Polynomials in One Variable: Formula and Examples

Polynomials in One Variable: Solving algebraic equations, or finding the roots of polynomials, is one of the oldest problems in mathematics, but we only developed the elegant and practical notation we use today in the 15th century. Algebra is the generalised form of arithmetic. In arithmetic, we use numbers like \(6,-10,0.01 .-0.66\), etc., each with a definite value, whereas, in algebra, we use letters \(x, y, z, a, b, p\), etc., along with numbers. The letters used in algebra are variables, literal numbers or simply literals. They do not have a fixed value.

When one or more, unlike terms, appear in an algebraic expression, it is called a polynomial. Our article begins with the study of a particular type of algebraic expression, the polynomial, and then moves on to understanding polynomials in one variable.

Polynomials in One Variable

In algebra, we come across two types of symbols, namely, constants and variables. A symbol with a fixed numerical value is called a constant, for example, \(10,112,-7, \frac{5}{7}\), whereas a symbol whose value changes with situations is called a variable, for instance, \(x, y, z, x^{2}\).

Now, in \(3 x, 3\) is a constant, and \(x\) is a variable, but, together, \(3 x\) is a variable because, as the value of \(x\) will change, the value of \(3 x\) will also vary accordingly. Similarly, \(3\) is constant, and \(x\) is a variable, but, together, each of \(x+3, x-3\) and \(x \div 3\) is a variable. Thus, we can conclude that every combination of a constant and a variable is always a variable.

Now consider a square of side \(4 \mathrm{~cm}\).

What is its perimeter? We know that the perimeter of a square is the sum of the lengths of its four sides. Thus, perimeter \(=4 \times 4=16 \mathrm{~cm}\).

Now, if each side of a square is \(11 \mathrm{~cm}\), what will be the perimeter??

The perimeter is \(11 \times 4=44 \mathrm{~cm}\). In case if the length of each side is \(y\) units, the perimeter is given by \(4 y\) units. So, as the length of the side of a square varies, the perimeter varies.

The area of the square having each side as \(y\) units will be \(y \times y\) square units or \(y^{2}\) square units. \(y^{2}\) is an algebraic expression. The other examples of algebraic expressions are \(2 x, x^{2}+3 x, x^{3}-4 x^{2}+10 x\). As we can see in the examples, all the algebraic expressions have only whole numbers as the exponents of the variable. Thus, expressions of such forms are called polynomials in one variable.

In the polynomial \(\left(x^{3}-4 x^{2}+10 x\right), x^{3}, 4 x^{2}\) and \(10 x\) are called the terms of the polynomial. Each term of a polynomial has a coefficient.

1. A polynomial can have a finite number of terms. The polynomial having only one term is called a monomial. For example, \(9,-11, x,-x y\).
2. A polynomial of two, unlike terms, is called a binomial. For example, \(x+3 y, 2-7 y z\).
3. A polynomial containing three unlike terms is called a trinomial. For example, \(x+y+2 z, a b-b c-c z\)

Degree of Polynomial in One Variable

When an algebraic expression is made of a variable only, it is called a polynomial in one variable. The degree of a polynomial on one variable is the greatest of the exponents or powers of its various terms. For example, in \(4 y^{2}+3 y^{5}+15 y^{6}\), the exponent of the term \(4 y^{2}=2\), the exponent of the term \(3 y^{5}=5\) and the exponent of the term \(15 y^{6}=6\). since the greatest exponent is \(6\), therefore the degree of the polynomial \(4 y^{2}+3 y^{5}+15 y^{6}\) is \(6\).

The degree of the polynomial \(100+x^{7}\) is \(7\), the degree of the polynomial \(12 y-5\) is \(1\), and so
on.

Observe the polynomial \(2, \frac{7}{5}\), etc. The only term here given is \(2\) or \(\frac{7}{5}\). We can write it as \(2 x^{0}\) or \(\frac{7}{5} x^{0}\). So, the exponent of \(x\) is \(0\). Therefore, the degree of the polynomial is \(0\).

Observe the polynomials, \(4 x+7,3 y, 5 t+1\). The degree of each of these polynomials is \(1\).

A polynomial of degree \(1\) is called a linear polynomial. A linear polynomial in one variable can at most have two terms. So, any linear polynomial in \(x\) will be in the form of \(a x+b\), where \(a\) and \(b\) are constants and \(x \neq 0\).

A polynomial with degree \(2\) is called a quadratic polynomial. For example, \(7 x^{2}-7 x \cdot\) A quadratic polynomial in one variable can at most have \(3\) terms, and, thus, any quadratic polynomial in \(x\) will be in the form of \(a x^{2}+b x+c\), where \(a, b\) and \(c\) are constants and \(x \neq 0\).

