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Exercise 1.11

The figure here shows the Venn diagram for the special sets $$\mathbb{N}, \mathbb{N}_0$$ and $$\mathbb{Z}$$.

Where does the number $$\text{2,13}$$ belong in the diagram?

$$\text{2,13}$$ is in its simplest form, therefore it is not in $$\mathbb{N}$$, $$\mathbb{N}_0$$ or $$\mathbb{Z}$$. It is in the space between the rectangle and $$\mathbb{Z}$$

In the following list, there are two false statements and one true statement. Which of the statements is true?

• Every natural number is an integer.
• Every whole number is a natural number.
• There are fractions in the integers.

Consider each statement:

• Integers are natural numbers and negative natural numbers. Therefore this statement is true.
• $$\text{0}$$ is not a natural number, therefore this statement is false.
• Integers are natural numbers and negative natural numbers, no fractions. Therefore this is false.

The only true statement is (i).

State whether the following numbers are real, non-real or undefined.

$$-\sqrt{-5}$$

This is the square root of a negative number and so is non-real.

$$\dfrac{\sqrt{8}}{0}$$

We are dividing by 0 and so this is undefined.

$$-\sqrt{15}$$

This is the square root of a positive number and so is real.

$$-\sqrt{7}$$

This is the square root of a positive number and so is real.

$$\sqrt{-1}$$

This is the square root of a negative number and so is non-real.

$$\sqrt{2}$$

This is the square root of a positive number and so is real.

State whether each of the following numbers are rational or irrational.

$$\sqrt[3]{4}$$

Irrational. It cannot be simplified to a fraction of integers.

$$45\pi$$

Irrational. It cannot be simplified to a fraction of integers

$$\sqrt{9}$$

$$\sqrt{9} = 3$$

Rational. Can be simplified to an integer

$$\sqrt[3]{8}$$

$$\sqrt[3]{8} = 2$$

Rational. Can be simplified to an integers.

If $$a$$ is an integer, $$b$$ is an integer and $$c$$ is irrational, which of the following are rational numbers?

$$\dfrac{-b}{a}$$

We have a fraction of integers and so this is rational.

$$c \div c$$

When we divide a number by itself we get 1 and so this is rational.

$$\dfrac{a}{c}$$

We are dividing an integer by an irrational number and so this is irrational. However if $$a = 0$$ then the fraction is equal to 0 and the number is rational.

$$\dfrac{1}{c}$$

We are dividing an integer by an irrational number and so this is irrational.

Consider the following list of numbers:

$\sqrt[3]{26} \; ; \; \frac{3}{2} \; ; \; \sqrt{-24} \; ; \; \sqrt{39} \; ; \; \text{7,}\dot{1}\dot{1} \; ; \; \pi^2 \; ; \; \frac{\pi}{2} \; ; \; \text{7,12} \; ; \; -\sqrt{24} \; ; \; \frac{\sqrt{2}}{0} \; ; \; 3\pi \; ; \; \sqrt{78} \; ; \; 9 \; ; \; \pi$

Which of the numbers are non-real numbers?

Only $$\sqrt{-24}$$ is non-real as it is the square root of a negative number.

Without using a calculator, rank all the real numbers in ascending order.

We exclude $$\sqrt{-24}$$ from the list as it is non-real. We also exclude $$\frac{\sqrt{2}}{0}$$ as it is undefined. Then we note that:

• $$\sqrt[3]{26}$$ lies between 2 and 8
• $$\frac{3}{2} = \text{1,5}$$
• $$\sqrt{39}$$ lies between 6 and 7
• $$\pi^{2} \approx \text{9,8696}$$
• $$\frac{\pi}{2} \approx \text{1,5708}$$
• $$-\sqrt{24}$$ lies between $$-\text{4}$$ and $$-\text{5}$$
• $$3\pi \approx \text{9,4248}$$
• $$\sqrt{78}$$ lies between 8 and 9
• $$\pi \approx \text{3,1416}$$

Therefore the ordering is: $$-\sqrt{24} \; ; \; \frac{3}{2} \; ; \; \frac{\pi}{2} \; ; \; \sqrt[3]{26} \; ; \; \pi \; ; \; \sqrt{39} \; ; \; \text{7,}\dot{1}\dot{1} \; ; \; \text{7,12} \; ; \; \sqrt{78} \; ; \; 9 \; ; \; 3\pi \; ; \; \pi^2$$

Which of the numbers are irrational numbers?

Any number that cannot be written as a fraction of integers is irrational. Therefore $$-\sqrt{24} \; ; \; \frac{\pi}{2} \; ; \; \sqrt[3]{26} \; ; \; \pi \; ; \; \sqrt{39} \; ; \; \sqrt{78} \; ; \; 3\pi \; ; \; \pi^2$$ are all irrational.

Which of the numbers are rational numbers?

All numbers that can be written as a fraction of integers are rational numbers. Therefore $$\frac{3}{2} \; ; \; \text{7,}\dot{1}\dot{1} \; ; \; \text{7,12} \; ; \; 9$$ are all rational numbers.

Which of the numbers are integers?

Only $$9$$ is an integer.

Which of the numbers are undefined?

Any fraction that has a denominator of 0 is undefined, therefore only $$\frac{\sqrt{2}}{0}$$ is undefined.

Write each decimal as a simple fraction.

$$\text{0,12}$$

\begin{align*} \text{0,12} &= \frac{1}{10} + \frac{2}{100} \\ &= \frac{12}{100} \\ &= \frac{3}{25} \end{align*}

$$\text{0,006}$$

\begin{align*} \text{0,006} &= \frac{6}{\text{1 000}} \\ &= \frac{3}{\text{500}} \end{align*}
$$\text{4,}\overline{\text{14}}$$
\begin{align*} x &= \text{4,141414}\ldots \\ 100x &= \text{414,141414}\ldots \\ 100x - x &= (\text{414,141414}\ldots) - (\text{4,141414}\ldots) \\ \text{99}x &= \text{410} \\ \therefore x &= \frac{\text{410}}{\text{99}} \end{align*}

$$\text{1,59}$$

\begin{align*} \text{1,59} &= 1 + \frac{5}{10} + \frac{9}{100} \\ &= 1\frac{59}{100} \end{align*}

$$\text{12,27}\dot{7}$$

\begin{align*} x & = \text{12,27}\dot{7} \\ 10x &= \text{122,}\dot{7} \\ 100x & = \text{1 227,}\dot{7} \\ \therefore 100x - 10x & = 90x = \text{1 105} \\ \therefore x &= \frac{\text{1 105}}{90} \\ &= \frac{221}{18} \end{align*}
$$\text{0,8}\dot{\text{2}}$$
\begin{align*} \text{0,8} \dot{\text{2}} &= \text{0,8} \text{2 222,..} \\ x &= \text{0,8222...} \\ 10x &= \text{8,222...} \\ 100x &= \text{82,222...} \\ 100x -10x& = \text{82,222} - \text{8,222...} \\ \text{90}x & = \text{74,000} \\ \text{90}x & = \text{74} \\ \therefore x & = \frac{\text{37}}{\text{45}} \end{align*}
$$\text{7,}\overline{\text{36}}$$
\begin{align*} x &= \text{7,363636...} \\ 100x &= \text{736,363636...} \\ 100x - x &= (\text{736,363636...}) - (\text{7,363636...}) \\ \text{99}x &= \text{729} \\ \therefore x &= \frac{\text{81}}{\text{11}} \end{align*}

Show that the decimal $$\text{3,21}\dot{1}\dot{8}$$ is a rational number.

\begin{align*} x & = \text{3,21}\dot{1}\dot{8}\\ \text{1 000}x & = \text{32 118,}\overline{18} \\ \therefore \text{10 000}x - x & = \text{9 999}x = \text{32 115} \\ \therefore x & = \frac{\text{32 115}}{\text{9 999}} \end{align*}

This is a rational number because both the numerator and denominator are integers.

