Literal Calculation

Developing and Factorizing Equations

For any real numbers:

• $a(b+c)=ab+ac$
• $a(b-c)=ab-ac$
• $(a+b)(c+d)=ac+ad+bc+bd$

Remarkable Identities

For any real numbers:

• Square of a sum: $(a+b{)}^2=a^2+2ab+b^2$
• Square of a difference: $(a-b{)}^2=a^2-2ab+b^2$
• Difference of two squares: $a^2-b^2=(a+b)(a-b)$

Powers and Exponents

For any non-null natural integer:

• $a^n=\underset{n}{\underbrace{a*a*a*\cdots *a}}$
• When $n=0;a^0=1$
• If n is a natural integer et a is a non-null real number: $a^{-n}=\frac{1}{a^n}$

Square Roots

If a is a positive real number, then there are two reals (one positive and one negative) for which the square is a.

• The positive one is written as:$\sqrt{a}$
• If a = 0, we can write $\sqrt{0}=0$
• The formula $\sqrt{a^2}$ is equal to a or -a
• For two non-null positive reals: $\sqrt{ab}=\sqrt{a}\sqrt{b}$
• If a is a positive real or null and b is positive: $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
• In particular: $\sqrt{\frac{1}{b}}=\frac{1}{\sqrt{b}}$
• In an expression like $\frac{A}{\sqrt{b}}$, you can make the denominator rational: $\frac{A}{\sqrt{b}}=\frac{A\sqrt{b}}{(\sqrt{b}{)}^2}=\frac{A\sqrt{b}}{b}$
• Thanks to the third remarkable identity, $(a+b\sqrt{n})(a-b\sqrt{n})=a^2-n{b}^2$
• In an expression like $\frac{A}{a+b\sqrt{2}}$you can simplify the denominator: $\frac{A}{a+b\sqrt{2}}=\frac{A(a-b\sqrt{2})}{a^2-2b^2}$

Equations

• If $ax+b=0$, if a is non-null, $x=-\frac{b}{a}$
• If A = 0 or B = 0, then AB = 0

Inequalities

For reals:

• If $a\le b$ then $a+c\le b+c$
• If $a\le b$ and $c\le d$ then $a+c\le b+c$

When you multiply two members of an inequality by the same number:

• You keep the sense of the inequality if the number is positive or null. If $a\le b$ and $c\ge 0$then $ac\le bc$
• You change the sense of the inequality if the number is negative or null. If $a\le b$ and $c\le 0$ then $ac\ge bc$

You can multiply, member by member the inequalities of the same sense when all the terms are positive or null:

• If $0\le a\le b$and $0\le c\le d$ then $ac\le bd$

When resolving an inequality like $ax+b\ge 0$ you look first at the sign of a and then resolve in three steps:

• If a > 0:
  • Algebraic resolution. The solutions are the reals x such that $x\ge -\frac{b}{a}$
  • You write the set of solutions as: $S=[-\frac{b}{a};+\mathrm{\infty }[$
  • You represent this set of the real number line as greater than $-\frac{b}{a}$
• If a < 0:
  • Algebraic resolution. The solutions are the reals x such that $x\le -\frac{b}{a}$
  • You write the set of solutions as: $S=]-\mathrm{\infty };-\frac{b}{a}]$
  • You represent this set of the real number line as smaller than $-\frac{b}{a}$