Algebraic Processes: JAMB Mathematics Notes

A friendly, step-by-step guide to algebra: letters, signs, brackets, fractions, equations, factorising, functions, variation, polynomials, inequalities, and matrices.

Algebraic expressions: read the expression before calculating

Algebra becomes much easier when you learn to read the expression first. Do not rush to calculate. First ask: What are the separate terms? Which terms are alike? What does each number belong to?

A term is one part of an expression. Terms are separated by plus or minus signs, and the sign belongs to the term that follows it.

Look at $5 x^{2} - 3 x + 7$ . Read it as three separate terms: $5 x^{2}$ , $- 3 x$ , and $+ 7$ .

A coefficient is the signed number multiplying a letter or a power of a letter.

In $5 x^{2}$ , the coefficient of $x^{2}$ is $5$ .
In $- 3 x$ , the coefficient of $x$ is $- 3$ . Keep the minus sign.
A constant is the number with no letter. In $5 x^{2} - 3 x + 7$ , the constant is $7$ .
A small raised number is a power. For example, $x^{2} = x \times x$ and $x^{3} = x \times x \times x$ .

Sometimes the number $1$ is not written, but it is still there.

$x = 1 x$ , so the coefficient of $x$ is $1$ .
$- x = - 1 x$ , so the coefficient is $- 1$ .
$x^{2} = 1 x^{2}$ , so the coefficient of $x^{2}$ is $1$ .
$- x^{2} = - 1 x^{2}$ , so the coefficient is $- 1$ .

Two terms are like terms only when the letters and their powers match exactly. For example, $3 x$ and $5 x$ are like terms, and $4 x^{2}$ and $- 2 x^{2}$ are like terms. But $3 x$ and $3 x^{2}$ are not like terms, and $2 x$ and $2 y$ are not like terms.

When you collect like terms, keep the letter and its power. Add or subtract only the coefficients.

Example 1: Name the parts of $5 x^{2} - 3 x + 7$ .

Step 1: Separate the expression into terms: $5 x^{2}$ , $- 3 x$ , and $+ 7$ .

Step 2: Look at the $x^{2}$ -term. The number beside $x^{2}$ is $5$ , so its coefficient is $5$ .

Step 3: Look at the $x$ -term. It is $- 3 x$ , so its coefficient is $- 3$ , not $3$ .

Step 4: Look for the number with no $x$ . That is $7$ , so $7$ is the constant.

The highest power of $x$ is $2$ .

Example 2: Find $2 x^{2} - x y + y$ when $x = - 2$ and $y = 3$ .

Step 1: Write down what the question gives you: $x = - 2$ and $y = 3$ .

Step 2: Replace every $x$ by $- 2$ and every $y$ by $3$ . Put a negative replacement in brackets: $2 (- 2)^{2} - (- 2) (3) + 3$ .

Step 3: Do the power first: $(- 2)^{2} = 4$ .

Step 4: Multiply: $2 (4) - (- 6) + 3$ .

Step 5: Simplify: $8 + 6 + 3 = 17$ .

So the value is 17.

Signs and the order of work

Signs are small, but they matter. Use BODMAS: Brackets, Orders or powers, Division and Multiplication, then Addition and Subtraction.

Same signs give a positive answer when multiplying: $(+) (+) = +$ and $(-) (-) = +$ .
Different signs give a negative answer when multiplying: $(+) (-) = -$ and $(-) (+) = -$ .
A minus before a bracket changes every sign inside when the bracket is removed.
Never join unlike terms. You may join $3 x + 2 x$ , but not $3 x + 2 y$ or $3 x + 2 x^{2}$ .

Example 3: Simplify $4 x + 3 y - 7 + 2 x - y + 5$ .

Step 1: Find the $x$ -terms: $4 x$ and $2 x$ . They are like terms.

Step 2: Find the $y$ -terms: $3 y$ and $- y$ . Remember that $- y = - 1 y$ .

Step 3: Find the constants: $- 7$ and $+ 5$ .

Step 4: Group the like terms: $(4 x + 2 x) + (3 y - y) + (- 7 + 5)$ .

Step 5: Add or subtract only the coefficients in each group: $6 x + 2 y - 2$ .

That is the simplest form.

Opening brackets

To open a bracket, multiply the item outside by every term inside. Think of sharing food: nobody inside the bracket should be missed.

Example 4: Expand $- 3 (2 x - 5)$ .

Multiply $- 3$ by $2 x$ : $- 3 \times 2 x = - 6 x$ .

Multiply $- 3$ by $- 5$ : $- 3 \times - 5 = + 15$ .