Write the following fractions as decimal numbers:

$$\dfrac{\text{1}}{\text{18}}$$
\begin{align*} \text{18}|\overline{\text{1,0000}} &= \text{0} \text{ remainder } \text{0} \\ \text{18}|\overline{\text{1,}^1\text{0 000}} &= \text{0} \text{ remainder } \text{0} \\ \text{18}|\overline{\text{1,}^{10}\text{0}^1\text{000}} &= \text{5} \text{ remainder } \text{10} \\ \text{18}|\overline{\text{1,}^{10}\text{0}^1\text{0}^1\text{00}} &= \text{5} \text{ remainder } \text{10} \\ \frac{\text{1}}{\text{18}} &= \text{0,05555...} \\ &= \text{0,0}\dot{\text{5}} \end{align*}
$$1\frac{\text{1}}{\text{2}}$$
\begin{align*} 1\frac{\text{1}}{\text{2}} = \frac{3}{2} \\ \text{2}|\overline{\text{3,0000}} &= \text{1} \text{ remainder } \text{1} \\ \text{2}|\overline{\text{3,}^1\text{0 000}} &= \text{5} \text{ remainder } \text{0} \\ &=\text{1,5} \end{align*}

Express $$\text{0,}\overline{78}$$ as a fraction $$\frac{a}{b}$$ where $$a, b \in \mathbb{Z}$$ (show all working).

\begin{align*} x & = \text{0,}\overline{78} \\ 100x & = \text{78,}\overline{78} \\ \therefore 100x - x & = 99 \\ \therefore x & = \frac{78}{99} \end{align*}

For each of the following numbers:

• write the next three digits;
• state whether the number is rational or irrational.

$$\text{1,11235...}$$

• The number does not terminate (this is shown by the $$\ldots$$). There is also no indication of a repeating pattern of digits since there is not dot or bar over any of the numbers. The next three digits could be any numbers.

• Irrational, there is no repeating pattern.

$$\text{1,}\dot{1}$$

• Since there is a dot over the 1 we know that the 1 repeats. The next three digits are: $$111$$
• Rational, there is a repeating pattern of digits.

Write the following rational numbers to 2 decimal places.

$$\frac{1}{2}$$

To write to two decimal places we must convert to decimal: $$\frac{1}{2}=\text{0,50}$$.

$$\text{1}$$

To write to two decimal places just add a comma and two $$\text{0}$$'s: $$\text{1,00}$$.

$$\text{0,11111}\bar{1}$$

We mark where the cut off point is, determine if it has to be rounded up or not and then write the answer. In this case there is a $$\text{1}$$ after the cut off point so we do not round up. The final answer is: $$\text{0,11111}\overline{1} \approx \text{0,11}$$.

$$\text{0,99999}\bar{1}$$

We mark where the cut off point is, determine if it has to be rounded up or not and then write the answer. In this case there is a $$\text{9}$$ after the cut off point so we round up. The final answer is: $$\text{0,99999}\overline{1} \approx \text{1,00}$$.

Round off the following irrational numbers to 3 decimal places.

$$\text{3,141592654...}$$

$$\text{3,142}$$ (round up as there is a $$\text{5}$$ after the cut off point).

$$\text{1,618033989...}$$

$$\text{1,618}$$ (no rounding as there is a $$\text{0}$$ after the cut off point).

$$\text{1,41421356...}$$

$$\text{1,414}$$ (no rounding as there is a $$\text{2}$$ after the cut off point).

$$\text{2,71828182845904523536...}$$

$$\text{2,718}$$ (no rounding as there is a $$\text{2}$$ after the cut off point).

Round off the number $$\text{1 523,00195593}$$ to $$\text{4}$$ decimal places.

$\text{1 523,00195593} \approx \text{1 523,0020}$

Round off the number $$\text{1 982,94028996}$$ to $$\text{6}$$ decimal places.

$\text{1 982,94028996} \approx \text{1 982,940290}$

Round off the number $$\text{101,52378984}$$ to $$\text{4}$$ decimal places.

$\text{101,52378984} \approx \text{101,5238}$

Use your calculator and write the following irrational numbers to 3 decimal places.

$$\sqrt{2}$$

$\sqrt{2} \approx \text{1,414213562...} \approx \text{1,414}$

$$\sqrt{3}$$

$\sqrt{3} \approx \text{1,732050808...} \approx \text{1,732}$

$$\sqrt{5}$$

$\sqrt{5} \approx \text{2,236067977...} \approx \text{2,236}$

$$\sqrt{6}$$

$\sqrt{6} \approx \text{2,449489743...} \approx \text{2,449}$

Use your calculator (where necessary) and write the following numbers to 5 decimal places. State whether the numbers are irrational or rational.

$$\sqrt{8}$$

$\sqrt{8} \approx \text{2,828427125...} \approx \text{2,82843}$

Irrational number.

$$\sqrt{768}$$

$\sqrt{768} \approx \text{27,71281292...} \approx \text{27,71281}$

Irrational number.

$$\sqrt{\text{0,49}}$$

$\sqrt{\text{0,49}} = \text{0,70000}$

Rational number.

$$\sqrt{\text{0,0016}}$$

$\sqrt{\text{0,0016}} = \text{0,04000}$

Rational number.

$$\sqrt{\text{0,25}}$$

$\sqrt{\text{0,25}} = \text{0,50000}$

Rational number.

$$\sqrt{36}$$

$\sqrt{36} = \text{6,00000}$

Rational number.

$$\sqrt{1960}$$

$\sqrt{1960} \approx \text{44,27188724...} \approx \text{44,27189}$

Irrational number.

$$\sqrt{\text{0,0036}}$$

$\sqrt{\text{0,0036}} = \text{0,06000}$

Rational number.

$$-8\sqrt{\text{0,04}}$$

$-8\sqrt{\text{0,04}} = -8(\text{0,20000}) = -\text{1,60000}$

Rational number.

$$5\sqrt{80}$$

$5\sqrt{80} \approx 5(\text{8,94427191...}) \approx \text{44,72136}$

Irrational number.

