Математика с д-р Атанас Илчев

Often in solving various problems we have to represent a polynomial as a product of factors. These multipliers can be either monomials or other polynomials. For example, it is well known that $a(b+c)=a.b+a.c$. If we write this equality in reverse order i.e. $a.b+a.c=a(b+c)$ we see that we have already represented the sum of the monomials $ab$ and $bc$ as a product. Let us consider some problems. 1 Problem Calculate rationally $14.85+14.15.$ Solution: The first warrant for solving this problem is to first calculate the product $14.85$ and the product $14.15$ and then add the resulting numbers. The way is not wrong of course, but it does not fit the word rational. Let us now instead consider the equality $a.b+a.c=a(b+c)$, where $a=14$, $b=85$ and $c=15$. We replace the letters with their corresponding equal numbers and get $14.85+14.15=14.(85+15)=14.100=1400.$  Problem 2 Decompose the polynomial $5t+5m$ into factors. Solution:  Notice that in the first and second addends we h...

1 Problem Find the numerical value of the expression $a^2+b^2$ if $a+b=5$ and $ab=9$. Solution: Let's recall the formula for truncated multiplication $(a+b)^2=a^2+2ab+b^2$, which we have already reviewed here. Notice that $a^2+b^2=(a+b)^2-2ab$, this is because $(a+b)^2-2ab=a^2+2ab+b^2-2ab$, and the ones complement $2ab$ and $-2ab$ are truncated. Thus, we have already figured out how to represent $a^2+b^2$ as the sum and product of $a$ and $b$, hence $a^2+b^2=(a+b)^2-2ab=5^2-2.9=25-18=7.$ Problem 2 Prove the identity $(ab+cd)^2=(a^2+c^2)(b^2+d^2)-(ad-cb)^2.$ Solution: We need to prove that the left side of this equality is equal to the right side. We will simplify both sides of the equality and compare them, hence $LS=a^2b^2+2abcd+c^2d^2$. Now we simplify the right-hand side of the equality  $LS=a^2b^2+a^2d^2+b^2c^2+c^2d^2-(a^2d^2-2adcb+c^2b^2)=$ $=a^2b^2+a^2d^2+b^2c^2+c^2d^2-a^2d^2+2adcb-c^2b^2=a^2b^2+c^2d^2+2abcd$.  Thus we get that $LS=RS$ and the equality is an identity...

We continue with the next and last of the abbreviated multiplication formulas $(a\pm b)(a^2\mp ab+b^2)=a^3\pm b^3$. We will look at some problems to show some applications of it. Problem 1 Perform the multiplication $(3-x)(9+3x+x^2).$ Solution: Notice that given the expression $(3-x)(9+3x+x^2)$, we can write it in the form $(3-x)(3^2+3x+x^2).$ We will apply the formula $(a-b)(a^2+ab+b^2)$, replacing $a$ with $3$ and $b$ with $x$, so we get $(3-x)(3^2+3x+x^2)=3^3-x^3.$ Problem 2 Perform the multiplication $(3t+2)(9t^2-6t+4).$ Solution: Given an expression, we can write it in the form $(3t+2)[(3t)^2-3t.2+2^2]$. Now it is easy to see that we can apply the formula $(a+b)(a^2-ab+b^2)$, where $a=3t$ and $b=2$, so we get $(3t+2)[(3t)^2-3t.2+2^2]=(3t)^3+2^3=27t^3+8.$ Problem 3 Simplify the expression $(x-2)(x^2+2x+4)-x(x-2)(x+2)-4(x-2).$ Solution: Apply the formulas $(a-b)(a^2+ab+b^2)=a^3-b^3$ and $(a-b)(a+b)=a^2-b^2$, hence $(x-2)(x^2+2x+4)-x(x-2)(x+2)-4(x-2)=x^3-2^3-x(x^2-4)-4x+8=x^3-8...

