Separate an expression from an equation
An algebraic expression represents a value; an equation states that two expressions are equal. The expression 3x² − 2x + 7 cannot be solved until a condition is provided. Its terms are 3x², −2x and 7. Like terms have the same variable part, so 4x² + 3x − x² simplifies to 3x² + 3x. Multiplying differs from adding. Label the operation before using an algebraic rule, particularly when both powers and brackets appear.
The distributive law gives. Therefore. A negative sign before a bracket changes every term's sign. Expanding two brackets requires each term of one bracket to multiply each term of the other. For example. Check by multiplying slowly rather than relying on the appearance of a memorised pattern.
Factor expressions and preserve restrictions
Factorisation reverses expansion. Take out a common factor first. Useful identities include and a² ± 2ab + b² = (a ± b)². A sum of two squares does not factor over the real numbers in the same way as a difference of squares. For a grouped expression such as ax + ay + bx + by, group to obtain.
An algebraic fraction can be simplified by cancelling common factors, but not terms joined by addition. Thus only for x ≠ 3. Factoring gives (x − 3)(x + 3)/(x − 3), which explains both the cancellation and the restriction. The original expression is undefined at x = 3 even though the simplified formula has a numerical value there. For 1/x + 1/y, use denominator xy to obtain (x + y)/(xy), with x ≠ 0 and y ≠ 0.
Use polynomial division and theorems
A polynomial in x has nonnegative whole-number powers of x. If P(x) is divided by x − a, the remainder is P(a). This remainder theorem follows from: substituting a leaves only r. If is a factor, which is the factor theorem. For, P(2) = 8 − 8 = 0, so x − 2 is a factor. In fact.
When dividing x³ − 1 by x − 1, write missing coefficients as zero so the powers stay aligned. The quotient is x² + x + 1 and remainder zero, which can be checked by multiplication. A cubic can have more than one real root, but a list of roots must still be checked against the original equation. If a factor theorem problem contains an unknown , substitute the stated root into P(x) first and solve the resulting simpler equation for that .
Solve inequalities with the correct direction
An inequality describes a set of values rather than one necessarily unique number. Adding or subtracting the same quantity preserves its direction. Multiplying or dividing both sides by a positive number also preserves it. Multiplying or dividing by a negative number reverses it: −2x < 6 gives x > −3. This reversal follows from the order of numbers on the number line. A closed endpoint represents ≤ or ≥; an open endpoint represents < or >.
For a compound inequality, apply operations to all three parts. From 2 < 3x + 5 ≤ 11, subtract 5 to get −3 < 3x ≤ 6, then divide by 3 to obtain −1 < x ≤ 2. If x must be an integer, the possible values are 0, 1 and 2. State a restriction such as “integer” only when the problem supplies it. The full real solution set otherwise contains every real number between the indicated endpoints.
and graphical inequalities
For (x − 1)(x − 4) < 0, the roots 1 and 4 divide the line into three intervals. Test a value in each interval. At x = 0 the product is positive; at x = 2 it is negative; at x = 5 it is positive. Therefore 1 < x < 4. For a non-strict inequality ≤ 0, include the roots. Simply writing x < 1 and x < 4 from the factors is not a valid use of the zero-product rule, which applies to equality.
In two variables, the boundary of is x + y = 6. Plot its intercepts (6, 0) and (0, 6), use a solid line because equality is allowed, and test a point away from the line. The origin satisfies, so the permitted half-plane contains it. Several inequalities require the common shaded region. If a simple linear objective is optimised over a closed polygonal feasible region, evaluate it at the vertices, while checking that every vertex satisfies all stated constraints.
Self-check: simplify 5x − 2(x + 3), solve −3x ≥ 12, and find the remainder when x² + 3x + 1 is divided by x − 2. The answers are 3x − 6, x ≤ −4, and 11. Substitution checks inequalities too: −4 is included in the second answer and gives equality. Preserve denominator restrictions, distinguish a factor from a term, and include endpoints only when the inequality permits them. These habits make a solution complete rather than merely plausible.
Factor a cubic and interpret its sign
For x³ − 2x² − x + 2, group the first two and last two terms. The roots are 2, 1 and −1. Expanding the result verifies all coefficients. The factor theorem can identify a first factor when grouping is not obvious, after which polynomial division reduces the degree.
A with positive leading and negative discriminant is positive for all real x. For instance, x² + 1 > 0 has every real x as its solution, while x² + 1 < 0 has no real solution. A sign chart should therefore begin by finding real roots, if any, and considering the leading . Do not assume every inequality gives an interval between two integer roots. If a rational inequality has a variable denominator, also mark forbidden denominator zeros on the sign chart and never include them as endpoints.