Round off:

$$\frac{\sqrt{2}}{2}$$ to the nearest $$\text{2}$$ decimal places.

\begin{align*} \frac{\sqrt{2}}{2} & \approx \text{0,7071...} \\ & \approx \text{0,71} \end{align*}

$$\sqrt{14}$$ to the nearest $$\text{3}$$ decimal places.

\begin{align*} \sqrt{14} & \approx \text{3,741657...} \\ & \approx \text{3,742} \end{align*}

Write the following irrational numbers to 3 decimal places and then write each one as a rational number to get an approximation of the irrational number.

$$\text{3,141592654...}$$

\begin{align*} \text{3,141592654...} & \approx \text{3,142} \\ & \approx 3 \frac{142}{\text{1 000}} \\ & \approx \frac{\text{1 571}}{\text{500}} \end{align*}

$$\text{1,618033989...}$$

\begin{align*} \text{1,618033989...} & \approx \text{1,618} \\ & \approx 1 \frac{618}{\text{1 000}} \\ & \approx \frac{\text{809}}{\text{500}} \end{align*}

$$\text{1,41421356...}$$

\begin{align*} \text{1,41421356...} & \approx \text{1,414} \\ & \approx 1 \frac{414}{\text{1 000}} \\ & \approx \frac{\text{707}}{\text{500}} \end{align*}

$$\text{2,71828182845904523536...}$$

\begin{align*} \text{2,71828182845904523536...} & \approx \text{2,718} \\ & \approx 2 \frac{718}{\text{1 000}} \\ & \approx \frac{\text{1 359}}{\text{500}} \end{align*}

Determine between which two consecutive integers the following irrational numbers lie, without using a calculator.

$$\sqrt{5}$$

$2 \text{ and } 3 ~(2^2 = 4 \text{ and } 3^2=9)$

$$\sqrt{10}$$

$3 \text{ and } 4 ~(3^2 = 9 \text{ and } 4^2=16)$

$$\sqrt{20}$$

$4 \text{ and } 5 ~(4^2 = 16 \text{ and } 5^2 = 25)$

$$\sqrt{30}$$

$5 \text{ and } 6 ~(5^2 = 25 \text{ and } 6^2 = 36)$

$$\sqrt[3]{5}$$

$1 \text{ and } 2 ~(1^3 = 1 \text{ and } 2^3=8)$

$$\sqrt[3]{10}$$

$2 \text{ and } 3 ~(2^3 = 8 \text{ and } 3^3 = 27)$

$$\sqrt[3]{20}$$

$2 \text{ and } 3 ~(2^3 = 8 \text{ and } 3^3=27)$

$$\sqrt[3]{30}$$

$3 \text{ and } 4 ~(3^3 = 27 \text{ and } 4^3 = 64)$

$$\sqrt{90}$$

$9 \text{ and } 10 ~(9^2 = 81 \text{ and } 10^2 = 100)$

$$\sqrt{72}$$

$8 \text{ and } 9 ~(8^2 = 64 \text{ and } 9^2 = 81)$
$$\sqrt[3]{58}$$
$3 \text{ and } 4 ~(3^3 = 27 \text{ and } 4^3 = 64)$
$$\sqrt[3]{118}$$
$4 \text{ and } 5 ~(4^3 = 64 \text{ and } 5^3 = 125)$

Estimate the following surds to the nearest $$\text{1}$$ decimal place, without using a calculator.

$$\sqrt{14}$$

$$\sqrt{14}$$ lies between 3 and 4. Since $$3^{2} = 9$$ and $$4^{2} = 16$$ it lies closer to 4 than to 3.

Therefore $$\text{3,7}$$ or $$\text{3,8}$$ are suitable estimates.

$$\sqrt{110}$$

$$\sqrt{110}$$ lies between 10 and 11. Since $$10^{2} = 100$$ and $$11^{2} = 121$$ it lies almost exactly between 10 and 11.

Therefore $$\text{10,5}$$ is a suitable estimate.

$$\sqrt{48}$$

$$\sqrt{48}$$ lies between 6 and 7. Since $$6^{2} = 36$$ and $$7^{2} = 49$$ it lies closer to 7 than to 6.

Therefore $$\text{6,9}$$ is a suitable estimate.

$$\sqrt{57}$$

$$\sqrt{57}$$ lies between 7 and 8. Since $$7^{2} = 49$$ and $$8^{2} = 64$$ it lies almost exactly between the two numbers.

Therefore $$\text{4,5}$$ or $$\text{4,6}$$ are suitable estimates.

Expand the following products:

$$(a + 5)^2$$

\begin{align*} (a + 5)^2 & = (a + 5)(a + 5) \\ &= a^2 + 5a + 5a + 25\\ &=a^2 + 10a + 25 \end{align*}

$$(n + 12)^2$$

\begin{align*} (n + 12)^2 & = (n + 12)(n + 12) \\ &= n^2 + 12n + 12n + 144 \\ &=n^2 + 24n +144 \end{align*}

$$(d - 4)^2$$

\begin{align*} (d - 4)^2 & = (d - 4)(d - 4) \\ &= d^2 - 4d - 4d + 16\\ &=d^2 - 8d + 16 \end{align*}

$$(7w + 2)(7w - 2)$$

\begin{align*} (7w + 2)(7w - 2) &= 49w^{2} - 14w + 14w - 4\\ &= 49 w^{2} - 4 \end{align*}

$$(12 q + 1)(12 q - 1)$$

\begin{align*} (12 q + 1)(12 q - 1) &= 144 q^{2} - 12q + 12q -1\\ &= 144 q^{2} - 1 \end{align*}

$$-(-x - 2)(x + 2)$$

\begin{align*} -(-x - 2)(x + 2) &=(x + 2)(x + 2) \\ & = x^{2} + 2x + 2x + 4\\ & = x^{2} + 4x + 4 \end{align*}

$$(5 k - 4)(5 k + 4)$$

\begin{align*} (5 k - 4)(5 k + 4) &= 25 k^{2} + \text{20} k - \text{20} k -16\\ &= 25 k^{2} - 16 \end{align*}

$$(5 f + 4)(2 f + 2)$$

\begin{align*} (5 f + 4)(2 f + 2) &= 10 f^{2} + 10 f + 8 f + 8\\ &= 10 f^{2} + 18 f + 8 \end{align*}

$$(3 n + 6)(6 n + 5)$$

\begin{align*} (3 n + 6)(6 n + 5) &= 18 n^{2} + 15 n + 36 n + 30 \\ &= 18 n^{2} + 51 n + 30 \end{align*}

$$(2 g + 6)(g + 6)$$

\begin{align*} (2 g + 6)(g + 6) &= 2 g^{2} + 12 g + 6 g + 36 \\ &= 2 g^{2} + 18 g + 36 \end{align*}

$$(4 y + 1)(4 y + 8)$$

\begin{align*} (4 y + 1)(4 y + 8) &= 16 y^{2} + 32 y + 4y +8 \\ &= 16 y^{2} + 36 y + 8 \end{align*}

$$(d - 3)(7 d + 2)$$

\begin{align*} (d - 3)(7 d + 2) &= 7 d^{2} + 2d - 21d - 6 \\ &= 7 d^{2} - 19 d - 6 \end{align*}

$$(6 z - 4)(z - 2)$$

\begin{align*} (6 z - 4)(z - 2) &= 6 z^{2} - 12 z - 4 z + 8 \\ &= 6 z^{2} - 16 z + 8 \end{align*}

$$(5w - 11)^2$$

\begin{align*} (5w - 11)^2 & = (5w - 11)(5w - 11) \\ & = 25w^{2} - 55w - 55w + 121 \\ &= 25w^{2} - 110w + 121 \end{align*}

$$(5s - 1)^2$$

\begin{align*} (5s - 1)^2 & = (5s - 1)(5s - 1) \\ & = 25s^{2} - 5s - 5s + 1 \\ &= 25 s^{2} - 10s + 1 \end{align*}

$$(3d -8)^2$$

\begin{align*} (3d - 8)^2 & = (3d - 8)(3d - 8) \\ & = 9d^2 - 24d - 24d + 64 \\ &= 9d^2 - 48d + 64 \end{align*}