So $- 3 (2 x - 5) = - 6 x + 15$ .

When two brackets touch, every term in the first bracket multiplies every term in the second.

Example 5: Expand $(x + 4) (x - 3)$ .

Multiply $x$ by both terms: $x (x - 3) = x^{2} - 3 x$ .

Multiply 4 by both terms: $4 (x - 3) = 4 x - 12$ .

Join everything: $x^{2} - 3 x + 4 x - 12$ .

Collect like terms: $x^{2} + x - 12$ .

Example 6: Simplify $2 x (x - 3) - (x + 2) (x - 2)$ .

Open the first part: $2 x (x - 3) = 2 x^{2} - 6 x$ .

The second pair is a difference of squares: $(x + 2) (x - 2) = x^{2} - 4$ .

Keep the minus before that whole bracket: $2 x^{2} - 6 x - (x^{2} - 4)$ .

Change both signs inside: $2 x^{2} - 6 x - x^{2} + 4 = x^{2} - 6 x + 4$ .

Factorising: putting brackets back

Factorising is the reverse of expanding. Your check is easy: open your final brackets. If you return to the starting expression, you are right.

First look for a common factor. Ask, “What number and letters divide every term?”

Example 7: Factorise $12 x^{3} y - 18 x^{2} y^{2}$ .

The highest number that divides 12 and 18 is 6.

Both terms contain at least $x^{2}$ and one $y$ .

Take out $6 x^{2} y$ : $12 x^{3} y - 18 x^{2} y^{2} = 6 x^{2} y (2 x - 3 y)$ .

Check: $6 x^{2} y \times 2 x = 12 x^{3} y$ and $6 x^{2} y \times - 3 y = - 18 x^{2} y^{2}$ .

A difference of two squares follows $a^{2} - b^{2} = (a - b) (a + b)$ . It must be subtraction, not addition.

Example 8: Factorise $81 a^{4} - 16 b^{4}$ completely.

Both parts are squares: $(9 a^{2})^{2} - (4 b^{2})^{2}$ .

First step: $(9 a^{2} - 4 b^{2}) (9 a^{2} + 4 b^{2})$ .

The first bracket is also a difference of squares: $9 a^{2} - 4 b^{2} = (3 a - 2 b) (3 a + 2 b)$ .

So the complete answer is $(3 a - 2 b) (3 a + 2 b) (9 a^{2} + 4 b^{2})$ .

For $x^{2} + b x + c$ , find two numbers whose product is $c$ and whose sum is $b$ .

Example 9: Factorise $x^{2} + 7 x + 12$ .

We need two numbers that multiply to 12 and add to 7.

The numbers are 3 and 4.

Therefore, $x^{2} + 7 x + 12 = (x + 3) (x + 4)$ .

If the number before $x^{2}$ is not 1, multiply the first and last numbers, then split the middle term.

Example 10: Factorise $6 x^{2} + 7 x - 3$ .

Multiply 6 by $- 3$ : the answer is $- 18$ .

We need two numbers with product $- 18$ and sum 7. They are 9 and $- 2$ .

Split $7 x$ : $6 x^{2} + 9 x - 2 x - 3$ .

Group the terms: $3 x (2 x + 3) - 1 (2 x + 3)$ .

Take out the shared bracket: $(3 x - 1) (2 x + 3)$ .

Learn these two special patterns too:

Perfect square: $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ .
Perfect square: $a^{2} - 2 ab + b^{2} = (a - b)^{2}$ .
Difference of cubes: $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ .
Sum of cubes: $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ .

Example 11: Factorise $k^{2} - 2 k p + p^{2}$ .

This matches $a^{2} - 2 ab + b^{2}$ .

Here $a = k$ and $b = p$ .

Therefore, $k^{2} - 2 k p + p^{2} = (k - p)^{2}$ .

Example 12: Factorise $27 x^{3} - 8 y^{3}$ .

Write each term as a cube: $27 x^{3} = (3 x)^{3}$ and $8 y^{3} = (2 y)^{3}$ .

Use the difference-of-cubes rule.

The result is $(3 x - 2 y) (9 x^{2} + 6 x y + 4 y^{2})$ .

Sometimes a long expression becomes easy when one repeated part is given a short name.

Example 13: Factorise $x^{2} + 2 x y + y^{2} + 3 x + 3 y - 18$ .

Notice that $x^{2} + 2 x y + y^{2} = (x + y)^{2}$ and $3 x + 3 y = 3 (x + y)$ .

Let $u = x + y$ . The expression becomes $u^{2} + 3 u - 18$ .

