Polynomial Formula

Polynomial Formula

A polynomial formula is a formula that expresses the polynomial expression. The polynomial an expression that has two or more than two terms(algebraic terms) is known as a polynomial expression. A repetitive summation or subtraction of binomials or monomials forms a polynomial expression. A polynomial can have both like as well as unlike terms in it. Like terms in polynomials are the terms which have the same variable and same power and the terms that have different variables and different powers are known as, unlike terms. Let us see the polynomial formula in the following section along with the solved examples.

What is Polynomial Formula?

The polynomial formula has variables with different power and the highest power of the variable on solving is known as the degree of the polynomial. The polynomial formula is also known as the standard form of the polynomial where the arrangement of the variables is according to the decreasing power of the variable in the formula. 

Polynomial Formula

Polynomial Formula is given by: 

(axn+bx{n−1}+cx{n−3}+……+rx+s)

Where

• a, b, c, …, s are coefficients
• x is the variable
• n is the degree of the polynomial

Some basic formulas associated with the polynomial expression given above are,

1.F(x) = a_n(xⁿ)

where

• a is the coefficient
• x is the variable 
• n is the exponent

2. F(x) = a_nxⁿ + a_n-1x^n-1 + a_n-2x^n-2 + …….. + a₁x +a₀ = 0

3. F(x)=a_nxⁿ+..+rx+s

•  n is a natural number

a_n−b_n=(a−b)(a_n−1+a_n−2b+…)

•  n is even number

a_n+b_n=(a+b)(a_n−1−a_n−2b+…)

•  n is odd number
• a_n+b_n=(a+b)(a_n−1−an−2b+…)

Applications of Polynomial Formula

It has applications in engineering, computer, management, business, and even in farming. Variables and constants are used to create expressions defining quantities that are known and unknown. 

The polynomial equations are formed with variables, exponents, and coefficients. Polynomials can be solved by factoring them into either in terms of the degree or variables present in the given equation.

Let’s take a quick look at a couple of examples to understand the polynomial formula better.

Examples Using Polynomial Formula

Example 1: Find the factors of the given polynomial formula (x²+12x+36).

Solution:

To find:

factors of the polynomial

(x²+12x+36)

(x²+2(6)x+6²)

(x+6)²

Answer: Factors of the polynomial (x²+12x+36) are (x+6) and (x+6).

Example 2: Find the factors of the given polynomial formula (x²+3x-28). 

Solution:

To find:

factors of the polynomial

(x²+3x-28)

(x²+7x-4x-28)

(x(x+7)-4(x+7)

(x-4)(x+7)

Answer: Factors of the polynomial (x²+11x+28) are (x-4) and (x+7)

Example 3: Calculate the factors of the polynomial x² – 6x + 9?

Solution:

x² – 6x + 9
= x² – 2(3x) + 3²
= x² – 2(3)(x) + 3²
= (x – 3)²

Answer: Factors of the polynomial x² – 6x + 9 are (x – 3) and (x – 3).

FAQs on Polynomial Formula

What Is the General Polynomial Formula?

xⁿ where a is the coefficient, x is the variable, n is the exponent. On the other hand, the polynomial formula in the expanded form is,

Types of Polynomial

Polynomial equation is of four types :

1. Monomial: This type of polynomial contains only one term. For example, x² , x, y, 3y, 4z
2. Binomial: This type of polynomial contains two terms. For example, x² – 10x
3. Trinomial: This type of polynomial contains three terms. For example, x² – 10x+9
4. Quadratic Polynomial: This type of polynomial contains four terms. For example, x³+2x² – 10x+7

The general Polynomial Formula is,

F(x) = a_nxⁿ + bx^n-1 + a_n-2x^n-2 + …….. + rx +s

• If n is a natural number: aⁿ – bⁿ = (a – b)(a^n-1 + a^n-2b +…+ b^n-2a + b^n-1)
• If n is even (n = 2a): xⁿ + yⁿ = (x + y)(x^n-1 – x^n-2y +…+ y^n-2x – y^n-1)
• If n is odd number: xⁿ + yⁿ = (x + y)(x^n-1 – x^n-2y +…- y^n-2x + y^n-1)

Polynomial Identities

1. (x + y)²= x² + 2xy + y²
2. (x – y)²= x² – 2xy + y²
3. x²– y² = (x + y)(x – y)
4. (x + a)(x + b) = x²+ (a + b)x + ab
5. (x + y + z)²= x² + y² + c² + 2xy + 2yz + 2zx
6. (x + y)³= x³ + y³ + 3xy (x + y)
7. (x – y)³= x³ – y³ – 3xy (x – y)
8. x³+ y³ = (x + y)(x² – xy + y²)
9. x³– y³ = (x – y)(x² + xy + y²)
10. x³+ y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
11. ax² + bx + c = 0 then x = \frac{( -b \pm \sqrt{(b² – 4ac)} )}{2a}

Solved Examples

Q1. Solve the equation: x² +16x + 64 = 0

Solution: Factors of x² +16x + 64

We the factors of the 64 are 2, 4, 8, 16, 32, 64. The sum of middle term is 8.