$$5f^2(3f + 5) + 7f(3f^2 + 7)$$

\begin{align*} 5f^2(3f + 5) + 7f(3f^2 + 7) &= 15f^3 + 25f^2 + 21f^3 + 49f \\ &= 36f^3 + 25f^2 + 49f \end{align*}

$$8d(4d^3 + 2) + 6d^2(7d^2 + 4)$$

\begin{align*} 8d(4d^3 + 2) + 6d^2(7d^2 + 4) &=32d^4 + 16d + 42d^4 + 24d^2\\ &= 74d^4 + 16d + 24d^2 \end{align*}

$$\text{5}{x}^2(\text{2}{x}+\text{2}) + \text{7}{x}(\text{7}{x}^2 + \text{7})$$

\begin{align*} \text{5}{x}^2(\text{2}{x} + \text{2}) + \text{7}{x}(\text{7}{x}^2 + \text{7}) &= \text{10}{x}^3 + \text{10}{x}^2 + \text{49}{x}^3 + \text{49}{x}\\ &= \text{59}{x}^3 + \text{10}{x}^2 + \text{49}{x} \end{align*}

Expand the following:

$$(y^4 + 3y^2 + y)(y+1)(y-2)$$

\begin{align*} (y^4 + 3y^2 + y)(y+1)(y-2) &= (y^4 + 3y^2 + y)(y^2 - y - 2) \\ & = y^6 - y^5 - 2y^4 + 3y^4 - 3y^3 - 6y^2 + y^3 - y^2 -2y \\ & = y^6 - y^5 + y^4 - 2y^3 - 7y^2 - 2y \end{align*}
$$(x+1)^2 - (x-1)^2$$
\begin{align*} (x+1)^2 - (x-1)^2 &= x^2 + 2x + 1 - (x^2 -2x + 1) \\ &= x^2 + 2x + 1 - x^2 + 2x - 1 \\ &= 4x \end{align*}
$$(x^2 + 2x + 1)(x^2 -2x + 1)$$
\begin{align*} (x^2 + 2x + 1)(x^2 -2x + 1) &= x^4 - 2x^3 + x^2 + 2x^3 - 4x^2 + 2x + x^2 - 2x + 1 \\ &=x^4 - 2x^2 + 1 \end{align*}
$$(4a - 3b)(16a^2 + 12ab + 9b^2)$$
\begin{align*} (4a - 3b)(16a^2 + 12ab + 9b^2) &= 64a^3 + 48a^2b + 36ab^2 - 48a^2b - 36ab^2 - 27b^3 \\ &=64a^3 - 27b^3 \end{align*}
$$2(x + 3y)(x^2 - xy - y^2)$$
\begin{align*} 2(x + 3y)(x^2 - xy - y^2) &= 2(x^3 - x^2y - xy^2 + 3x^2y - 3xy^2 - 3y^3) \\ &=2x^3 + 4x^2y - 8xy^2 - 6y^3 \end{align*}
$$(3a - 5b)(3a + 5b)(a^2 + ab - b^2)$$
\begin{align*} (3a - 5b)(3a + 5b)(a^2 + ab - b^2) &= (9a^2 - 25b^2)(a^{2} + ab - b^2) \\ & = 9a^4 + 9a^3 - 9a^2b^2 - 25a^2b^2 + 25ab^3 - 25b^4 \\ & = 9a^4 + 9a^3 - 34a^2b^2 + 25ab^3 - 25b^4 \end{align*}
$$\left(y - \dfrac{1}{y}\right)\left(y + \dfrac{1}{y}\right)$$
\begin{align*} \left(y - \frac{1}{y}\right)\left(y + \frac{1}{y}\right) &= y^2 + 1 -1 + \frac{1}{y^2} \\ &= y^2 - \frac{1}{y^2} \end{align*}
$$\left(\dfrac{a}{3} - \dfrac{3}{a}\right)\left(\dfrac{a}{3} + \dfrac{3}{a}\right)$$
\begin{align*} \left(\frac{a}{3} - \frac{3}{a}\right)\left(\frac{a}{3} + \frac{3}{a}\right) &= \frac{a^2}{9} + 1 - 1+ \frac{3}{a^2} \\ &= \frac{a^2}{9} - \frac{3}{a^2} \end{align*}
$$\dfrac{1}{3}(12x - 9y) + \dfrac{1}{6}(12x + 18y)$$
\begin{align*} \frac{1}{3}(12x - 9y) + \frac{1}{6}(12x + 18y) &= 4x - 3y + 2x + 3y \\ &=6x \end{align*}
$$(x+2)(x-2) - (x+2)^2$$
\begin{align*} (x+2)(x-2) - (x+2)^2 &= x^2 - 4 - (x^2 + 4x + 4) \\ &= -4x - 8 \end{align*}

What is the value of $$e$$ in $$(x - 4)(x + e) = x^2 - 16$$?

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

From the constant term we see that $$4e = 16$$, therefore $$e = 4$$.

In $$(x+2)(x+k) = x^2 + bx + c$$:

For which of these values of $$k$$ will $$b$$ be positive?

$$-6 \; ; \; -1 \; ; \; 0 \; ; \; 1 \; ; \; 6$$
\begin{align*} (x + 2)(x + k) & = x^2 + kx + 2x + 2k \\ & = x^2 + (k + 2)x + 2k \end{align*}

The $$b$$ term is $$k + 2$$ and so any value greater than $$-\text{2}$$ will make the $$b$$ term positive.

Therefore $$-1 \; ; \; 0 \; ; \; 1 \; ; \; 6$$

For which of these values of $$k$$ will $$c$$ be positive?

$$-6 \; ; \; -1 \; ; \; 0 \; ; \; 1 \; ; \; 6$$

From above we see that the $$c$$ term is $$2k$$. Therefore any positive value of $$k$$ will make $$c$$ positive.

Therefore $$0 \; ; \; 1 \; ; \; 6$$

For what values of $$k$$ will $$c$$ be positive?

From above we see that the $$c$$ term is $$2k$$. Therefore any positive value of $$k$$ will make $$c$$ positive.

Therefore $$k > 0$$

For what values of $$k$$ will $$b$$ be positive?

From above we see that any value greater than $$-\text{2}$$ will make the $$b$$ term positive.

Therefore $$k > -2$$.