Factorise it: $u^{2} + 3 u - 18 = (u + 6) (u - 3)$ .

Put $x + y$ back: $(x + y + 6) (x + y - 3)$ .

Algebraic fractions

Treat an algebraic fraction like an ordinary fraction. Factorise first. You may cancel factors, but you may not cancel pieces joined by plus or minus signs.

Example 14: Simplify $\frac{x ^{3} + 3 x ^{2} - 10 x}{2 x ^{2} - 8}$ .

Factorise the top: $x^{3} + 3 x^{2} - 10 x = x (x^{2} + 3 x - 10) = x (x + 5) (x - 2)$ .

Factorise the bottom: $2 x^{2} - 8 = 2 (x^{2} - 4) = 2 (x - 2) (x + 2)$ .

Cancel the common factor $(x - 2)$ .

The answer is $\frac{x ( x + 5 )}{2 ( x + 2 )}$ , where $x \neq = 2$ and $x \neq = - 2$ .

To add fractions, first make their denominators the same.

Example 15: Simplify $\frac{1}{x + 1} + \frac{1}{x - 1}$ .

The common denominator is $(x + 1) (x - 1) = x^{2} - 1$ .

The numerator becomes $(x - 1) + (x + 1)$ .

Join like terms: $x - 1 + x + 1 = 2 x$ .

So the answer is $\frac{2 x}{x ^{2} - 1}$ , where $x \neq = 1$ and $x \neq = - 1$ .

When solving an equation with fractions, multiply every term by the lowest common denominator. This removes the fractions.

Example 16: Solve $\frac{2}{2 r - 1} - \frac{5}{3} = \frac{1}{r + 2}$ .

The common denominator is $3 (2 r - 1) (r + 2)$ .

Multiply every term by it: $6 (r + 2) - 5 (2 r - 1) (r + 2) = 3 (2 r - 1)$ .

Open and simplify: $- 10 r^{2} - 9 r + 22 = 6 r - 3$ .

Move everything to one side: $2 r^{2} + 3 r - 5 = 0$ .

Factorise: $(2 r + 5) (r - 1) = 0$ .

Therefore, $r = - \frac{5}{2}$ or $r = 1$ . Neither answer makes an original denominator zero.

Linear equations: keep the balance

An equation is like a balanced scale. Whatever you do to one side, do the same to the other side. Your goal is to leave the unknown alone.

Example 17: Solve $5 x - 7 = 18$ .

Add 7 to both sides: $5 x = 25$ .

Divide both sides by 5: $x = 5$ .

Check: $5 (5) - 7 = 25 - 7 = 18$ . It works.

Example 18: Solve $4 x + 3 = 2 x + 15$ .

Subtract $2 x$ from both sides: $2 x + 3 = 15$ .

Subtract 3 from both sides: $2 x = 12$ .

Divide by 2: $x = 6$ .

Example 19: Solve $\frac{3}{4} t + \frac{1}{3} (21 - t) = 11$ .

The common denominator of 4 and 3 is 12, so multiply every term by 12.

This gives $9 t + 4 (21 - t) = 132$ .

Open the bracket: $9 t + 84 - 4 t = 132$ .

So $5 t = 48$ , and $t = \frac{48}{5}$ .

Making a letter the subject

The subject is the letter standing alone. Undo operations in reverse order. If the wanted letter appears in more than one term, bring those terms together and factor it out.

Example 20: Make $Q$ the subject of $p = \frac{M}{5} (X + Q) + 1$ .

Subtract 1: $p - 1 = \frac{M}{5} (X + Q)$ .

Multiply by 5: $5 (p - 1) = M (X + Q)$ .

Divide by $M$ : $\frac{5 ( p - 1 )}{M} = X + Q$ .

Subtract $X$ : $Q = \frac{5 ( p - 1 )}{M} - X$ .

Example 21: Make $x$ the subject of $y = \frac{3 x - 9 c}{4 x + 5 d}$ .

Multiply both sides by $4 x + 5 d$ : $y (4 x + 5 d) = 3 x - 9 c$ .

Open the bracket: $4 x y + 5 d y = 3 x - 9 c$ .

Put the $x$ -terms together: $4 x y - 3 x = - 9 c - 5 d y$ .

Factor out $x$ : $x (4 y - 3) = - (9 c + 5 d y)$ .

Divide: $x = \frac{9 c + 5 d y}{3 - 4 y}$ .

Example 22: Make $v$ the subject of $r = \frac{3 v}{π h}$ .

Square both sides: $r^{2} = \frac{3 v}{π h}$ .

Multiply by $π h$ : $π h r^{2} = 3 v$ .