We continue with the next of the abbreviated multiplication formulas $(a\pm b)^3=a^3\pm 3a^2b+3ab^2\pm b^3$. Let's look at some problems to illustrate some of its applications. Problem 1 Perform the grading $(x+2)^3$. Solution: To solve this problem we will apply the formula $(a+b)^3=a^3+3a^2b+3ab^2+b^3$, where in our case $a=x$ and $b=2$, so we get $(x+2)^3=x^3+3x^2.2+3x.2^2+2^3=x^3+6x^2+12x+8$. Problem 2 Perform the $(2m-3n)^3$ grading. Solution:  $(2m)^2.3n+3.(2m).(3n)^2-(3n)^3=8m^3-36m^2n+54mn^2-27n^3.$ 3 Problem Perform the grading $(3t+z^2)^3$ Solution: To solve this problem we will apply the formula $(a+b)^3=a^3+3a^2b+3ab^2+b^3$, where $a=3t$ and $b=z^2$, hence $(3t+z^2)^3=(3t)^3+3. (3t)^2.z^2+3.(3t)(z^2)^2+(z^2)^3=27t^3+27t^2z^2+9z^4t+z^6.$ 4 Problem Simplify the expression $(x+1)^3-2(x-1)^2.$ Solution: Notice that the expression we need to simplify involves two of the shortcut multiplication formulas $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ and $(a-b)^2=a^2-2ab+b^2$, the latter...

We continue with the next of the abbreviated multiplication formulas $(a-b)(a+b)=a^2-b^2$. Let's look at some problems to illustrate its applications. Problem 1 Perform the multiplication $(x+y)(x-y)$. Solution: Now we apply the formula $(a-b)(a+b)=a^2-b^2$, where $a=x$ and $b=y$, hence $(x-y)(x+y)=x^2-y^2$.  Problem 2 Perform the multiplication $(3x-4y)(3x+4y)$. Solution: Apply the formula $(a-b)(a+b)=a^2-b^2$, where $a=3x$ and $b=4y$, hence $(3x-4y)(3x+4y)=(3x)^2-(4y)^2=9x^2-16y^2$. 3 Problem Perform the multiplication $(x^2-z)(x^2+z)$. Solution: Apply the formula $(a-b)(a+b)=a^2-b^2$, where $a=x^2$ and $b=z$, hence $(x^2-z)(x^2+z)=(x^2)^2-(z)^2=x^4-z^2$. Let us recall the power grading property, i.e. $(a^n)^m=a^{n.m}$. 4 Problem Calculate $17.23$ in a rational way. Solution: Represent the product $17.23$ in the following way $17.23=(20-3)(20+3)$ and apply the formula $(a-b)(a+b)=a^2-b^2$, so $17.23=(20-3)(20+3)=20^2-3^2=400-9=391$. 5 Problem Simplify the expression $(3x-...

We will give definitions of some of the basic concepts, which we will explain with concrete examples. Definition 1: A rational expression that has no variables in the denominator is called an integer rational expression. Definition 2:  Monomial, we will call an integer rational expression, which is a product of letters and numbers. One's are also any number, variable or parameter. Definition 3: We will say that a monomial is in normal form when it is written with only one numerical multiplier, which stands in first place and is called a quotient, and any product of ones is written as a power. Example: Let's consider the monomial $3x.x.y.4.y.z$. Obviously, this monomial is not in normal form because in its notation we have two numerical factors $3$ and $4$, and also, products of equal letters that are not written as a power. The normal form of the monomial would be $3x.x.y.4.y.z=3.4.x.x.y.y.z=12x^2y^2z.$ As you can see, $12x^2y^2z$ is obviously in normal form because it satisf...

With this and the following lessons, we will try together to overcome the difficulties in solving various problems in which these formulas are applied. Let us recall them before we begin: 1. $(a \pm b)^{2}=a^{2}\pm 2ab+b^{2}$; 2. $(a-b)(a+b)=a^{2}-b^{2}$; 3. $(a \pm b)^{3}=a^{3}\pm 3a^{2}b+3ab^{2}\pm b^{3}$; 4. $(a \pm b)(a^{2}\pm ab+b^{2})=a^{3}\pm b^{3}$. In this article, we consider the formulas $(a\pm b)^{2}=a^{2}\pm 2ab+b^{2}$. Let's solve a few problems to show how we will apply them: Problem 1 Perform the grading $(2x+y)^{2}.$ Solution: Let's consider the formula $(a+b)^{2}=a^{2}+2ab+b^{2}$. In our expression, $2x$ plays the role of $a$ and $y$ plays the role of $b$. Let us now write $2x$ instead of $a$ and $y$ instead of $b$. Thus we get that $(2x+y)^{2}=(2x)^{2}+2.2x.y+y^{2}$. Now we need to exponentiate $(2x)^{2}$. Let's recall the following property learned in 6th grade $(a.b)^{n}=a^{n}.b^{n}$, hence $(2x)^{2}=2^{2}.x^{2}=4x^{2}$. Let us now, having made this c...