Expand: $$\left(3a - \dfrac{1}{2a}\right)^2$$

\begin{align*} \left(3a - \frac{1}{2a}\right)^2 &= 9a^2 + 3 + \frac{1}{4a^2} \end{align*}

Expand: $$\left(3a - \dfrac{1}{2a}\right)\left(9a^2 + \dfrac{3}{2} + \dfrac{1}{4a^2}\right)$$

\begin{align*} \left(3a - \frac{1}{2a}\right)\left(9a^2 + \frac{3}{2} + \frac{1}{4a^2}\right) &= 27a^3 + \frac{9}{2}a + \frac{3}{4a} - \frac{9}{2}a - \frac{3}{4a} - \frac{1}{8a^3} \\ &=27a^3 - \frac{1}{8a^3} \end{align*}

Given that $$3a - \dfrac{1}{2a} = 7$$, determine the value of $$27a^3 - \dfrac{1}{8a^3}$$ without solving for $$a$$.

\begin{align*} 27a^3 - \frac{1}{8a^3} &= \left(3a - \frac{1}{2a}\right)\left(9a^2 + \frac{3}{2} + \frac{1}{4a^2}\right) \\ &= 7\left(9a^2 + \frac{3}{2} + \frac{1}{4a^2}\right) \\ \\ 9a^2 + \frac{3}{2} + \frac{1}{4a^2} &= \left(3a - \frac{1}{2a}\right)^2 + \frac{9}{2} \\ &= 7^2 + \frac{9}{2} \\ \\ 27a^3 - \frac{1}{8a^3} &= 374\frac{1}{2} \end{align*}

Solve by factorising:

$$17^2 - 15^2$$
\begin{align*} 17^2 - 15^2 &= (17-15)(17+15) \\ &=2(32) \\ &=64 \end{align*}
$$13^2 -12^2$$
\begin{align*} 13^2 -12^2 &= (13-12)(13+12) \\ &= 25 \end{align*}
$$120045^2 - 120035^2$$
\begin{align*} 120045^2 - 120035^2 &= (120045 - 120035)(120045 + 120035) \\ &= 10(240080) \\ &= 2400800 \end{align*}

$$26^2 - 24^2$$

\begin{align*} 26^2 - 24^2 &= (26-24)(26+24) \\ &=2(50) \\ &= 100 \end{align*}

Represent the following as a product of its prime factors:

$$143$$
\begin{align*} 143 &= 144- 1 \\ &=(12-1)(12+1) \\ &=11 \times 13 \end{align*}
$$168$$
\begin{align*} 168 &= 169 - 1 \\ &=(13-1)(13+1) \\ &=12(14) \\ &=3 \times 2^2 \times 2 \times 7 \\ &=2^3 \times 3 \times 7 \end{align*}
$$899$$
\begin{align*} 899 &= 900 - 1 \\ &=(30- 1)(30+1) \\ &=29 \times 31 \end{align*}
$$99$$
\begin{align*} 99 &= 100 - 1 \\ &=(10-1)(10+1) \\ &=3^2 \times 11 \end{align*}
$$1599$$
\begin{align*} 1599 &= 1600 -1 \\ &=(40-1)(40+1) \\ &=39(41)\\ &=3 \times 13 \times 41 \end{align*}

Factorise:

$$a^{2} - 9$$

${a}^{2} - 9 = (a - 3)(a + 3)$

$$9b^{2} - 81$$

\begin{align*} 9b^{2} - 81 & = 9(b^{2} - 9)\\ & = 9(b - 3)(b + 3) \end{align*}

$$m^{2} - \frac{1}{9}$$

$m^{2} - \frac{1}{9} = \left(m - \frac{1}{3}\right)\left(m + \frac{1}{3}\right)$

$$5 - 5a^{2}b^{6}$$

\begin{align*} 5 - 5a^{2}b^{6} & = 5(1 - a^{2}b^{6})\\ & = 5(1 - ab^{3})(1 + ab^{3}) \end{align*}

$$16ba^{4} - 81b$$

\begin{align*} 16ba^{4} - 81b & = b(16a^{4} - 81)\\ & = b(4a^{2} - 9)(4a^{2} + 9) \\ & = b(2a - 3)(2a + 3)(4a^{2} + 9) \end{align*}

$$a^{2} - 10a + 25$$

$a^{2} - 10a + 25 = (a - 5)(a - 5)$

$$16b^{2} + 56b + 49$$

$16b^{2} + 56b + 49 = (4b + 7)(4b + 7)$

$$-4b^{2} - 144b^{8} + 48b^{5}$$

\begin{align*} -4b^{2} - 144b^{8} + 48b^{5} & = -4b^{2}(1 + 36b^{6} - 12b^{3}) \\ & = -4b^{2}(6b^{3} - 1)(6b^{3} - 1)\\ & = -4b^{2}(6b^{3} - 1)^{2} \end{align*}

$$16 - x^{4}$$

\begin{align*} 16 - x^{4} & = (4 - x^{2})(4 + x^{2}) \\ & = (4 + x^{2})(2 + x)(2 - x) \end{align*}

$$7x^{2} - 14x + 7xy - 14y$$

\begin{align*} 7x^{2} - 14x + 7xy - 14y & = 7(x^{2} - 2x + xy - 2y)\\ & = 7(x(x - 2) + y(x - 2)) \\ & = 7(x - 2)(x + y) \end{align*}

$$y^{2} - 7y - 30$$

$y^{2} - 7y - 30 = (y - 10)(y + 3)$

$$1 - x - x^{2} + x^{3}$$

\begin{align*} 1 - x - x^{2} + x^{3} & = (1 - x) - x^{2}(1 - x)\\ & = (1 - x)(1 - x^{2}) \\ & = (1 - x)(1 - x)(1 + x) \\ & = (1 - x)^{2}(1 + x) \end{align*}

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

\begin{align*} -3(1 - p^{2}) + p + 1 & = -3(1 - p)(1 + p) + (1 + p)\\ & = (1 + p)[-3(1 - p) + 1] \\ & = (1 + p)(-2 + 3p) \end{align*}

$$x^{2} - 2x + 1 - y^{4}$$

\begin{align*} x^{2} - 2x + 1 - y^{4} & = x(x - 2) + (1 - y^2)(1 + y^2)\\ &= x(x - 2) + (1 + y)(1 - y)(1 + y^2) \end{align*}

$$4b(x^{3} - 1) + x(1 - x^{3})$$

\begin{align*} 4b(x^{3} - 1) + x(1 - x^{3}) & = (x^3 - 1)(4b - x)\\ &= (x - 1)(x^2 + x + 1)(4b - x) \end{align*}

$$3m(v-7)+19(-7 + v)$$

\begin{align*} 3 m (v-7) +19 (-7 + v) & = 3 m (v-7) +19 (v-7) \\ & = (v-7)(3m+19) \end{align*}

$$3f(z+3)+19(3 + z)$$

\begin{align*} 3 f (z+3) +19 (3 + z) & = 3 f (z+3) +19 (z+3) \\ &= (3f+19)(z+3) \end{align*}

$$3p^{3} - \frac{1}{9}$$

$3p^{3} - \frac{1}{9} = 3(p - \frac{1}{3})(p^2 + \frac{p}{3} + \frac{1}{9})$

$$8x^{6} - 125y^{9}$$

$8x^{6} - 125y^{9} = (2x^2 - 5y^3)(4x^4 + 10x^2y^3 + 25y^6)$

$$(2 + p)^{3} - 8(p + 1)^{3}$$

\begin{align*} (2 + p)^{3} - 8(p + 1)^{3} & = [(p + 2) - 2(p + 1)][(p + 2)^2 + 2(p + 2)(p + 1) + 4(p + 1)^2] \\ & = [p + 2 - 2p - 2][p^2 + 4p + 4 + 2p^2 + 6p + 4 + 4p^2 + 8p + 4] \\ & = (-p)(12 + 18p + 7p^2) \end{align*}
$$\dfrac{1}{3}a^3 - a^2b + 2a^2b - 6ab^2 +3ab^2 - 9b^3$$
\begin{align*} \frac{1}{3}a^3 - a^2b + 2a^2b - 6ab^2 +3ab^2 - 9b^3 &= \frac{1}{3}a^{2}(a - 3b) + 2ab(a -3b) + 3b^2(a -3b) \\ &=(\frac{1}{3}a^{2} + 2ab + 3b^2)(a-3b) \\ &=\frac{(a^2+ 6ab + 9b^2)(a -3b)}{3} \\ &=\frac{(a + 3b)^2(a -3b)}{3} \end{align*}