Divide by 3: $v = \frac{π h r ^{2}}{3}$ .

Example 23: Make $T$ the subject of $P = (\frac{Q ( R - T )}{15})^{\frac{1}{3}}$ .

Cube both sides: $P^{3} = \frac{Q ( R - T )}{15}$ .

Multiply by 15 and divide by $Q$ : $\frac{15 P ^{3}}{Q} = R - T$ .

Move the terms carefully: $T = R - \frac{15 P ^{3}}{Q}$ .

Example 24: Make $q$ the subject of $t = \frac{pq}{r} - r^{2} q$ .

Square both sides: $t^{2} = \frac{pq}{r} - r^{2} q$ .

Factor out $q$ : $t^{2} = q (\frac{p}{r} - r^{2})$ .

Write the bracket as one fraction: $\frac{p}{r} - r^{2} = \frac{p - r ^{3}}{r}$ .

Divide by the bracket: $q = \frac{r t ^{2}}{p - r ^{3}}$ .

Be careful when undoing a square. A square can come from a positive or negative number. For example, $S = t^{2} - 4 t + 4 = (t - 2)^{2} = ∣ t - 2∣$ , so $t = 2 + S$ or $t = 2 - S$ , unless the question gives a condition that removes one answer.

Simultaneous equations

Two equations can work together to find two unknowns. Use elimination to make one letter disappear, or substitution to replace one letter with an equal expression.

Example 25: Solve $x + y = 9$ and $x - y = 3$ .

Add the equations. The $+ y$ and $- y$ cancel: $2 x = 12$ .

So $x = 6$ .

Put $x = 6$ into $x + y = 9$ : $6 + y = 9$ .

Therefore, $y = 3$ .

Example 26: Solve $3 x - 5 y = 9$ and $6 x - 4 y = 12$ .

Multiply the first equation by 2: $6 x - 10 y = 18$ .

Subtract the second equation: $(6 x - 10 y) - (6 x - 4 y) = 18 - 12$ .

This gives $- 6 y = 6$ , so $y = - 1$ .

Put $y = - 1$ into $3 x - 5 y = 9$ : $3 x + 5 = 9$ .

Therefore, $x = \frac{4}{3}$ and $y = - 1$ .

Example 27: Solve $\frac{x}{2} - \frac{y}{5} = 1$ and $y - \frac{x}{3} = 8$ .

Clear the fractions in the first equation by multiplying by 10: $5 x - 2 y = 10$ .

Clear the fractions in the second by multiplying by 3: $3 y - x = 24$ .

From the second equation, $x = 3 y - 24$ .

Put this into the first: $5 (3 y - 24) - 2 y = 10$ .

So $13 y = 130$ , giving $y = 10$ , and then $x = 6$ .

Some pairs contain squares or products. Factorise or substitute until they become ordinary equations.

Example 28: Solve $x - y = 2$ and $x^{2} - y^{2} = 8$ .

Factorise the second equation: $x^{2} - y^{2} = (x - y) (x + y)$ .

Use $x - y = 2$ : $2 (x + y) = 8$ , so $x + y = 4$ .

Now solve $x - y = 2$ and $x + y = 4$ .

Add them: $2 x = 6$ , so $x = 3$ , and then $y = 1$ .

Example 29: Solve $3 x + y = 21$ and $x y = 30$ .

From the first equation, $y = 21 - 3 x$ .

Put this into $x y = 30$ : $x (21 - 3 x) = 30$ .

Rearrange: $x^{2} - 7 x + 10 = 0$ .

Factorise: $(x - 5) (x - 2) = 0$ , so $x = 5$ or $x = 2$ .

The two pairs are $(x, y) = (5, 6)$ and $(x, y) = (2, 15)$ .

Quadratic equations and roots

A quadratic equation has highest power $2$ . Before solving it, first put it in the standard form $a x^{2} + b x + c = 0$ .

Do not use the quadratic formula until you have identified $a$ , $b$ , and $c$ . Read the equation first.

Finding $a$ : $a$ is the signed number multiplying $x^{2}$ .

$5 x^{2}$ means $a = 5$ .
$- 3 x^{2}$ means $a = - 3$ .
$x^{2}$ means $1 x^{2}$ , so $a = 1$ .
$- x^{2}$ means $- 1 x^{2}$ , so $a = - 1$ .

Finding $b$ : $b$ is the signed number multiplying $x$ .

$7 x$ means $b = 7$ .
$- 4 x$ means $b = - 4$ .
$x$ means $1 x$ , so $b = 1$ .
$- x$ means $- 1 x$ , so $b = - 1$ .
If there is no $x$ -term, then $b = 0$ .

