Algebra Theorem Basic

The basic Algebra will perform the following four basic operations. Addition, subtraction, multiplication and division. Algebra uses for variables, invariable, coefficients, exponents, words and expressions. The basic concept of the algebra is equalize the algebraic equations on both sides. In algebra theorem we can use the following properties such as commutative, associative, identities and inverse. In algebra, some theorems and example problems are given below.

Basic algebra theorems:

Theorem 1:

Remainder Theorem:

Let P(x) denoted by any polynomial equation of degree greater than or equal to 1 and let a be any real number. When P(x) is divided by the binomial (x–a) the remainder is P(a).

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Proof:

When P(x) is divided by (x – a), let the quotient be q(x) and the remainder be r(x).

Then we have p(x) = (x–a) q(x) + r(x) where degree of r(x) < degree of divisor or r(x) = 0.

Since degree of x – a is 1, r(x) is a constant say r, as discussed earlier. Hence for all values of x,

P(x) = (x – a) q(x) + r.

In particular for x = a, we have from above

P(a) = (a – a) q(a) + r = 0 × q(a) + r  P(a) = r.

Theorem 2:

Factor Theorem:

If p(x) is a polynomial of degree n > 1 and a is any real number then

(i) (x–a) is a factor of p(x) if p(a) = 0 and

(ii) p(a) = 0 if (x–a) is a factor of p(x).

Theorem 3:

Fundamental theorem:

A number a is a solution to the equation P(x) = 0 if substituting a for x makes it identity: P(a) = 0. The coefficients are assumed to belong to a correct set of numbers where we also look for a solution.

The polynomial form is very general and also frequently studying P(x) = Q(x) is more suitable.

Example problems by using Basic algebra theorem:

Example problem 1:

Factoring a given polynomial: f (x) = x4 – x2 using fundamental theorem in algebra.

Solution:

The factorization for f could be done in this way,

f (x) = x4 – x2

We can pull out common terms x2 :

x4- x2 = x2 (x2 – 1).

The final answer is

= x2 ( x2 - 1 )

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Example problem 2:

Find the remainder when 3x3 + 4x2 – 5x + 8 is divided by x + 2.

Solution:

When P(x) is divided by (x + 2), the remainder is P(–2).

The remainder P(–2) = 3(–2)3 + 4(–2)2 – 5(–2) + 8

= 3(–8) + 4(4) + 10 + 8 = 10