$$6a^{2}-17a+5$$

$6a^{2}-17a+5 = (2a-5)(3a-1)$

$$s^{2}+2s-15$$

$s^{2}+2s-15 = (s-3)(s+5)$

$$16v +24h +2j^{5}v +3j^{5}h$$

\begin{align*} 16v +24h +2j^{5}v +3j^{5}h &= 8 (2v +3h) +j^{5} (2v +3h) \\ &= (2v +3h) (8 +j^{5}) \end{align*}

$$18h -45g +2m^{3}h -5m^{3}g$$

\begin{align*} 18h -45g +2m^{3}h -5m^{3}g &= 9 (2h -5g) +m^{3} (2h -5g) \\ &= (2h -5g) (9 +m^{3}) \end{align*}

$$63d -18s +7u^{2}d -2u^{2}s$$

\begin{align*} 63d -18s +7u^{2}d -2u^{2}s &= 9 (7d -2s) +u^{2} (7d -2s) \\ &= (7d -2s) (9 +u^{2}) \end{align*}

Factorise the following:

$$6 a^{2} + 14 a + 8$$

\begin{align*} 6 a^{2} + 14 a + 8 & = \text{2} (3 a^{2} + 7 a + 4)\\ & = \text{2} \left(a + 1 \right) \left( 3 a + 4 \right) \end{align*}

$$6 g^{2} - 15 g - 9$$

\begin{align*} 6 g^{2} - 15 g - 9 & = \text{3} (2 g^{2} - 5 g - 3)\\ & = \text{3} \left(g - 3 \right) \left( 2 g + 1 \right) \end{align*}

$$125g^3 - r^3$$

$125g^3 - r^3 = (5g -r)(25g^2 +5gr + r^2)$

$$8r^3 + z^3$$

$8r^3 + z^3 = (2r +z)(4r^2 -2rz + z^2)$

$$14m-4n+7jm-2jn$$

\begin{align*} 14m-4n+7jm-2jn &= 2(7m-2n)+j(7m-2n)\\ &= (7m-2n)(2+j) \end{align*}

$$25d-15m+5yd-3ym$$

\begin{align*} 25d-15m+5yd-3ym &= 5(5d-3m)+y(5d-3m)\\ &= (5d-3m)(5+y) \end{align*}

$$g^\text{3}-\text{27}$$

$g^\text{3}-\text{27} = (g-\text{3})(g^\text{2}\text{+3}g+9)$

$$z^\text{3}\text{+125}$$

$z^\text{3}\text{+125} = (z\text{+5})(z^\text{2}-\text{5}z+25)$

$$b^2 - (3a - 2b)^2$$
\begin{align*} b^2 - (3a - 2b)^2 &= (b - (3a - 2b))(b + 3a - 2b) \\ &= (3b - 3a)(3a -b) \\ &= 3(b-a)(3a - b) \end{align*}
$$9y^2 - (4x + 2y)^2$$
\begin{align*} 9y^2 - (4x + 2y)^2 &= (3y + 4x + 2y)(3y - (4x + 2y)) \\ &=(4x + 5y)(y - 4x) \end{align*}
$$16x^6 - 3y^8$$
\begin{align*} 16x^6 - 3y^8 &= 4(4x^6 - 9y^8) \\ &=4(4x^6 - 9y^8) \\ &=4(4x^3 - 3y^4)(4x^3 + 3y^4) \end{align*}
$$\dfrac{1}{6}a^2 - 24b^4$$
\begin{align*} \frac{1}{6}a^2 - 24b^4 &= \frac{1}{6}\left(a^2 - 144b^4\right) \\ &= \frac{1}{6}\left(a - 12b^2\right)\left(a + 12b^2\right) \end{align*}
$$4(a - 3) - 81x^2(a - 3)$$
\begin{align*} 4(a - 3) - 81x^2(a - 3) &= (a-3)(4- 81x^2)\\ &=(a-3)(2-9x)(2+9x) \end{align*}
$$(2+b)^2 - 11(2+b) - 12$$
\begin{align*} (2+b)^2 - 11(2+b) - 12 &= ((2+b) + 1)((2+b) - 12) \\ & = (b+3)(b-10) \end{align*}
$$2x^2 + 7xy + 5y^2$$
$2x^2 + 7xy + 5y^2 = (2x +5y)(x +y)$
$$x^2 - 2xy - 15y^2$$
$x^2 - 2xy - 15y^2 = (x-5y)(x + 3y)$
$$4x^4 + 11x^2 + 6$$
$4x^4 + 11x^2 + 6 = (4x^2 + 3)(x^2 + 2)$
$$6x^4 - 38x^2 + 40$$
\begin{align*} 6x^4 - 38x^2 + 40 &= 2(3x^4 - 19x^2 + 20) \\ &=2(3x^2 - 4)(x^2 - 5) \end{align*}
$$9a^2x + 9a^{2}y + 27a^2 - b^2x - b^2y - 3b^2$$
\begin{align*} 9a^2x + 9a^{2}y + 27a^2 - b^2x - b^2y - 3b^2 &= (9a^2 - b^2)(x + y + 3) \\ &= (3a+ b)(3a -b)(x +y +3) \end{align*}
$$2(2y^2 - 5y) - 24$$
\begin{align*} 2(2y^2 - 5y) - 24 &= 2(2y^2 - 5y) - 2(12) \\ &=2(2y^2 - 5y - 12) \\ &=2(2y + 3)(y-4) \end{align*}
$$\dfrac{1}{2}x^3 - \frac{9}{2}x - 2x^2 + 18$$
\begin{align*} \frac{1}{2}x^3 - \frac{9}{2}x - 2x^2 + 18 &= \frac{x^3 - 9x - 4x^2 +36}{2} \\ &=\frac{x^2(x-4) - 9(x - 4)}{2} \\ &=\frac{(x-4)(x^2 - 9)}{2} \\ &=\frac{(x-4)(x - 3)(x + 3)}{2} \end{align*}

$$27r^3s^3 - 1$$

$27r^3s^3 - 1 = (3rs -1)(9r^2s^2 +3rs + 1)$

$$\dfrac{1}{125h^3} + r^3$$

$\frac{1}{125h^3} + r^3 = \left(\frac{1}{5h} +r\right)\left(\frac{1}{25h^2} -\frac{r}{5h} + r^2\right)$

$$j(64n^3 - b^3) + k(64n^3 - b^3)$$

\begin{align*} j(64n^3 - b^3) + k(64n^3 - b^3) &= (j+k)(64n^3 - b^3) \\ &= (j + k)(4n - b)(16n^2 +4nb + b^2) \end{align*}