We call a polynomial with degree \(3\) a cubic polynomial. A few examples of a cubic polynomial in \(y\) are \(3 y^{3}, 10 y^{3}+5 y^{2}, y^{3}+6 y^{2}-6 y+5\). A cubic polynomial in one variable can have at most \(4\) terms. The standard form of a cubic polynomial is \(a x^{3}+b x^{2}+c x+d\) where \(a \neq 0\) and \(a, b, c\) and \(d\) are constants.

We have seen what a polynomial of degree \(1\), degree \(2\), or degree \(3\) looks like. Hence, the generalised form of a polynomial in one variable of degree \(n\) can be written as

\(a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots . .+a_{1} x^{1}+a_{0}\)

Factorisation of Polynomial in One Variable

Factorisation Using Identities:

The three identities are:

1. \((a+b)^{2}=a^{2}+2 a b+b^{2}\)
2. \((a-b)^{2}=a^{2}-2 a b+b^{2}\)
3. \((a-b)(a+b)=a^{2}-b^{2}\)

Let’s have a look at the examples using each identity to factorise the polynomial in one variable.

Example 1: Factorise the polynomial \(x^{2}+10 x+25\).
Solution: \(x^{2}+10 x+25=(x)^{2}+2(x)(5)+5^{2}\)
\(=(x)^{2}+2(x)(5)+5^{2}\)
\(=(x+5)^{2}\)
Hence, we can factorise \(x^{2}+10 x+25\) as \((x+5)(x+5)\).

Example 2: Factorise the polynomial \(49 t^{2}-14 t+1\).
Solution: \(49 t^{2}-14 t+1=(7 t)^{2}-2(7 t)(1)+(1)^{2}\)
\(=(7 t-1)^{2}\)
Hence, we can factorise \(49 t^{2}-14 t+1\) as \((7 t-1)(7 t-1)\).

Example 3: Factorise the polynomial \(3 y^{2}-75\).
Solution: \(3 y^{2}-75\)
When there is a common factor, we should take it out first.
Therefore, \(3 y^{2}-75=3\left(y^{2}-25\right)\)
\(3\left(y^{2}-25\right)=3\left[(y)^{2}-(5)^{2}\right]\)
\(=3(y+5)(y-5)\)
Hence, we can factorise \(3 y^{2}-75\) as \(3(y+5)(y-5)\).

Sum or Difference of Two Cubes

We shall use the following identities:

1. \(a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\)
2. \(a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\)

Let us solve an example to understand better the concept:

Example 1: Resolve the algebraic expression \(64 x^{3}-\frac{343}{x^{3}}\) into factors.
Solution: \(64 x^{3}-\frac{343}{x^{3}}=(4 x)^{3}-\left(\frac{7}{x}\right)^{3}\)
\(=\left(4 x-\frac{7}{x}\right)\left[(4 x)^{2}+4 x \times \frac{7}{x}+\left(\frac{7}{x}\right)^{2}\right]\)
\(=\left(4 x-\frac{7}{x}\right)\left[16 x^{2}+28+\frac{49}{x^{2}}\right]\)

Hence, we can factorise \(64 x^{3}-\frac{343}{x^{3}}\) as \(\left(4 x-\frac{7}{x}\right)\left[16 x^{2}+28+\frac{49}{x^{2}}\right]\).

Factorisation by Splitting the Middle Term

Splitting the middle term can be applied on the quadratic equation of the form of \(a x^{2}+b x+c\), where \(a, b\) and \(c\) are constants and \(x \neq 0\).

The rule to factorise \(a x^{2}+b x+c\), where \(a, b\) and \(c\) are real numbers is split \(b\) (the coefficient of \(x\)) into two real numbers such that the algebraic sum of these two numbers is \(b\), and their product is \(a c\). Then factorise by grouping method.

Although it is not always possible to factorise a polynomial in one variable \(a x^{2}+b x+c \); the following rule can save a lot of time.

For the expression \(a x^{2}+b x+c\), work out \(b^{2}-4 a c\). If it is a perfect square, then the given expression will factorise. Otherwise, not.

For example, factorise \(x^{2}-8 x+12\).

The middle term is \(-8 x\), we have to split in such a way that the obtained numbers when multiplied together gives \(12 x^{2}\).

Thus, \(-8 x\) can be written as \(-2 x-6 x\).

