2.2 Algebraic manipulation

Learning objectives

2.2A Evaluate expressions by substituting numerical valuesevaluate expressions by substituting numerical values for letters 2.2B Collect like termscollect like terms 2.2C Expanding a single term over a bracketmultiply a single term over a bracket 2.2D Taking out common factorstake out common factors 2.2E Products of two linear expressionsexpand the product of two simple linear expressions 2.2F Factorising simple quadratic expressionsunderstand the concept of a quadratic expression and factorise expressions limited to x² + bx + c

Substitution means replacing each symbol by its stated numerical value, then evaluating the resulting numerical expression.

Step	Action
1	copy the expression and identify every symbol
2	replace each symbol with its value in brackets
3	follow the usual operation order
4	check the sign and approximate size

If $a=-3$ and $b=4$ , then $2a^2-b=2(-3)^2-4=18-4=14$ . The brackets ensure that the square applies to the whole value $-3$ .

Use the same value every time a symbol occurs. If $x=1/2$ , then $4x^2+3x=4(1/2)^2+3(1/2)=5/2$ .

Do not replace $x^2$ by $-3^2$ when $x=-3$ : without brackets, $-3^2$ means $-(3^2)$ .

Like terms have exactly the same variable part, including the same letters raised to the same powers. Their numerical coefficients can be added or subtracted.

Terms	Like?	Reason
$7x$ and $-2x$	yes	both have variable part $x$
$3x^2$ and $8x^2$	yes	both have variable part $x^2$
$4xy$ and $-xy$	yes	$xy=yx$
$x$ and $x^2$	no	powers differ
$2x$ and $2y$	no	letters differ

$12x-7y-5x+2y=(12-5)x+(-7+2)y=7x-5y$ .

Constants are also like terms with each other. Keep unlike groups separate and write the simplified expression in a clear order.

Combining changes only coefficients: $3x+5x=8x$ , not $8x^2$ ; $3x+5y$ cannot be combined.

The distributive law multiplies the term outside a bracket by every term inside it: $a(b+c)=ab+ac$ .

Form	Expansion
$x(2x+5)$	$2x^2+5x$
$3y(y-4)$	$3y^2-12y$
$-2(a+6)$	$-2a-12$
$-m(3m-7)$	$-3m^2+7m$

Draw or mentally track one multiplication for each term in the bracket, multiply coefficients and letter factors separately, then simplify.

The number of expanded terms should initially match the number inside the bracket; substituting a simple value can check equivalence.

A negative multiplier changes every sign. Expanding $-2(x-3)$ gives $-2x+6$ , not $-2x-6$ .

Factorising reverses expansion. Taking out a common factor writes an expression as a product of that factor and a bracket.

Part	Greatest common factor
coefficients	their highest common factor
each letter	the lowest power present in every term
bracket	each original term divided by the common factor

$12x^3y-18x^2y^2=6x^2y(2x-3y)$ . Expanding the result reproduces both original terms.

For $m^2+7m$ , both terms contain $m$ , so $m^2+7m=m(m+7)$ .

Only take out factors shared by every term. A factorised answer is complete only when the bracket has no further common factor.

To expand a product of two linear brackets, multiply every term in the first bracket by every term in the second, then collect like terms.

Product from $(y+9)(y-4)$	Result
$y imes y$	$y^2$
$y imes(-4)$	$-4y$
$9 imes y$	$9y$
$9 imes(-4)$	$-36$

Adding the four products gives $y^2-4y+9y-36=y^2+5y-36$ .

$(ax+b)(cx+d)=acx^2+(ad+bc)x+bd$ . This is a check on the leading, middle and constant terms, not a shortcut for omitting products.

The middle coefficient comes from two cross-products. Multiplying only the first and last terms loses essential terms.

For a monic quadratic $x^2+bx+c$ , find two numbers whose product is $c$ and whose sum is $b$ ; they become the constants in two brackets.

Signs in $x^2+bx+c$	Factor signs
$c>0$ , $b>0$	both positive
$c>0$ , $b<0$	both negative
$c<0$	opposite signs; larger magnitude gives sign of $b$

For $x^2-5x-36$ , the required numbers are $4$ and $-9$ : their product is $-36$ and sum is $-5$ . Hence $(x+4)(x-9)$ .

Re-expand the brackets: the outer and inner products must combine to the original middle term.

A pair with the correct product is not enough; it must also give the exact middle coefficient.