Finding $c$ : $c$ is the constant, the number with no $x$ . If there is no constant term, then $c = 0$ .

For example, in $2 x^{2} + 7 x - 4 = 0$ , we have $a = 2$ , $b = 7$ , $c = - 4$ . In $- x^{2} + 6 = 0$ , we have $a = - 1$ , $b = 0$ , $c = 6$ . In $3 x^{2} - 5 x = 0$ , we have $a = 3$ , $b = - 5$ , $c = 0$ .

The sign belongs to the coefficient. If the term is $- 4 x$ , write $b = - 4$ , not $b = 4$ .

If the equation is not equal to zero, rearrange it first. For example, $x^{2} + 3 x = 10$ becomes $x^{2} + 3 x - 10 = 0$ . Only then choose $a$ , $b$ , and $c$ .

Example 30: Solve $x^{2} - 5 x + 6 = 0$ .

Step 1: Read the equation. The $x^{2}$ -term is $x^{2} = 1 x^{2}$ , so $a = 1$ . The $x$ -term is $- 5 x$ , so $b = - 5$ . The constant is $+ 6$ , so $c = 6$ .

Step 2: This quadratic factorises easily. We need two numbers whose product is $6$ and whose sum is $- 5$ .

Step 3: The numbers are $- 2$ and $- 3$ , because $(- 2) (- 3) = 6$ and $- 2 + (- 3) = - 5$ .

Step 4: Factorise: $(x - 2) (x - 3) = 0$ .

Step 5: Set each factor equal to zero: $x - 2 = 0$ or $x - 3 = 0$ .

Therefore, $x = 2$ or $x = 3$ .

If the factors are not easy to see, use the quadratic formula. The formula is $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ .

Use the formula in this order:

First write the equation as $a x^{2} + b x + c = 0$ .
Then write $a = \dots$ , $b = \dots$ , and $c = \dots$ , keeping all signs.
Write the quadratic formula before inserting any numbers.
Wherever the formula has $a$ , replace it with the value you found for $a$ . Wherever it has $b$ , replace it with your $b$ . Wherever it has $c$ , replace it with your $c$ .
Put negative values in brackets. For example, if $b = - 4$ , then $b^{2} = (- 4)^{2}$ and $- b = - (- 4)$ .
Work out the number under the square root, then use the $+$ sign for one answer and the $-$ sign for the other.

Example 31: Solve $3 x^{2} - 4 x - 5 = 0$ .

Step 1: The equation is already in the form $a x^{2} + b x + c = 0$ .

Step 2: Look at the number beside $x^{2}$ : it is $3$ . Therefore, $a = 3$ .

Step 3: Look at the number beside $x$ : the term is $- 4 x$ . Therefore, $b = - 4$ .

Step 4: Look at the number with no $x$ : it is $- 5$ . Therefore, $c = - 5$ .

Step 5: Write the formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ .

Step 6: Replace every $a$ by $3$ , every $b$ by $- 4$ , and every $c$ by $- 5$ : $x = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( 3 ) ( - 5 )}{2 ( 3 )}$ .

Step 7: Simplify carefully. $- (- 4) = 4$ , $(- 4)^{2} = 16$ , and $- 4 (3) (- 5) = + 60$ . So the number under the square root is $16 + 60 = 76$ .

Therefore, $x = \frac{4 \pm 76}{6} = \frac{4 \pm 2 19}{6}$ .

So the two answers are $x = \frac{2 + 19}{3}$ or $x = \frac{2 - 19}{3}$ .

Quadratic functions and graphs

For a quadratic graph $y = a x^{2} + b x + c$ , first look at $a$ , the signed coefficient of $x^{2}$ . If $a > 0$ , the graph opens upward and has a minimum. If $a < 0$ , the graph opens downward and has a maximum.

Example 32: Find the minimum value of $y = 2 x^{2} - 8 x + 11$ .

Step 1: Read the $x^{2}$ -term: $2 x^{2}$ , so $a = 2$ . Because $a > 0$ , the graph opens upward, so we expect a minimum.

Step 2: Take 2 from the first two terms: $y = 2 (x^{2} - 4 x) + 11$ .

Step 3: Complete the square: $x^{2} - 4 x = (x - 2)^{2} - 4$ .

Step 4: Substitute that back: $y = 2 [(x - 2)^{2} - 4] + 11 = 2 (x - 2)^{2} + 3$ .

Step 5: A square cannot be below zero. Its smallest value is 0, reached when $x = 2$ .

Therefore, the minimum value of $y$ is 3.

