\displaystyle = \frac{-11 \pm \sqrt{109}}{6}$

$\displaystyle = \frac{-11 + \sqrt{109}}{6}, \frac{-11 - \sqrt{109}}{6}$

$\therefore 3x^{2} + 11x + 1 = a(x - \alpha)(x - \beta)$

$\displaystyle = 3\left[x - \left(\frac{-11 + \sqrt{109}}{6}\right)\right]\left[x - \left(\frac{-11 - \sqrt{109}}{6}\right)\right]$

Answer: $\displaystyle 3\left[x - \left(\frac{-11 + \sqrt{109}}{6}\right)\right]\left[x - \left(\frac{-11 - \sqrt{109}}{6}\right)\right]$

$4z^{3} \div 2z = 2z^{2}$; $2z^{2} \times (2z - 1) = 4z^{3} - 2z^{2}$; subtract: $26z^{2} + 17z - 15$

$26z^{2} \div 2z = 13z$; $13z \times (2z - 1) = 26z^{2} - 13z$; subtract: $30z - 15$

$30z \div 2z = 15$; $15 \times (2z - 1) = 30z - 15$; subtract: $0$

$\therefore 4z^{3} + 24z^{2} + 17z - 15 = (2z - 1)(2z^{2} + 13z + 15)$

$2 \times 15 = 30 = 3 \times 10$ and $3 + 10 = 13$

$2z^{2} + 13z + 15 = 2z^{2} + 3z + 10z + 15$

$= z(2z + 3) + 5(2z + 3)$

$= (2z + 3)(z + 5)$

$\therefore 4z^{3} + 24z^{2} + 17z - 15 = (2z - 1)(2z + 3)(z + 5)$

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

Let $f(x) = 3x^{3} - 4x^{2} - 9x + 10$.

$f(1) = 3 - 4 - 9 + 10 = 0$

$\therefore x - 1$ is a factor

Write the coefficients and $1$ on the left: $1 \;\big|\; 3 \quad -4 \quad -9 \quad 10$

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

Second row: $0 \quad 3 \quad -1 \quad -10$

Bottom row: $3 \quad -1 \quad -10 \;\big|\; 0$

$\therefore f(x) = (x - 1)(3x^{2} - x - 10)$

$3x^{2} - x - 10 = 3x^{2} + 5x - 6x - 10$

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

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

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

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

$x^{7} - 128 = x^{7} - 2^{7}$

$f(2) = (2)^{7} - 128 = 0$

So $x - 2$ is a factor.

Answer: Yes, $f(2) = 0$

$\displaystyle \frac{9}{2}a^{3} \div \frac{3}{2}a = 3a^{2}$; $\displaystyle 3a^{2} \times (\frac{3}{2}a - x) = \frac{9}{2}a^{3} - 3a^{2}x$; subtract: $6a^{2}x + 2ax^{2} - 4x^{3}$

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

$\displaystyle 6ax^{2} \div \frac{3}{2}a = 4x^{2}$; $\displaystyle 4x^{2} \times (\frac{3}{2}a - x) = 6ax^{2} - 4x^{3}$; subtract: $0$

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

Answer: Quotient $= 3a^{2} + 4ax + 4x^{2}$