Therefore, \(x^{2}-8 x+12=x^{2}-2 x-6 x+12\)
\(=x(x-2)-6(x-2)\)
\(=(x-6)(x-2)\)
Hence, we can factorise \(x^{2}-8 x+12\) as \((x-6)(x-2)\).

Let us take one more example for a better understanding.

Example 2 : Factorise \(x^{2}+9 x+18\)
Solution: To factorise \(x^{2}+9 x+18\), we want to find two real numbers whose sum is \(9\) and the product is \(18\). By trial, we see that \(3+6=9\) and \(3 \times 6=18\)
Therefore, \(x^{2}+9 x+18=x^{2}+3 x+6 x+18\)
\(=x(x+3)+6(x+3)\)
\(=(x+3)(x+6)\)
Hence, we can factorise \(x^{2}+9 x+18\) as \((x+3)(x+6)\).

Solved Examples – Polynomial in One Variable

Q.1. What is the degree of the following polynomial.
a) \(3 a-5 a^{2}+7\)
b) \(12 x^{3}+4 x\)
Ans: a) In the polynomial \(3 a-5 a^{2}+7,-5 a^{2}\) has the highest power of \(2\). Thus, the degree is
\(2\).
b) In the polynomial \(12 x^{3}+4 x, 12 x^{3}\) has the highest power of \(3\). Thus, the degree is \(3\).

Q.2. Write the degree of the polynomial, \(1-100 a^{100}\).
Ans: In the polynomial \(1-100 a^{100}\), the degree is \(100\).

Q.3. Write the degree of the polynomial, \(x^{3}+x^{7}-x^{9}\).
Ans: As the highest power of variable \(x\) in the given polynomial is \(9\), the degree of the polynomial \(x^{3}+x^{7}-x^{9}\) is \(9\).

Q.4. Write the number of terms in the following given polynomials.
a) \(3 x-5 x^{5}+7\)
b) \(7 y^{3}+9 y^{2}-8 y+5\)
Ans: a) In the given polynomial \(3 x-5 x^{5}+7\), the number of terms is \(3\), and they are \(3 x,-5 x^{5}\) and \(7\).
b) In the given polynomial \(7 y^{3}+9 y^{2}-8 y+5\), the number of terms is \(4\), and they are \(7 y^{3}, 9 y^{2}, 8 y\) and \(5\).

Q.5. Find the degree of the polynomial \(2\).
Ans: \(2\) is a polynomial, and we can express it as \(2 x^{0}\). The degree of the polynomial \(2\) is \(0\).

Summary

This article discussed the polynomial and then mainly emphasises our study on the polynomial in one variable. We also learned to find the polynomial degree and learned about the quadratic polynomial and cubic polynomial. We now know to write the generalised form of linear, quadratic, and cubic polynomial and the polynomial with degree \(n\).

Frequently Asked Questions (FAQs) – Polynomial in One Variable

Let’s look at some of the commonly asked questions about polynomial in one variable:

Q.1. What is polynomial in one variable?
Ans: The algebraic expressions having only whole numbers as the exponents of the variable are polynomials. Thus, expressions of such forms in only one variable are called polynomials in one variable. For example, \(2 x, x^{2}+3 x, x^{3}-4 x^{2}+10 x\)

Q.2. How to identify polynomial in one variable?
Ans: Polynomials in one variable can be identified if they contain only one type of variable. For example, \(2 x+4\) has only one variable, i.e., \(x\)

Q.3. Write the generalised form of polynomial in one variable of degree \(n\).
Ans: We can write the generalised form of a polynomial in one variable of degree \(n\) as
\(a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots . .+a_{1} x^{1}+a_{0}\)

Q.4. How to check whether expressions are polynomials in one variable?
Ans: If the polynomial contains only one variable in it, then we can say that the expression is a polynomial in one variable. For example, \(y^{100}+y^{99}\). But if the expression contains more than one variable in it, then it is not a polynomial in one variable. For example, \(x^{2}-y z+a^{2} y\).

Q.5. Write the generalised form of the quadratic polynomial and a cubic polynomial.
Ans: The expression of generalised form for a quadratic polynomial is \(a x^{2}+b x+c\), where \(a, b\) and \(c\) are constants and \(x \neq 0\), whereas the expression of generalised form for a cubic polynomial is \(x^{3}+b x^{2}+c x+d\) where \(a \neq 0\) and \(a, b, c\) and \(d\) are constants.

We hope this detailed article on polynomials in one variable helped you in your studies. If you have any doubts or queries regarding this topic, feel to ask us in the comment section and we will be more than happy to assist you.