If the roots are $α$ and $β$ , then $α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$ . Again, identify $a$ , $b$ , and $c$ before substituting.

Example 33: The roots of $3 x^{2} + 5 x - 2 = 0$ are $α$ and $β$ . Find $\frac{1}{α} + \frac{1}{β}$ .

Step 1: Read the coefficients: $a = 3$ , $b = 5$ , $c = - 2$ .

Step 2: Use $α + β = - \frac{b}{a}$ : $α + β = - \frac{5}{3}$ .

Step 3: Use $α β = \frac{c}{a}$ : $α β = - \frac{2}{3}$ .

Step 4: Put the two fractions over one denominator: $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}$ .

Step 5: Substitute the values: $\frac{- 5/3}{- 2/3} = \frac{5}{2}$ .

A cubic may have an easy whole-number root. Test factors of the constant, then divide by the matching linear factor.

Example 34: Solve $x^{3} - 19 x - 30 = 0$ .

Try factors of 30. At $x = 5$ , $5^{3} - 19 (5) - 30 = 125 - 95 - 30 = 0$ .

So $(x - 5)$ is a factor.

Dividing gives $x^{2} + 5 x + 6 = (x + 2) (x + 3)$ .

Thus the roots are $x = 5$ , $x = - 2$ , and $x = - 3$ .

Linear inequalities

Solve a linear inequality like an equation, with one special rule: when you multiply or divide by a negative number, reverse the sign. Thus $<$ becomes $>$ , and $\leq$ becomes $\geq$ .

Example 35: Solve $2 x + 3 > 5 x + 8$ .

Subtract $2 x$ : $3 > 3 x + 8$ .

Subtract 8: $- 5 > 3 x$ .

Divide by positive 3: $- \frac{5}{3} > x$ .

Write the answer in the usual order: $x < - \frac{5}{3}$ .

Quadratic inequalities

For a quadratic inequality, find the roots, split the number line into regions, and test one value in each region.

Example 36: Solve $x^{2} + 3 x - 4 \leq 0$ .

Factorise: $(x + 4) (x - 1) \leq 0$ .

The boundary points are $- 4$ and 1.

Test $x = 0$ , which lies between them: $(0 + 4) (0 - 1) = - 4$ , so the middle region works.

Because the sign includes equality, include both endpoints.

Therefore, $- 4 \leq x \leq 1$ .

Functions

A function is like a machine. Put a number or expression into the input, follow the rule, and get one output. Replace every $x$ , and use brackets around the new input.

Example 37: If $f (x) = 3 x^{3} + 4 x^{2} + x - 8$ , find $f (- 2)$ .

Replace every $x$ with $- 2$ : $f (- 2) = 3 (- 2)^{3} + 4 (- 2)^{2} + (- 2) - 8$ .

Work out the powers: $3 (- 8) + 4 (4) - 2 - 8$ .

So $f (- 2) = - 24 + 16 - 2 - 8 = - 18$ .

Example 38: If $g (x) = x^{2} + 3 x$ , simplify $g (x + 1) - g (x)$ .

First, $g (x + 1) = (x + 1)^{2} + 3 (x + 1)$ .

Open and collect: $g (x + 1) = x^{2} + 5 x + 4$ .

Now subtract $g (x)$ : $(x^{2} + 5 x + 4) - (x^{2} + 3 x)$ .

Therefore, $g (x + 1) - g (x) = 2 x + 4$ .

Polynomials, division, and remainders

A polynomial contains whole-number powers of a variable. Arrange its terms from the highest power to the lowest. Put in a zero placeholder for any missing power when dividing.

Example 39: Expand $(2 x^{2} - x + 1) (3 - 2 x)$ .

Multiply by 3: $6 x^{2} - 3 x + 3$ .

Multiply by $- 2 x$ : $- 4 x^{3} + 2 x^{2} - 2 x$ .

Collect like terms: $- 4 x^{3} + 8 x^{2} - 5 x + 3$ .

Example 40: Divide $4 x^{3} - 3 x + 1$ by $2 x - 1$ .

Write the missing term: $4 x^{3} + 0 x^{2} - 3 x + 1$ .

First term: $4 x^{3} \div 2 x = 2 x^{2}$ . Multiply back and subtract, leaving $2 x^{2} - 3 x + 1$ .

Next: $2 x^{2} \div 2 x = x$ . Multiply back and subtract, leaving $- 2 x + 1$ .

Next: $- 2 x \div 2 x = - 1$ . Multiply back and subtract, leaving 0.

The quotient is $2 x^{2} + x - 1$ , with no remainder.