Simplify the following:

$$(a - 2)^{2} - a(a + 4)$$

\begin{align*} (a - 2)^{2} - a(a + 4) & = a^{2} - 4a + 4 - a^{2} - 4a\\ &= -8a + 4 \end{align*}

$$(5a - 4b)(25a^{2} + 20ab + 16b^{2})$$

\begin{align*} (5a - 4b)(25{a}^{2} + 20ab + 16{b}^{2}) & = 125a^{3} + 100a^{2}b + 80ab^{2} - 100a^{2}b - 80ab^{2} - 64b^{3}\\ & = 125a^3 - 64b^3 \end{align*}

$$(2m - 3)(4m^{2} + 9)(2m + 3)$$

\begin{align*} (2m - 3)(4m^{2} + 9)(2m + 3) & = (4m^{2} - 9)(4m^{2} + 9)\\ & = 16m^{4} - 81 \end{align*}

$$(a + 2b - c)(a + 2b + c)$$

\begin{align*} (a + 2b - c)(a + 2b + c) & = a^{2} + 2ab + ac + 2ab + 4b^{2} + 2bc - ac - 2bc - c^{2}\\ & = a^{2} + 4ab + 4b^{2} - c^{2} \end{align*}

$$\dfrac{m^2+11m+18}{4(m^2-4)} \div \dfrac{3m^2+27m}{24m^2-48m}$$

\begin{align*} \frac{m^2+11m+18}{4(m^2-4)} \div \frac{3m^2+27m}{24m^2-48m} &= \frac{m^2+11m+18}{4(m^2-4)} \times \frac{24m^2-48m}{3m^2+27m}\\ &= \frac{(m+9)(m+2)}{4(m-2)(m+2)} \times \frac{24m(m-2)}{3m(m+9)}\\ &= \frac{1}{4} \times \frac{24}{3}\\ &= 2 \end{align*}

$$\dfrac{t^2+9t+18}{5(t^2-9)} \div \dfrac{4t^2+24t}{100t^2-300t}$$

\begin{align*} \frac{t^2+9t+18}{5(t^2-9)} \div \frac{4t^2+24t}{100t^2-300t} &= \frac{t^2+9t+18}{5(t^2-9)} \times \frac{100t^2-300t}{4t^2+24t}\\ &= \frac{(t+6)(t+3)}{5(t-3)(t+3)} \times \frac{100t(t-3)}{4t(t+6)}\\ &= \frac{1}{5} \times \frac{100}{4}\\ &= 5 \end{align*}

$$\dfrac{4 - b^2}{3b - 6}$$
\begin{align*} \frac{4 - b^2}{3b - 6} &= \frac{(2-b)(2+b)}{3(b-2)} \\ &= -\frac{2+b}{3} \end{align*}
$$\dfrac{x^2 + 2x + 4}{x^3 - 8}$$
\begin{align*} \frac{x^2 + 2x + 4}{x^3 - 8} &= \frac{x^2 + 2x + 4}{(x-2)(x^2 + 2x + 4)} \\ &= \frac{1}{x-2} \end{align*}
$$\dfrac{x^2 - 5x - 14}{3x +6}$$
\begin{align*}\frac{x^2- 5x - 14}{3x +6} &= \frac{(x-7)(x+2)}{3(x+2)} \\ &= \frac{x-7}{3} \end{align*}

$$\dfrac{d^2+23d+132}{5(d^2-121)} \div \dfrac{4d^2+48d}{100d^2-1100d}$$

\begin{align*} \frac{d^2+23d+132}{5(d^2-121)} \div \frac{4d^2+48d}{100d^2-1100d} &= \frac{d^2+23d+132}{5(d^2-121)} \times \frac{100d^2-1100d}{4d^2+48d}\\ &= \frac{(d+12)(d+11)}{5(d-11)(d+11)} \times \frac{100d(d-11)}{4d(d+12)}\\ &= \frac{1}{5} \times \frac{100}{4}\\ &= 5 \end{align*}

$$\dfrac{a-2}{a^2 + 4a + 3} \div \dfrac{(a-1)(a+1)}{a-1} \times \dfrac{a^2 - 2a - 15}{a-2}$$
\begin{align*} & \frac{a-2}{a^2 + 4a + 3} \div \frac{(a-1)(a+1)}{a-1} \times \frac{a^2 - 2a - 15}{a-2}\\ &= \frac{a-2}{(a+1)(a+3)} \div \frac{(a-1)(a+1)}{a-1} \times \frac{(a+3)(a-5)}{a-2} \\ &=\frac{a-2}{(a+1)(a+3)} \times \frac{a-1}{(a-1)(a+1)} \times \frac{(a+3)(a-5)}{a-2} \\ &= \frac{a-5}{(a + 2)^{2}} \end{align*}
$$\dfrac{a+6}{a^2 + 12a + 11} \times \dfrac{a^2 + 14a + 33}{a+3} \div \dfrac{a^3 +216}{a+1}$$
\begin{align*} & \frac{a+6}{a^2 + 12a + 11} \times \frac{a^2 + 14a + 33}{a+3} \div \frac{a^3 +216}{a+1} \\ &= \frac{a+6}{(a+11)(a+1)} \times \frac{(a+11)(a+3)}{a+3} \times \frac{a+1}{(a+6)(a^2 + 6a + 36)} \\ &= \frac{1}{a^2 + 6a + 36} \end{align*}
$$2 \div \dfrac{a + b}{a + 2b} \times \dfrac{b^2 - ba -6a^2}{a^2 - 4b^2} \times \dfrac{a^2 - b - 2b^2}{3a -b}$$
\begin{align*} & 2 \div \frac{a + b}{a + 2b} \times \frac{b^2 - ba -6a^2}{a^2 - 4b^2} \times \frac{a^2 - b - 2b^2}{3a -b} \\ & = 2 \times \frac{a + 2b}{a+b} \times \frac{(b-3a)(b +2a)}{(a-2b)(a+2b)} \times \frac{(a-2b)(a+b)}{3a -b} \\ &= -2 (2a + b) \end{align*}

$$\dfrac{st+sb+ 31t+31b}{t+b}$$

\begin{align*} \frac{st+sb + 31t+31b} {(t+b)} &= \frac{s(t+b)+ 31(t+b)}{(t+b)}\\ &= \frac{(t+b)(s+31)}{(t+b)}\\ &= s+31 \end{align*}

$$\dfrac{ny+nq + 8y+8q}{y+q}$$

\begin{align*} \frac{ny+nq + 8y+8q}{(y+q)} &= \frac{n(y+q)+ 8(y+q)}{(y+q)}\\ &= \frac{(y+q)(n+8)}{(y+q)}\\ &= n+8 \end{align*}

$$\dfrac{p^{2} - q^{2}}{p} \div \dfrac{p + q}{p^{2} - pq}$$

\begin{align*} \frac{{p}^{2} - {q}^{2}}{p} \div \frac{p + q}{{p}^{2} - pq} & = \frac{(p - q)(p + q)}{p} \times \frac{p(p - q)}{p + q}\\ & = (p - q)^2\\ & = p^2 - 2pq +q^2 \end{align*}

$$\dfrac{2}{x} + \dfrac{x}{2} - \dfrac{2x}{3}$$

\begin{align*} \frac{2}{x} + \frac{x}{2} - \frac{2x}{3} & = \frac{12 + 3x^2 - 4x^2}{6x}\\ & = \frac{12 - x^2}{6x} \end{align*}

$$\dfrac{1}{a + 7} - \dfrac{a + 7}{a^{2} - 49}$$

\begin{align*} \frac{1}{a + 7} - \frac{a + 7}{a^2 - 49} &= \frac{1}{a + 7} - \frac{a + 7}{(a + 7)(a - 7)}\\ &= \frac{-14}{(a + 7)(a - 7)} \end{align*}

