2.1 Algebra

|f(x)|=a with a≥0 means f(x)=a or f(x)=−a. If a<0 there is no real solution. For inequalities, the sign of the comparison determines whether the solution lies inside or outside an interval.

Solve each branch, intersect with any domain, and test boundary points. Nested moduli require a second case split rather than one global square.

|2x−3|=5 gives 2x−3=5 or −5, so x=4 or x=−1.

Squaring can introduce solutions when the right side is not known non-negative; always substitute candidates into the original modulus equation.

Divide a polynomial by x−a using long division or synthetic division. The remainder theorem gives remainder P(a), and P(a)=0 means x−a is a factor.

Write descending powers, include zero coefficients, and verify by divisor×quotient+remainder. Use a known factor to reduce the degree before solving the remaining polynomial.

For P(x)=x³−2x²−x+2, P(1)=0, so x−1 is a factor; division gives x²−x−2=(x−2)(x+1).

A zero remainder proves divisibility by the chosen factor only; it does not make every nearby value a root.

The factor theorem says x−a is a factor of P(x) exactly when P(a)=0. The remainder theorem says division by x−a leaves remainder P(a).

Test simple candidates first, divide out a confirmed factor, then solve the lower-degree quotient. A repeated root also requires P′(a)=0.

For P(x)=x³−4x²+x+6, P(2)=0, so x−2 is a factor; division reduces the cubic before finding the remaining roots.

P(a)=0 identifies a factor x−a, not x+a; substitute the sign carefully.