The remainder theorem says the remainder when $P (x)$ is divided by $x - a$ is $P (a)$ . The factor theorem says $x - a$ is a factor exactly when $P (a) = 0$ .

Example 41: Find the remainder when $6 p^{3} - p^{2} - 47 p + 30$ is divided by $p - 3$ .

Here $p - 3 = p - a$ , so $a = 3$ .

Calculate $P (3) = 6 (3)^{3} - (3)^{2} - 47 (3) + 30$ .

This is $162 - 9 - 141 + 30 = 42$ .

The remainder is 42.

Example 42: Find the remainder when $x^{3} - 2 x^{2} + 3 x - 3$ is divided by $x^{2} + 1$ .

The divisor is quadratic, so use polynomial division, not the simple remainder theorem.

First quotient term: $x^{3} \div x^{2} = x$ . Subtract $x (x^{2} + 1)$ , leaving $- 2 x^{2} + 2 x - 3$ .

Next quotient term: $- 2 x^{2} \div x^{2} = - 2$ . Subtract $- 2 (x^{2} + 1)$ , leaving $2 x - 1$ .

The quotient is $x - 2$ , and the remainder is $2 x - 1$ .

Example 43: If $y - 1$ is a factor of $y^{3} + 4 y^{2} + k y - 6$ , find $k$ .

Because $y - 1$ is a factor, put $y = 1$ and make the result zero.

So $1^{3} + 4 (1)^{2} + k (1) - 6 = 0$ .

This gives $k - 1 = 0$ , so $k = 1$ .

Variation

Variation questions hide a constant called $k$ . Write the correct rule, use the first information to find $k$ , then answer the new question.

Direct variation: $y = k x$ , or $y = k x^{n}$ if a power is stated.
Inverse variation: $y = \frac{k}{x}$ , or $y = \frac{k}{x ^{n}}$ .
Joint variation: for example, $y = k x z$ .
Partial variation: one part is constant and another varies, for example $y = a + b x$ .

Example 44: $A$ varies directly as $b^{2}$ . If $A = 4$ when $b = 1$ , find $A$ when $b = 2$ .

Write the rule: $A = k b^{2}$ .

Use the first pair: $4 = k (1)^{2}$ , so $k = 4$ .

When $b = 2$ , $A = 4 (2)^{2} = 4 (4) = 16$ .

Example 45: $y$ varies inversely as $x$ . If $y = 6$ when $x = 4$ , find $y$ when $x = 8$ .

Write the rule: $y = \frac{k}{x}$ .

Use the first pair: $6 = \frac{k}{4}$ , so $k = 24$ .

When $x = 8$ , $y = \frac{24}{8} = 3$ .

Example 46: $P$ is partly constant and partly varies as $Q$ . If $P = 32$ when $Q = 16$ , and $P = 20$ when $Q = 12$ , find $P$ when $Q = 28$ .

Write the rule: $P = a + b Q$ .

Use the two pairs: $32 = a + 16 b$ and $20 = a + 12 b$ .

Subtract the equations: $12 = 4 b$ , so $b = 3$ .

Put $b = 3$ into $20 = a + 12 b$ : $20 = a + 36$ , so $a = - 16$ .

When $Q = 28$ , $P = - 16 + 3 (28) = 68$ .

Binary operations

A symbol such as $*$ or $\circ$ may be given a new meaning. Do not treat it as ordinary multiplication. Copy the rule and substitute the first and second inputs into their correct places.

Example 47: If $a * b = \frac{a + b}{ab} + (a - b)$ , find $3 * 2$ .

Put $a = 3$ and $b = 2$ : $3 * 2 = \frac{3 + 2}{3 ( 2 )} + (3 - 2)$ .

So $3 * 2 = \frac{5}{6} + 1 = \frac{11}{6}$ .

Example 48: If $a * b = a + b - 4$ , find the identity element.

An identity $e$ leaves any input unchanged, so $a * e = a$ .

Use the rule: $a + e - 4 = a$ .

Subtract $a$ from both sides: $e - 4 = 0$ .

Therefore, the identity element is 4.

Matrices: numbers arranged in boxes

A matrix is a rectangular arrangement of numbers. Its order is rows by columns. A $2 \times 3$ matrix has 2 rows and 3 columns.

Add or subtract matrices of the same order, position by position.
Multiply every entry when a number is placed before a matrix.
A transpose, $A^{T}$ , changes rows into columns.
The identity matrix $I$ behaves like 1: $A I = I A = A$ .
Matrix multiplication works only when the inside sizes match: $(m \times n) (n \times p)$ gives an $m \times p$ matrix.

