Математика с д-р Атанас Илчев

 Each of the abbreviated multiplication formulas we have looked at so far is an example of representing a polynomial as a product, for example $a^2-b^2=(a-b)(a+b)$, on the left hand side we have the polynomial $a^2-b^2$ and on the right hand side the product of factors $(a-b)(a+b)$. In the same way, $a^2\pm 2ab+b^2=(a\pm b)^2$. Clearly we see that again we have a polynomial on the left hand side of the equality and a product on the right hand side. Let's write down the other formulas with the left and right parts swapped $a^3\pm 3a^2b+3ab^2\pm b^3=(a\pm b)^3$ and $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2).$ We'll see below how we can apply these equalities to specific problems. Problem 1 Decompose the polynomial $4x^2-y^2$ into factors.  Solution:  In our case $a=2x$ and $b=y$, therefore $4x^2-y^2=(2x-y)(2x+y).$ Problem 2 Decompose the polynomial $p^2-(x+y)^2$ into factors.$ Solution: Again apply $a^2-b^2=(a-b)(a+b)$, in this case $a=p$, $b=x+y$, hence $[p-(x+y)](p+x+y)=(p-x-...

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-...

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...