3.1 Algebra
- Syllabus
- 9709–2028–2029
- Topic
- 3.1
- Level
- A2
Learning objectives
3.1.1Modulus equations• understand the meaning of |x|, sketch the graph of y = |ax + b| and use relations such as |a| = |b| ⇔ a2 = b2 and |x - a| < b ⇔ a - b < x < a + b when solving equations and inequalities Graphs of y = |f(x)| and y = f(|x|) for non-linear functions f are not included. e.g. |3x - 2| = |2x + 7|, 2x + 5 < |x + 1|.3.1.2Polynomial division• divide a polynomial, of degree not exceeding 4, by a linear or quadratic polynomial, and identify the quotient and remainder (which may be zero)3.1.3Factor and remainder theorems• use the factor theorem and the remainder theorem e.g. to find factors and remainders, solve polynomial equations or evaluate unknown coefficients. Including factors of the form (ax + b) in which the coefficient of x is not unity, and including calculation of remainders.3.1.4Partial fractions• recall an appropriate form for expressing rational functions in partial fractions, and carry out the decomposition, in cases where the denominator is no more complicated than - (ax + b)(cx + d)(ex + f) - (ax + b)(cx + d)2 - (ax + b)(cx2 + d) Excluding cases where the degree of the numerator exceeds that of the denominator.3.1.5Binomial expansion• use the expansion of (1 + x)n, where n is a rational number and x 11. Finding the general term in an expansion is not included. Adapting the standard series to expand e.g. x2 2 1 1 - - `j is included, and determining the set of values of x for which the expansion is valid in such cases is also included.
Use the linear modulus graph and sign-consistent cases
y=∣ax+b∣ is a V-graph with vertex (−b/a,0) for $a
e0,rangey\ge0,andbranchgradients-|a|then|a|.Itrepresentsdistancefromthezeroofax+b$.
| Form | Equivalent condition |
|---|---|
| ∣u∣=∣v∣ | u=v or u=−v |
| ∣x−a∣<b, b>0 | a−b<x<a+b |
| ∣x−a∣>b, b>0 | x<a−b or x>a+b |
For ∣f(x)∣=g(x) require g(x)≥0, then solve both sign branches and check the original. For variable-side inequalities, split at zeros of both sides and intersect each result with its sign interval.
$|3x-2|=|2x+7|$ gives $3x-2=2x+7$ or $3x-2=-(2x+7)$, so $x=9$ or $x=-1$.
Endpoint inclusion follows < versus ≤. Non-linear graphs y=∣f(x)∣ and y=f(∣x∣) are excluded.
Control polynomial division by the remainder degree
P(x)=D(x)Q(x)+R(x),\qquad \deg R<\deg D.A linear divisor leaves a constant remainder; a quadratic divisor may leave $mx+c$.
Write descending powers with zero coefficients for gaps. Repeatedly divide leading terms, multiply the whole divisor, subtract, and stop only when the remaining degree is lower than the divisor degree.
x^4+2x^2+3=(x^2+1)(x^2+1)+2.Thus quotient $x^2+1$ and remainder $2$.
Handle dividends of degree at most 4 and either linear or quadratic divisors. Synthetic division is only a shortcut for suitable linear divisors.
Always report both quotient and remainder, including zero, and verify DQ+R=P.
Convert divisor statements into polynomial evaluations
| Statement | Evaluation |
|---|---|
| remainder on division by x−c is k | P(c)=k |
| x−c is a factor | P(c)=0 |
| ax+b is a factor | P(−b/a)=0 |
Turn every factor or remainder statement into an equation. Solve simultaneous equations for unknown coefficients where needed; divide out confirmed factors before solving the lower-degree polynomial.
For $P(x)=x^3+kx+6$, if $2x-1$ is a factor then $P(1/2)=0$, giving $1/8+k/2+6=0$ and $k=-49/4$.
If division by x+2 leaves remainder 5, use P(−2)=5, not zero. Use full polynomial division when the quotient is also required.
The zero of ax+b is −b/a; keep the sign and non-zero remainder exactly as stated.
Write one numerator term for every permitted denominator factor
| Denominator structure | Partial-fraction form |
|---|---|
| (ax+b)(cx+d)(ex+f) | A/(ax+b)+B/(cx+d)+C/(ex+f) |
| (ax+b)(cx+d)2 | A/(ax+b)+B/(cx+d)+C/(cx+d)2 |
| (ax+b)(cx2+d) | A/(ax+b)+(Bx+C)/(cx2+d) |
Factor the denominator, write the complete matching template, multiply through by the original denominator, then substitute convenient roots and/or equate coefficients to solve all constants.
rac1{x(x+1)}= rac1x- rac1{x+1}.Multiplication by $x(x+1)$ verifies $1=(x+1)-x$.
Substitute the solved coefficients back and recombine as a check before using the decomposition in later algebra or integration.
A repeated factor needs every power, and a quadratic factor needs a linear numerator. Cases where numerator degree exceeds denominator degree are excluded here; do not add an improper-division method.
Expand rational powers and transform the validity condition
For rational $n$ and $|x|<1$:(1+x)^n=1+nx+ rac{n(n-1)}{2!}x^2+ rac{n(n-1)(n-2)}{3!}x^3+\cdots.Useonlyasmanyinitialtermsasrequested.
Rewrite the expression as a constant multiple of (1+u)n, substitute u into the displayed initial terms, expand and collect powers. Do not seek a general term; it is excluded.
(1-2x)^{-1/2}=1+x+ rac32x^2+ rac52x^3+\cdots,obtained with $n=-1/2$ and $u=-2x$.
Transform the convergence condition with the same substitution: ∣u∣<1. In the example, ∣−2x∣<1, so ∣x∣<1/2. For a shifted/scaled u(x), solve the resulting inequality and report the full set.
For non-integer rational n the expansion is generally infinite, not a finite P1 binomial. The standard ∣x∣<1 condition applies to the inner series variable u, not automatically to the original x.