With three or more factors, expand two factors at a time, simplify that result, then multiply it by the next factor.

Stage for $4n(n-3)(n+5)$	Result
expand the two brackets	$(n-3)(n+5)=n^2+2n-15$
multiply by $n$	$n^3+2n^2-15n$
multiply by $4$	$4n^3+8n^2-60n$

Keep brackets around each intermediate polynomial and align like powers before collecting terms. A grid is useful when an intermediate factor has three or more terms.

The highest-degree term comes from multiplying all highest-degree terms; the constant term comes from multiplying all constants when none of the factors is just a variable.

Do not expand all factors in one uncontrolled step. A factor outside the brackets must multiply every term of the intermediate polynomial.

General quadratic factorisation reverses the product of two linear expressions. First remove any common factor, then choose a method suited to the structure.

Structure	Route
$A^2-B^2$	$(A-B)(A+B)$
$x^2+bx+c$	find product $c$ , sum $b$
$ax^2+bx+c$	find product $ac$ , sum $b$ ; split the middle term and group

$6x^2+11x+3$ : since $ac=18$ and $9+2=11$ , write $6x^2+9x+2x+3=3x(2x+3)+1(2x+3)=(3x+1)(2x+3)$ .

$4c^2-9d^2=(2c)^2-(3d)^2=(2c-3d)(2c+3d)$ .

A sum of squares does not use the real difference-of-squares identity. Always expand the proposed factors to verify all three coefficients.

An algebraic fraction follows ordinary fraction rules. Factor expressions before cancelling, and record values that make any original denominator zero.

Task	Method
simplify	factor numerator and denominator, then cancel common factors
add or subtract	use a common denominator, combine numerators, then simplify
multiply	factor and cancel across factors before multiplying
divide	multiply by the reciprocal, then simplify

$5/3-(x+2)/(2x)=[10x-3(x+2)]/(6x)=(7x-6)/(6x)$ , where $x
e0$.

Cancellation is division by a common non-zero factor: $(x^2-9)/(x^2+3x)=[(x-3)(x+3)]/[x(x+3)]=(x-3)/x$ , with $x
e0,-3$.

Cancel factors, not terms joined by addition. In $(x+3)/x$ , the $x$ is not a factor of the whole numerator and cannot cancel.

Completing the square rewrites a quadratic as a squared linear expression plus or minus a constant, making its turning point visible.

Starting form	Completed-square form
$x^2+bx+c$	$(x+b/2)^2+c-b^2/4$
$ax^2+bx+c$	$a(x+b/(2a))^2+c-b^2/(4a)$

$3x^2-12x+7=3(x^2-4x)+7=3[(x-2)^2-4]+7=3(x-2)^2-5$ .

From $a(x-h)^2+k$ , the turning point is $(h,k)$ and the line of symmetry is $x=h$ . If $a>0$ the minimum value is $k$ ; if $a<0$ the maximum is $k$ .

When $a
e1 $, first factor$ a $from both the$ x^2 $and$ x$ terms. Do not factor it from the standalone constant unless the whole expression is being factored.

An algebraic proof represents every case allowed by a statement with general variables, transforms the expression logically, and ends by connecting the final form to the claim.

Stage	What to write
define	state an integer, consecutive numbers or another general form
form	translate the claimed quantity algebraically
transform	expand, simplify or factor using valid identities
classify	show the result has the required form
conclude	state why this proves the claim for all permitted values

Let $T_n=n(n+1)/2$ . Then $T_n+T_{n+1}=n(n+1)/2+(n+1)(n+2)/2=(n+1)[n+n+2]/2=(n+1)^2$ . Therefore the sum of two consecutive triangular numbers is a square.

Useful definitions include consecutive integers $n,n+1$ , even integers $2n$ , odd integers $2n+1$ , and multiples of $k$ as $kn$ .

Checking several numerical examples supports a conjecture but does not prove it. The variable argument must cover every permitted case and explicitly justify the conclusion.

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—2.3 Expressions and formulaeRead

2.2 Algebraic manipulation

Learning objectives

Substitute values into expressions

Collect like terms

Expand a single term over a bracket

Take out common factors

Expand two linear expressions

Factorise simple quadratic expressions

Expand products of multiple linear expressions

Factorise general quadratic expressions

Simplify and combine algebraic fractions

Complete the square

Construct an algebraic proof

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