3 Algebra
- Syllabus
- 2016
- Topic
- 3
- Level
- —
Learning objectives
3A Basic algebra processesThe basic processes of algebra Collecting like terms, using the four operations, the rules of indices, with integers and fractional powers3B FormulaeThe construction, interpretation and use of formulae and their manipulation To include change of subject of a formula and substitution3C FactorisationThe factorisation of simple algebraic expressions3D Factor theoremUse of the factor theorem Including application to cubics and factors of the form (ax + b) or (ax – b)3E Algebraic divisionAlgebraic division of a cubic by a linear factor3F Algebraic fractionsThe manipulation of simple algebraic fractions, the denominators being numerical, linear or quadratic Simple cases involving sum, difference, product and quotient of algebraic fractions3G EquationsSolution of equations of 1st, 2nd and 3rd degree containing one unknown quantity Solution of quadratics to include solution by factorisation, by graph, by completing the square or by formula Problems that result in the solution of such equations may also be set3H Linear simultaneous equationsSolution of linear simultaneous equations in two unknowns Simple questions may be set requiring the graphical solution of simultaneous linear equations3I Linear and quadratic simultaneous equationsSolve simultaneous equations in two unknowns, one equation being linear and the other being quadratic3J Linear inequalitiesSolution of linear inequalities, and the representations of solutions on the number line and two-dimensional space Simple questions may be set requiring the graphical solution of simultaneous linear inequalities No questions will be set on linear programming3K Quadratic inequalitiesSolve quadratic inequalities in one unknown and represent the solution set on a number line3L SequencesThe idea of a sequence Being able to recognise sequences with a common difference or common integer sequences, and to continue a given sequence
Manipulate algebra without changing its value
Algebraic manipulation preserves value by applying arithmetic and index laws to terms with compatible structure. Like terms have the same variable part, including the same powers.
| Move | Valid example |
|---|---|
| collect like terms | 3x2−5x+2x2+x=5x2−4x |
| expand | 3(2x−5)=6x−15 |
| multiply powers | x3/2x1/2=x2 |
| divide powers | x5/x2=x3 for $x |
| e0$ |
(2x+1)^2-(2x-2)(2x+1)=(2x+1)igl[(2x+1)-(2x-2)igr]=6x+3
Terms such as x and x2 are not like terms. A negative or fractional power follows the same index laws, but division and negative powers require a non-zero base.
Construct, use and rearrange formulae
A formula states how quantities are related. Define every symbol and its units, substitute with brackets, and rearrange by performing inverse operations on both sides.
To change the subject when it appears more than once: remove fractions, expand, collect every term containing the new subject on one side, factor it out, then divide by its coefficient.
u= rac{5-4t}{3+6t}\Rightarrow u(3+6t)=5-4t\Rightarrow t(6u+4)=5-3u\Rightarrow t= rac{5-3u}{6u+4}
The rearranged formula inherits restrictions from the original: here $3+6t
e0,andthefinaldivisionalsorequires6u+4
e0$. Do not change a sign merely because a term crosses the equals sign; apply the same operation to both sides.
Factorise by exposing a common structure
Factorising rewrites a sum or difference as a product. It is the reverse of expansion, so multiplying the factors back is a direct check.
| Structure | Factorising move |
|---|---|
| common factor | take the greatest common numerical and algebraic factor outside brackets |
| four terms | group pairs to create a repeated bracket |
| difference of squares | a2−b2=(a−b)(a+b) |
| quadratic | find two terms whose product and sum match the quadratic |
2x^3-6xz+x^2z-3z^2=2x(x^2-3z)+z(x^2-3z)=(2x+z)(x^2-3z)
A factor must divide every term in the part from which it is extracted. 'Completely factorised' means no remaining factor can be factorised further over the required number system.
Use the factor theorem to test linear factors
(x-r) ext{ is a factor of }f(x)\iff f(r)=0
For a factor ax−b, set it equal to zero first: its corresponding root is x=b/a. Substitution gives the remainder without carrying out full division.
f(x)=x^3-6x^2-7x+60,\quad f(5)=125-150-35+60=0\ f(x)=(x-5)(x^2-x-12)=(x-5)(x-4)(x+3)
Testing f(r) concerns the factor (x−r), not (x+r). A zero remainder proves a factor; a non-zero value is the remainder and disproves that proposed factor.
Divide a cubic by a linear expression
Polynomial division mirrors numerical long division: order terms by descending powers, include zero coefficients for missing powers, divide the leading terms, multiply back, subtract, and repeat.
rac{x^3-4x^2+x+6}{x+1}=x^2-5x+6\quad ext{because}\quad(x+1)(x^2-5x+6)=x^3-4x^2+x+6
When the divisor is a known factor, the remainder is zero and the quadratic quotient can often be factorised further: x2−5x+6=(x−2)(x−3).
Keep place value aligned: if the cubic has no x2 or x term, write a zero placeholder. If a remainder remains, report quotient plus remainder divided by the divisor.
