Maths Resources - Algebric Formulae & Identities
Algebraic Identities
Polynomials
$$(a + b + c) x = ax + bx + cx $$
$$\frac { a + b + c }{x} = \frac{a}{x} + \frac{b}{x} + \frac{c}{x}$$
On the Basis of number of terms:
- Monomial
- Having one term only e.g. 2x, 5z2
- Binomial
- Having two terms e.g. x + 4, x2 - 3
- Trinomial
- Having three terms. e.g. x2 - 3x - 4
On the basis of degree:
- Constant Polynomial
- Polynomial of degree zero. e.g. $$ p(x)_ = 5, g(x) = \frac{3}{4} $$
- Linear Polynomial
- A polynomial of degree 1. e.g. $$f(x) = -3x + 7 $$ In general, $p(x) = ax + b$, where a ≠ 0
- Quadratic Polynomial
- A polynomial of degree 2. e.g. $$f(x) = 4x^2 - 20y + 25$$ In general $p(x) = ax^2 + bx +c $ where a ≠ 0 is a quadratic polynomial.
- Cubic Polynomial
- A polynomial of degree 3. e.g. $$f(x) = x^3 + 4x^2 - 20y + 25$$ In general $p(x) = ax^3 + bx^2 +cx + d $ where a ≠ 0 is a cubic polynomial.
Fractions
$$\frac {a}{b} + \frac {c}{d} = \frac{ad + bc}{bd}$$
$$\frac {a}{b} - \frac {c}{d} = \frac{ad - bc}{bd}$$
$$\frac {a}{b} \times \frac {c}{d} = \frac{ac}{bd}$$
Identities
$$(a + b)^2 = a^2 + b^2 + 2ab $$
$$(a - b)^2 = a^2 + b^2 - 2ab $$
$$(a^2 - b^2) = (a + b)(a - b)$$
$$ (x + a)(x +b) = x^2 + (a + b)x + ab $$
$$(a + b + c)^2 = a^2 + b^2 + c ^2 + 2ab + 2bc + 2ca $$
$$(a + b - c)^2 = a^2 + b^2 + c ^2 + 2ab - 2bc - 2ca $$
$$(a - b - c)^2 = a^2 + b^2 + c ^2 - 2ab + 2bc - 2ca $$
$$(a + b)^3 = a^3 + b^3 + 3a^b + 3ab^2 $$
$$(a - b)^3 = a^3 - b^3 - 3a^b + 3ab^2 $$
$$ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 -ab -bc -ca) $$
$$ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $$
$$ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $$
Exponents
$$ a^m \times a^n = a^{m+n}$$
$$ (a.b)^m = a^m . a^n$$
$$ a^0 = 1, a \neq 0$$
$$ (a^m)^n = a^{m.n}$$
$$ \frac{a^m}{a^n} = a^{m - n}$$
$$ (\frac{a}{b})^m = \frac{a^m}{b^m}$$
$$ a^{-m} = \frac{1}{a^m}a \neq 0$$
$$a^{\frac{1}{n}} = \sqrt [ n ]{ a } $$
$$a^{\frac{m}{n}} = \sqrt [ n ]{ a^m } $$
Roots
$$\sqrt [ n ]{ a } = a^{\frac{1}{n}}$$
$$\sqrt [ n ]{ a^m } = a^{\frac{m}{n}}$$
$$\sqrt[n]{a.b} = \sqrt[n]{a}.\sqrt[n]{a}$$
$$\frac{\sqrt[n]{a}}{\sqrt[m]{b}} = \sqrt[nm]{\frac{a^m}{b^n}}$$