$$\dfrac{x + 2}{2x^{3}} + 16$$

\begin{align*} \frac{x + 2}{2x^3} + 16 &= \frac{(x + 2) + 16(2x^3)}{2x^3}\\ &= \frac{32x^3 + x + 2}{2x^3} \end{align*}

$$\dfrac{1 - 2a}{4a^{2} - 1} - \dfrac{a - 1}{2a^{2} - 3a + 1} - \dfrac{1}{1 - a}$$

\begin{align*} \frac{1 - 2a}{4a^{2} - 1} - \frac{a - 1}{2a^{2} - 3a + 1} - \frac{1}{1 - a} & = \frac{1 - 2a}{(2a - 1)(2a + 1)} - \frac{a - 1}{(2a - 1)(a - 1)} + \frac{1}{a - 1}\\ & = -\frac{(2a - 1)}{(2a - 1)(2a + 1)} - \frac{1}{2a - 1} + \frac{1}{a - 1}\\ &=\frac{4a - 1}{(2a + 1)(2a - 1)(a - 1)} \end{align*}
$$\dfrac{1}{2}x + \dfrac{x-2}{3} + 4$$
\begin{align*} \frac{1}{2}x + \frac{x-2}{3} + 4 &= \frac{3x + 2(x-2) + (2)(3)(4)}{6} \\ &= \frac{3x + 2x - 4 + 24}{6} \\ &= \frac{5x + 20}{6} \end{align*}
$$\dfrac{1}{x^2 + 2x} + \dfrac{4x^2 - x - 3}{x^2 + 2x - 3}$$
\begin{align*} \frac{1}{x^2 + 2x} + \frac{4x^2 - x - 3}{x^2 + 2x - 3} &= \frac{1}{x(x + 2)} + \frac{(4x+3)(x-1)}{(x-1)(x+3)} \\ &= \frac{1}{x(x + 2)} + \frac{4x+3}{x+3} \\ &= \frac{x+3 + x(4x+3)(x+2)}{x(x + 2)(x+3)} \\ &= \frac{x+3 +x(4x^2 + 11x + 6)}{x(x + 2)(x+3)} \\ &= \frac{4x^3 + 11x^2 + 7x + 3}{x(x + 2)(x+3)} \end{align*}
$$\dfrac{b^2 + 6b + 9}{b^2-9} + \dfrac{b^2 - 6b + 8}{(b-2)(b+3)} + \dfrac{1}{b+3}$$
\begin{align*} \frac{b^2 + 6b + 9}{b^2-9} + \frac{b^2 - 6b + 8}{(b-2)(b+3)} + \frac{1}{b+3} &= \frac{(b+3)^2}{(b+3)(b-3)} + \frac{(b-4)(b-2)}{(b-2)(b+3)} + \frac{1}{b+3} \\ &= \frac{b+3}{b-3} + \frac{b-4}{b+3} + \frac{1}{b+3} \\ &= \frac{(b+3)^2 + (b-3)(b-4) + b -3}{(b-3)(b+3)} \\ &= \frac{b^2 + 6b + 9 + b^2 - 7b + 12 + b -3}{(b-3)(b+3)} \\ &=\frac{2b^2 + 18}{(b-3)(b+3)}\\ &= \frac{2(b^{2} + 9)}{(b-3)(b+3)} \end{align*}

$$\dfrac{x^{2} + 2x}{x^{2} + x + 6} \times \dfrac{x^{2} + 2x + 1}{x^{2} + 3x + 2}$$

\begin{align*} \frac{x^2 + 2x}{x^2 + x + 6} \times \frac{x^2 + 2x + 1}{x^2 + 3x + 2} &=\frac{x(x + 2)}{x^2 + x + 6} \times \frac{(x + 1)(x + 1)}{(x + 2)(x + 1)}\\ &= \frac{x(x + 1)}{x^2 + x + 6} \end{align*}

$$\dfrac{12}{z+12} + \dfrac{5}{z-5}$$

\begin{align*} \frac{12}{z+12} + \frac{5}{z-5} &= \frac{12(z-5) + 5(z+12)}{(z+12)(z-5)}\\ &= \frac{12z-60 + 5z+60}{(z+12)(z-5)}\\ &= \frac{17z}{(z+12)(z-5)} \end{align*}

$$\dfrac{11}{w-11} - \dfrac{4}{w-4}$$

\begin{align*} \frac{11}{w-11} - \frac{4}{w-4} \\ &= \frac{11(w-4) - 4(w-11)}{(w-11)(w-4)}\\ &= \frac{11w-44 - 4w+44}{(w-11)(w-4)}\\ &= \frac{7w}{(w-11)(w-4)} \end{align*}

Show that $$(2x - 1)^{2} - (x - 3)^{2}$$ can be simplified to $$(x + 2)(3x - 4)$$.

\begin{align*} (2x - 1)^{2} - (x - 3)^{2} & = (2x - 1)(2x - 1) - (x - 3)(x - 3) \\ &= 4x^2 - 2x - 2x + 1 - (x^2 - 3x - 3x - 9)\\ &= 3x^2 + 2x - 8\\ &=(3x - 4)(x + 2) \end{align*}

What must be added to $$x^{2} - x + 4$$ to make it equal to $$(x + 2)^{2}$$?

Suppose $$A$$ must be added to the expression to get the desired result.

\begin{align*} \therefore (x^2 - x + 4) + A &= (x + 2)^2\\ \therefore A &= (x + 2)(x + 2) - (x^2 - x + 4)\\ & = x^2 + 2x + 2x + 4- x^2 + x - 4\\ & = 5x \end{align*}

Therefore $$5x$$ must be added.

Evaluate $$\dfrac{x^{3} + 1}{x^{2} - x + 1}$$ if $$x = \text{7,85} \text{ without using a calculator. Show your work.}$$

First simplify the expression:

\begin{align*} \frac{x^{3} + 1}{x^{2} - x + 1} & = \frac{(x + 1)(x^2 - x + 1)}{x^2 - x + 1} \\ & = x + 1 \end{align*}

Now substitute the value of $$x$$: $$\text{7,85} + 1 = \text{8,85}$$.

With what expression must $$(a - 2b)$$ be multiplied to get a product of $$(a^{3} - 8b^{3})$$?

$(a - 2b)(a^2 + 2ab + 4b^2) = a^3 - 8b^3$

So, the expression is $$a^2 + 2ab + 4b^2$$.

With what expression must $$27x^{3} + 1$$ be divided to get a quotient of $$3x + 1$$?

\begin{align*} 27x^3 + 1 & = (3x + 1)(9x^2 - 3x + 1)\\ \frac{(3x + 1)(9x^2 - 3x + 1)}{9x^2 - 3x + 1} &= 3x + 1 \end{align*}

Therefore the expression is $$9x^2 - 3x + 1$$.

What are the restrictions on the following?

$$\dfrac{4}{3x^2 + 2x - 1}$$
\begin{align*} \frac{4}{3x^2 + 2x - 1} &= \frac{4}{(3x - 1)(x+1)} \\ x \ne \frac{1}{3} &\text{ and } x \ne -1 \end{align*}
$$\dfrac{a}{3(b-a) + ab - a^2}$$
\begin{align*} \frac{a}{3(b-a) + ab - a^2} &= \frac{a}{3(b-a) + a(b-a)} \\ &=\frac{a}{(b-a)(a+3)} \\ a \ne b &\text { and } a \ne -3 \end{align*}