Operate safely with algebraic fractions
| Operation | Reliable move |
|---|---|
| simplify | factor numerator and denominator, then cancel common factors |
| add/subtract | use a common denominator, combine numerators, then factor |
| multiply | factor first, cancel, then multiply |
| divide | multiply by the reciprocal of the second fraction |
rac6{x-2}+ rac4{x+3}= rac{6(x+3)+4(x-2)}{(x-2)(x+3)}= rac{10(x+1)}{(x-2)(x+3)}
State excluded values from every original denominator. In the worked expression, $x
e2,-3$, even if later simplification were to cancel one of those factors.
Only factors cancel; terms joined by addition or subtraction do not. For example, x cannot be cancelled from (x+3)/x.
Select a method for linear, quadratic and cubic equations
| Degree/form | Useful solution route |
|---|---|
| linear | collect the unknown terms and isolate the unknown |
| quadratic | factorise, complete the square, use the formula, or read graph intersections |
| cubic | find a linear factor/root, divide to a quadratic, then solve the quadratic |
ax^2+bx+c=0\Rightarrow x= rac{-b\pm\sqrt{b^2-4ac}}{2a}\quad(a
e0)
2x^2-17x-33=0\Rightarrow x= rac{17\pm\sqrt{553}}4\Rightarrow xpprox10.1 ext{ or }-1.63;(3 ext{ s.f.})
Substitute each solution into the original equation, not only a rearranged form. Clear fractions only after recording denominator restrictions, and reject a root that makes an original denominator zero.
Solve two linear equations together
A simultaneous solution is one ordered pair (x,y) that satisfies both linear equations. Algebraically, eliminate one unknown or substitute an expression for it; graphically, it is the intersection of the two lines.
2x+3y=2.5,\quad4x+2y=7\ 4x+6y=5\Rightarrow4y=-2\Rightarrow y=-0.5,\quad x=2
For elimination, first scale one or both equations so one pair of coefficients is equal or opposite. Add or subtract the complete equations, solve the remaining one-variable equation, then substitute back.
Parallel distinct lines have no solution; the same line written twice has infinitely many solutions. Always check the pair in both original equations.
Solve a linear and quadratic equation simultaneously
Use the linear equation to express one variable in terms of the other, substitute into the quadratic equation, solve the resulting quadratic, then find the matching second coordinate for every valid root.
x^2+y^2=26,\quad2x+y=9\Rightarrow y=9-2x\ x^2+(9-2x)^2=26\Rightarrow5x^2-36x+55=0\ (x,y)=(5,-1) ext{ or }\left( rac{11}{5}, rac{23}{5}
ight)
The pairs are the intersection points of a line and a quadratic curve. There may be zero, one or two real pairs, so do not stop after finding the first root.
Keep each x value paired with the y obtained from it. Substitution must replace every occurrence of the chosen variable, including squared occurrences.
Solve and represent linear inequalities
| Situation | Representation rule |
|---|---|
| multiply/divide by a negative | reverse the inequality sign |
| < or > on a number line | open endpoint |
| ≤ or ≥ on a number line | closed endpoint |
| strict boundary in 2D | dashed line |
| inclusive boundary in 2D | solid line |
-13\le5x-3<12\Rightarrow-10\le5x<15\Rightarrow-2\le x<3
For a two-dimensional inequality, draw its boundary line, test a point not on the line, and shade the half-plane whose test point makes the inequality true. For simultaneous inequalities, keep only the overlap.
Do not reverse the sign when adding or subtracting. This syllabus may ask for simultaneous graphical inequalities, but not linear programming or optimisation.
Solve a quadratic inequality by sign intervals
Move every term to one side, factorise or find the roots, place the critical values on a number line, then determine where the quadratic has the required sign.
2x^2+9x+7<0\Rightarrow(2x+7)(x+1)<0\Rightarrow- rac72<x<-1
The roots split the number line into intervals on which the sign cannot change without passing through a root. A positive-leading quadratic is negative between two distinct real roots and positive outside them.
Include a root only for ≤ or ≥, never for < or >. Do not solve the inequality by treating the inequality sign as an equals sign and reporting only the roots.
Recognise and continue number sequences
| Pattern | Recognition | Continuation example |
|---|---|---|
| arithmetic | constant first difference | 4,9,14,19,… adds 5 |
| geometric | constant multiplier | −729,243,−81,27,… multiplies by −1/3 |
| square numbers | 12,22,32,… | 1,4,9,16,25,… |
| triangular numbers | add 2,3,4,… | 1,3,6,10,15,… |
Compare consecutive terms first. If differences are constant, continue by adding that common difference. If ratios are constant and defined, continue by multiplying. Otherwise test a familiar integer pattern or alternating rule.
A pattern must explain every supplied transition, not only the last two terms. More than one rule can fit a short list, so use the simplest rule consistent with the stated context.