Many members of
Kapitoly: Many members of, Division of polynomials, The root of a polynomial, Decomposing polynomials
A polynomial is an expression that contains the variable x and the standard operations of addition, multiplication, and exponentiation on an integer exponent. These polynomials can then also be added, subtracted, multiplied, divided and multiplied. Polynomials are also called polynomials.
Basic relationships
An example of a simple polynomial is a polynomial:
$$2x^2+5x-12$$
In general, we could write a polynomial as follows
$$a_nx^n + a_{n-1}x^{n-1}+\ldots+a_1x^1+a_0x^0,$$
where the real numbers before x, i.e. an are called the coefficients and n is called the degree of the polynomial. The number n corresponds to the highest power of the polynomial, where an≠0. If an were equal to zero, then we would essentially cancel the variable x to which the coefficient belongs, since x to anything is zero:
$$0^3=0.$$
So in the case of the former, the degree of the polynomial is two, since the highest non-zero power of the variable x is two. In the case of the other polynomial
$$7x^2+5x^4+15x$$
it would be true that the degree of the polynomial is 4 (the highest exponent is 4). We usually write polynomials in descending order with respect to the exponents used, so that we have the member with the highest power first. So we can rewrite the previous polynomial as follows:
$$5x^4+7x^2+15x.$$
We see that the polynomial is missing a term whose variable would have exponent three and zero. That doesn't matter at all, we simply imagine that the coefficient on these terms is zero. In fact, we can write
$$5x^4+0x^3+7x^2+15x^1 + 0x^0.$$
We may have all the terms already there, but the polynomial is unnecessarily opaque. So we usually leave out the zero terms.
Adding and subtracting polynomials
Adding and subtracting polynomials is quite simple. We always just add or subtract the coefficients of terms with the same exponent. Thus
$$ax^n+bx^n=(a+b)x^n.$$
A simple example:
$$3x^2+5x^2=(3+5)x^2=8x^2$$
If we want to add longer polynomials, we always have to pick members with the same exponent at the variable.
$$\begin{eqnarray} (7x^3+5x^2+x) + (2x^3-3x^2+9x) =\\ =(7+2)x^3+(5-3)x^2+(1+9)x. \end{eqnarray}$$
If a term with a given exponent does not appear in the polynomial, it remains unchanged (imagine that there is such a term in the other polynomial, but with zero coefficients).
$$(x^3+2x)+(4x^2+5x+3)=x^3+4x^2+7x+3$$
Subtraction of polynomials works the same way as addition, except that we multiply the second polynomial by minus one, i.e. we swap all the signs. An example follows:
$$(3x^2+6x)-(2x^2+14x)=(3x^2+6x)+(-2x^2-14x)$$
And this is already a classic addition of polynomials, so it can be added as always:
$$(3x^2+6x)+(-2x^2-14x)=x^2-8x$$
Recall that if there is a negative term in the polynomial, it turns positive after this modification.
$$\begin{eqnarray} &&(x^2+2x)-(3x^2-10x)=\\ &=&(x^2+2x)+(-3x^2+10x)=\\ &=&-2x^2+12x \end{eqnarray}$$
Multiplication of polynomials
When multiplying polynomials, we multiply each member of the first polynomial by each member of the second polynomial. We multiply the coefficients normally, like classical real numbers. In contrast, we just add the exponents of the variables according to the rules for counting with powers. So the following is an example:
$$(3x^2+4x)\cdot 5x^2=(3\cdot5)x^{2+2}+(4\cdot5)x^{1+2}=15x^4+20x^3$$
This example was still working with a polynomial of one term, so it did not illustrate multiplication across all the terms of the polynomial. A slightly larger example follows:
$$\begin{eqnarray} &&(x^3+2x)\cdot(4x^2+7x)=\\ &=&(1\cdot4)x^{3+2}+(1\cdot7)x^{3+1}+(2\cdot4)x^{1+2}+(2\cdot7)x^{1+1} \end{eqnarray}$$
If a minus sign appears in a polynomial, then it will normally show up in the multiplication.
$$\begin{eqnarray} &&(x^3-2x)\cdot(4x^2-7x)=\\ &=&(1\cdot4)x^{3+2}+(1\cdot(-7))x^{3+1}+(-2\cdot4)x^{1+2}+\\ &+&(-2\cdot(-7))x^{1+1}=4x^5-7x^4-8x^3+14x^2 \end{eqnarray}$$
Modifying polynomials
When we edit polynomials, we classically want to edit the polynomial to make it simpler. To do this we use things like expanding, truncating, extruding, applying formulas, and so on. These formulas would be useful to know. A list of useful formulas can be found elsewhere. We will use some of them here, which I will mention in due course. The classic formula that is used is the formula
$$(a+b)^2=a^2+2ab+b^2$$
If you are wondering how this formula came about, just multiply it out.
$$\begin{eqnarray} (a+b)^2&=&(a+b)(a+b)\\ &=&a^2+ab+ba+b^2\\ &=&a^2+2ab+b^2 \end{eqnarray}$$
We'll start with a simple example.
$$\begin{eqnarray} &&(a+b)^2+(a+2b)^2-3a\cdot2b=\\ &=&a^2+2ab+b^2+a^2+4ab+4b^2-6ab=\\ &=&2a^2+5b^2 \end{eqnarray}$$
We multiplied the two left parentheses by the formula I gave above. I remind you that we've decomposed the second parenthesis as
$$(a+2b)^2=a^2+4ab+4b^2.$$
Another classic formula is
$$a^2-b^2=(a+b)(a-b).$$
Here's a second example, this time with a fraction There are polynomials in both the numerator and denominator, and we'll try to modify them so that the whole fraction can be truncated nicely. In the first step, we apply the first formula in the numerator, only in reverse. In the denominator we apply the second formula. At the end, we just truncate (a + b).
$$\begin{eqnarray} \frac{a^2+2ab+b^2}{a^2-b^2}&=&\frac{(a+b)^2}{(a+b)(a-b)}\\ &=&\frac{(a+b)(a+b)}{(a+b)(a-b)}\\ &=&\frac{a+b}{a-b} \end{eqnarray}$$
In the third example, we'll try applying the exclamation. In the numerator, we first extract 3a3 and then decompose the parenthesis (4a2 − 1) according to the formula a2 − b2, taking advantage of the fact that
$$1^2=1^1\rightarrow(a^2-1^2)=(a+1)(a-1).$$
In the denominator, we then first simply multiply the first two terms, also output 3a3, and finally truncate everything.
$$\begin{eqnarray} \frac{12a^5-3a^3}{2a^2\cdot3a^2+3a^3}&=&\frac{3a^3(4a^2-1)}{6a^4+3a^3}\\ &=&\frac{3a^3(2a+1)(2a-1)}{3a^3(2a+1)}\\ &=&2a-1 \end{eqnarray}$$