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2.1 Algebra

Syllabus
9709–2028–2029
Topic
2.1
Level
AS

Learning objectives

2.1.1Modulus 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|.2.1.2Polynomial 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)2.1.3Factor 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.

Use distance and sign regions for linear modulus problems

∣u∣|u| is the non-negative distance of uu from 00. Thus y=∣ax+b∣y=|ax+b| is a V-shaped graph with vertex where ax+b=0ax+b=0, slopes −∣a∣-|a| then ∣a∣|a|, and range y≥0y\ge0.

Form Equivalent relation
∣u∣=c|u|=c, c≥0c\ge0 u=cu=c or u=−cu=-c
∣u∣=∣v∣|u|=|v| u2=v2u^2=v^2, equivalently u=vu=v or u=−vu=-v
∣x−a∣<b|x-a|<b, b>0b>0 a−b<x<a+ba-b<x<a+b
∣x−a∣>b|x-a|>b, b>0b>0 x<a−bx<a-b or x>a+bx>a+b

If the other side contains xx, locate every zero of both expressions, split the number line into sign regions, remove the modulus with the correct sign in each region, solve and intersect with that region.

$|3x-2|=|2x+7|$ gives $3x-2=2x+7$ or $3x-2=-(2x+7)$, so $x=9$ or $x=-1$.

Check equality endpoints according to << or ≤\le. Non-linear graphs y=∣f(x)∣y=|f(x)| and y=f(∣x∣)y=f(|x|) are excluded from this objective.

Divide polynomials and control the remainder degree

For dividend $P(x)$ and non-zero divisor $D(x)$:P(x)=D(x)Q(x)+R(x),\qquad \deg R<\deg D.

Divisor Possible remainder
linear constant
quadratic linear mx+cmx+c (including zero/special cases)

Write every polynomial in descending powers and insert zero coefficients for missing powers. Divide leading terms, multiply the divisor, subtract the whole row, bring down the next term and repeat until the remainder degree is smaller than the divisor degree.

x^4+2x^2+3=(x^2+1)(x^2+1)+2.The quotient is $x^2+1$ and the remainder is $2$, whose degree is below $2$.

Verify divisor imesimes quotient ++ remainder equals the original dividend. Synthetic division is only a shortcut for suitable linear divisors, not a quadratic divisor.

Turn factors and remainders into evaluation equations

Division statement Evaluation statement
remainder on division by x−cx-c is kk P(c)=kP(c)=k
x−cx-c is a factor P(c)=0P(c)=0
ax+bax+b is a factor, $a
e0∣|P(-b/a)=0$

Translate every factor or stated remainder into a substitution equation. If coefficients are unknown, solve the resulting simultaneous equations; then divide out confirmed factors and solve any lower-degree quotient when roots are required.

Let $P(x)=x^3+kx+6$. If $2x-1$ is a factor, $P(1/2)=0$: rac18+ rac{k}{2}+6=0\quad\Rightarrow\quad k=- rac{49}{4}.

For a non-zero remainder, keep the right side: if division by x+2x+2 leaves 55, then P(−2)=5P(-2)=5, not 00. Use polynomial division when the full quotient is needed.

The zero of ax+bax+b is −b/a-b/a. A factor statement and a remainder statement are different equations; no repeated-root derivative test is needed for this objective.