Instruction

1

A polynomial is a fundamental concept for the solution of algebraic equations and representations of exponential, rational and other functions. This structure is the most common in school course subject

**to the square of**the nth equation.2

Often as simplifying the cumbersome expressions, there is a need to build a

**trinomial**into**the square**. This is no ready-made formula, but there are several methods. For example, imagine**a square****trinomial**and the product of two identical terms.3

Consider this example: construct in

**square****trinomial**3•x2 + 4•x – 8.4

Change the entry (3•x2 + 4•x – 8)2 (3•x2 + 4•x – 8)•( 3•x2 + 4•x – 8) and use the rule of multiplication of polynomials, which consists in sequential computation works. First, multiply the first component of the first parentheses to each term of the second, then do the same with the second and finally the third:(3•x2 + 4•x – 8)•( 3•x2 + 4•x – 8) = 3•x2•(3•x2 + 4•x - 8) + 4•x•(3•x2 + 4•x – 8) – 8•(3•x2 + 4•x – 8) = 9•x^4 + 12•X3 – 24•x2 + 12•X3 + 16•x2 – 32•x – 24•x2 – 32•x + 64 = 9•x^4 + 24•X3 – 32•x2 – 64•x + 64.

5

The same result can come, if we remember that the result of multiplying two

**trinomial**s remains the sum of six items, three of which are**square**, AMI each term, and the other three are their various pairwise works in the doubled form. This simple elementary formula looks like this:(a + b + c)2 = a2 + b2 + c2 + 2•a•b + 2•a•c + 2•b•c.6

Apply it to your example:(3•x2 + 4•x - 8)2 = (3•x2 + 4•x + (-8))2 =(3•x2)2 + (4•x)2 + (-8)2 + 2•(3•x2)•(4•x) + 2•(3•x2)•(-8) + 2•(4•x)•(-8) = 9•x^4 + 16•x2 + 64 + 24•X3 – 48•x2 – 64•x = 9•x^4 + 24•X3 - 32•x2 - 64•x + 64.

7

As you can see, the answer was the same, but the manipulation took less than. This is especially important when the terms are themselves complex structures. This method is applicable for

**the trinomial**and any degree and any number of variables.# Advice 2 : How to put a fraction into a square

In the solution of arithmetic and algebraic tasks is sometimes required to build

**a fraction**in**a square**. The easiest way to do it when**the fraction**to a decimal is fairly simple calculator. However, if**the fraction**of the ordinary or combined, in the construction of such number in**the square**may experience some difficulties.You will need

- calculator, computer, Excel.

Instruction

1

To build a decimal

**fraction**in**the square**, take a scientific calculator, type it erected in**the square of****the fraction**, and press the construction in the second degree. On most calculators this button is labeled as "x2". On the standard Windows calculator function calculates the**square**looks like "x^2". For example,**square**a decimal, is equal to 3,14: 3,142 = 9,8596.2

To build in

**the square**of a decimal**fraction**on a normal (accounting) a calculator, multiply this number with itself. By the way, some models of calculators the possibility of raising the number in**the square**even in the absence of a special button. So please check the instruction manual for the specific calculator. Sometimes examples of "tricky" exponentiation is given on the back cover or on the box of the calculator. For example, many calculators for the erection of a number in**a square**is enough to press button "x" and "=".3

For raising to

**the square**fractions (consisting of numerator and denominator), erected in**the square**separately the numerator and denominator of this fraction. That is, use the following rule:(b / h)2 = P2 / Z2, where h is the numerator, b – denominator.Example: (3/4)2 = 32/42 = 9/16.4

If erected in

**the square of****the fraction**– combined (composed of a whole part and fractions), you must bring her to the ordinary mind. That is, apply the following formula:(C h/z)2 = ((u*h+h) / h)2 = (p*z+h)2 / Z2, where C is the integer part of mixed fraction.Example: (3 2/5)2 = ((3*5+2) / 5)2 = (3*5+2)2 / 52 = 172 / 52 = 289/25 = 11 14/25.5

If you build in

**a square**ordinary (not decimal) fractions you have to constantly, use the program MS Excel. To do this, enter in one cell the following formula: =DEGREE(A2;2) where A2 is the cell address which will be entered erected in**the square****a fraction**.To tell the program that input numbers must be treated as an ordinary**fraction**(i.e., not to convert it to decimal form), dial before you**roll**the first digit "0" and "gap". That is, for input, for example, the fraction 2/3, you need to enter: "0 2/3" (and press Enter). While in the entry line displays the decimal representation of fractions introduced. The meaning and representation of fractions directly in the cell preserved in its original form. In addition, when using mathematical functions, whose arguments are fractions the result will also be presented in the form of fractions. Hence**the square**of the fraction 2/3 would be represented as 4/9.# Advice 3 : How to be a square of binomials

The method of allocation of square binomials used in simplifying the cumbersome expressions, and solving quadratic equations. In practice it is usually combined with other techniques, including factorization, grouping, etc.