For example, if $A = (142536)$ , then its rows become the columns of $A^{T} = 123456$ .

Example 49: Let $P = (12 3 - 5)$ and $Q = (31 - 7 2)$ . Find $P + 2 Q$ .

First double every entry of $Q$ : $2 Q = (62 - 14 4)$ .

Now add matching positions.

Therefore, $P + 2 Q = (74 - 11 - 1)$ .

For multiplication, multiply a row of the first matrix by a column of the second, then add those products.

Example 50: Let $A = 221132$ and $B = (3422)$ . Find $A B$ .

The first entry is row 1 times column 1: $2 (3) + 1 (4) = 10$ .

The next entry is row 1 times column 2: $2 (2) + 1 (2) = 6$ .

Repeat for the other rows.

Thus $A B = 1018116106$ .

Order matters: usually $A B \neq = B A$ , and sometimes $B A$ is not possible.

Example 51: If $P = (1223)$ , find $P^{2} - 4 P - I$ .

First multiply: $P^{2} = P P = (58813)$ .

Also, $4 P = (48812)$ and $I = (1001)$ .

Subtract position by position.

The answer is $(0000)$ , the zero matrix.

Determinants

For $A = (a c b d)$ , the determinant is $∣ A ∣ = a d - b c$ . Multiply down the main diagonal, then subtract the product of the other diagonal.

Example 52: Find $5634$ .

Multiply the main diagonal: $5 \times 4 = 20$ .

Multiply the other diagonal: $3 \times 6 = 18$ .

Subtract: $20 - 18 = 2$ .

For a $3 \times 3$ determinant, expand along a row or column. Choose one with zeros when possible because zeros remove work. Remember the sign pattern $+, -, +$ . Along the first row, $a d g b e h c f i = a (e i - f h) - b (d i - f g) + c (d h - e g)$ .

Example 53: Find $010375284$ .

Expand down the first column. Only the middle entry is non-zero, and its sign is negative.

So the determinant is $- 13524$ .

This is $- 1 (3 \times 4 - 2 \times 5) = - 1 (12 - 10) = - 2$ .

Inverse matrices and matrix equations

A $2 \times 2$ inverse is $A^{- 1} = \frac{1}{a d - b c} (d - c - b a)$ . Swap the main-diagonal entries, change the signs of the other two, and divide by the determinant. If the determinant is zero, there is no inverse.

Example 54: Find the inverse of $N = (2134)$ .

Find the determinant: $∣ N ∣ = 2 (4) - 3 (1) = 5$ .

Swap 2 and 4, then change the signs of 3 and 1.

Therefore, $N^{- 1} = \frac{1}{5} (4 - 1 - 3 2)$ .

A matrix equation can be read as simultaneous equations, or solved with an inverse.

Example 55: Solve $(52 - 6 - 7) (x y) = (7 - 11)$ .

Read the rows as equations: $5 x - 6 y = 7$ and $2 x - 7 y = - 11$ .

Multiply the first by 2: $10 x - 12 y = 14$ .

Multiply the second by 5: $10 x - 35 y = - 55$ .

Subtract: $23 y = 69$ , so $y = 3$ .

Then $5 x - 18 = 7$ , so $x = 5$ .

Example 56: Find $A$ if $A (02 1 - 1) = (21 - 1 0)$ .

Call the middle matrix $C$ and the right matrix $D$ . Since $A C = D$ , multiply on the right by $C^{- 1}$ : $A = D C^{- 1}$ .

Here $C^{- 1} = (\frac{1}{2} 1 \frac{1}{2} 0)$ .

Multiply: $A = (21 - 1 0) (\frac{1}{2} 1 \frac{1}{2} 0)$ .

Therefore, $A = (0 \frac{1}{2} 1 \frac{1}{2})$ .

A simple exam plan

You do not need to see the whole answer at once. Take one small step, then the next.

Copy the question carefully and keep every negative sign.
Ask what type it is: simplify, factorise, solve, make a subject, find a remainder, or use a matrix.
Write the matching rule before putting in numbers.
Use brackets when substituting negative values.
Show one operation per line. This makes mistakes easy to find.
Never divide by an expression without checking when it is zero.
For an equation, substitute your answer back into the original question.
For factorising, expand your brackets to check.
For a matrix product, check the sizes before multiplying.
If you feel stuck, do the first safe step: open a bracket, clear fractions, collect like terms, or factorise.

You can learn algebra. It is not magic; it is a collection of small rules. Practise one kind at a time, check your work, and the long questions will begin to look like short puzzles joined together.

Algebraic Processes