Instruction

1

The method of singling out a complete square of binomials is based on the use of two formulas of the reduced multiplication of polynomials. These formulas are special cases of the Binomial theorem for the second degree and allow you to simplify the search expression so that it was possible to conduct further reduction or decomposition on the multipliers:

(m + n)2 = m2 + 2·m·n + n2;

(m - n)2 = m2 - 2·m·n + n2.

(m + n)2 = m2 + 2·m·n + n2;

(m - n)2 = m2 - 2·m·n + n2.

2

According to this method from the original polynomial is required to allocate the squares of the two monomials and the sum/difference of their double works. The application of this method makes sense if the high-degree terms not less than 2. Suppose the task is to factorize a lowering of the degree the following expression:

4·y^4 + z^4

4·y^4 + z^4

3

To solve the problem you need to use the method of allocation of the full square. Thus, the term consists of two variables with monomials of even degree. Therefore, it is possible to identify each of them using m and n:

m = 2·y2; n = z2.

m = 2·y2; n = z2.

4

Now we need to give initial expression to the form (m + n)2. It already contains the squares of these terms, but not enough double work. You need to add it artificially, and then subtract:

(2·y2)2 + 2·2·y2·z2 + (z2)2 - 2·2·y2 ·z2 = (2·y2 + z2)2 – 4·y2·z2.

(2·y2)2 + 2·2·y2·z2 + (z2)2 - 2·2·y2 ·z2 = (2·y2 + z2)2 – 4·y2·z2.

5

In the resulting expression, you can see the formula difference of squares:

(2·y2 + z2)2 – (2·y·z)2 = (2·y2 + z2 – 2·y·z)· (2·y2 + z2 + 2·y·z).

(2·y2 + z2)2 – (2·y·z)2 = (2·y2 + z2 – 2·y·z)· (2·y2 + z2 + 2·y·z).

6

So, the method consists of two stages: allocation of the monomials of a complete square of m and n, the addition and subtraction of their double works. Selection method full square binomials can be used not only independently but also in combination with other methods: making the brackets common factor, change of variable, grouping terms, etc.

7

Example 2.

Select the full square in the expression:

4·y2 + 2·y·z + z2.

Solution.

4·y2 + 2·y·z + z2 =[m = 2·y, n = z] = (2·y)2 + 2·2·y·z + (z) 2 – 2·y·z = (2·y + z)2 – 2·y·z.

Select the full square in the expression:

4·y2 + 2·y·z + z2.

Solution.

4·y2 + 2·y·z + z2 =[m = 2·y, n = z] = (2·y)2 + 2·2·y·z + (z) 2 – 2·y·z = (2·y + z)2 – 2·y·z.

8

The method used for finding roots of a quadratic equation. The left part of the equation is a trinomial of the form a·y2 + b·y + c, where a, b and c are numbers and a ≠ 0.

a·y2 + b·y + c = a·(y2 + (b/a)·y) + c = a·(y2 + 2·(b/(2·a))·y) + c = a·(y2 + 2·(b/(2·a))·y + b2/(4·a2)) + c – b2/(4·a) = a·(y + b/(2·a)) 2 – (b2 – 4·a·c)/(4·a).

a·y2 + b·y + c = a·(y2 + (b/a)·y) + c = a·(y2 + 2·(b/(2·a))·y) + c = a·(y2 + 2·(b/(2·a))·y + b2/(4·a2)) + c – b2/(4·a) = a·(y + b/(2·a)) 2 – (b2 – 4·a·c)/(4·a).

9

These calculations lead to the notion of discriminant which is equal to (b2 – 4·a·c)/(4·a) and the roots of the equation are equal:

y_1,2 = ±(b/(2•a)) ± √ ((b2 – 4·a·c)/(4·a)).

y_1,2 = ±(b/(2•a)) ± √ ((b2 – 4·a·c)/(4·a)).