Now, x² + (8 + 8) x + 8 ͯ 8

=x² +8x +8x + 8 ͯ 8

= x(x + 8) + 8 (x + 8) = (x + 8) (x + 8)

Q.2. Solve x³ – 7x² + 12x = 0

Solution: To factor x³−7x²+12x

= x (x²−5x+6)

Now, we factorise x²−7x+12

We the factors of the 12 = 3, 4

Now, x² − (3 + 4) x + 3 ͯ 4

=x² -3x -4x + 3 ͯ 4

= x(x – 3) – 4 (x – 3) = (x−3) (x−4)

Factors of x³−7x²+12x  are  x (x−3) (x−4)

Question: What are the factors of the polynomial x² – 10x + 25?
Solution:

x² – 10x + 25
= x² – 2(5x) + 5²
= x² – 2(5)(x) + 5²
= (x – 5)²

Hence, the factors are (x – 5) and (x – 5).

Types of Polynomial Equation

Polynomial equations are classified upon the degree of the polynomial. For practical reasons, we distinguish polynomial equations into four types.

1. Monomial/Linear Equation

A polynomial equation with only one variable term is called a monomial equation. It is also called a linear equation. The algebraic form of a linear equation is of the form:

ax + b=0, where a is the coefficient, b is the constant and the degree of the polynomial is 1.

Examples:

2x + 10 = 0

x – 5 = 0

2. Binomial/Quadratic Equation

A polynomial with two variable terms is called a binomial equation. It can also be called a quadratic equation. The algebraic form of a quadratic equation is of the form:

ax² + bx + c = 0, where a and b are coefficients, c is the constant and the degree of the polynomial is 2.

Examples:

2x² + 2x + 2 = 0

x² – 4=0

3. Trinomial/Cubic Equation

A polynomial with three variable terms is called a trinomial equation. It is also called a cubic equation. The algebraic form of a quadratic equation is of the form:

ax³ + bx² + cx + d = 0, where a, b and c are coefficients, d is the constant and the degree of the polynomial is 3.

Examples:

x³ + 2x² + 3x – 5 = 0

2x³ – 5x = 0

4. Polynomial Equation

A polynomial with more than three variable terms is called a polynomial equation. It is of the form

anxn + an-1xn-1 + an-2xn-2 + . . . + a1x + a1 = 0.

Examples,

4x⁴ + 2x³ + x² + 5 = 0

10x⁵ + 2x – 10 = 0

Solving Polynomial Equations

Polynomial equations are generally solved with the hit and trial method. We put in the value of the independent variable and try to get the value of the expression equal to zero.

In case of a linear equation, obtaining the value of the independent variable is simple. We solve the equation for the value of zero.

For the polynomial, 2x – 4 = 0

2x = 4

x = 2

However, this solution is not easily applicable in higher degrees of polynomials, therefore, we go with the hit and trial method.

Application of Polynomial Equations in Real Life

Polynomials, polynomial equations and polynomial functions are used to graph curves in the real world. Professionals from various careers who carry out complex calculations to determine the nature of curves in real-life scenarios utilise polynomial equations and functions.

Civil Engineering requires their use to determine the curvature of roads, flyovers, bridges and other complex spatial structures. Global Positioning System (GPS) calculates a complex system of polynomial equations to locate your position. 

Polynomial functions prove to be important mechanisms to predict traffic patterns to gauge your estimated time of arrival to your given destination on your public transport apps like Uber and Ola. These equations allow governments to efficiently implement traffic control measures like planning the correct placement of traffic lights to ensure smooth roads.

Robotics employs polynomial equations to predict the trajectory of projectiles. These equations and functions are instrumental in computational biochemistry to model interacting proteins. 

The banking industry and accountants use them for tax and loan calculations. Economists and businessmen use polynomial equations to model market patterns and cost analysis. Statisticians, data analysers and researchers also use them among many other expressions in Mathematics.

People aspiring to be in the nursing industry are also expected to know polynomial equations. They use it to monitor parameters like the depleting or increasing presence of a drug in small quantities in the bloodstream of the patient and the medication dosage required in accordance with the changing health status over time in the hospital.

Even environmental conservationists, nature enthusiasts, the weather department and the forestry department put polynomial functions to good use to predict how natural resources are changing (increasing or depleting) over a period of time-based on the previous data gathered earlier. 

Damage Control teams and Crisis Management teams use them to monitor the curving trajectory of cyclones and hurricanes on the basis of a bunch of variables like wind speed.

To summarise, polynomials, polynomial equations and polynomial functions may seem like Mathematical jargon on paper, but in reality, they are used in most scientific applications. They are crucial for the success of entire industries that help science make human lives more advanced, efficient and sustainable.

Polynomial Equation Formula

F(x) = a_n xⁿ + a_n-1 x^n-1 + a_n-2 x^n-2 +……… + a₁ x + a₀

Polynomial Equation

A polynomial equation is an equation that has varied terms and generally includes variables coefficient and exponent. Polynomials can retain various exponents. The degree of a polynomial is considered as the greatest exponent. The degree of a polynomial states the number of roots that can be present in a polynomial equation.

A polynomial equation is generally a polynomial expression which has been fixed to make the expression equals to zero.

For Example-If the greatest exponent is 3, then we can say that the polynomial equation has 3 roots. The polynomial equation can be easily written if we are aware of the number of roots.

Polynomials can be solved by factoring them in respect of degrees and variables that exist in the polynomial equation.

Polynomial Equation Example

4x² + 6x + 21=0 where 4x²+6x+21 is considered as a polynomial expression which is written on the left side and has been fixed to make the polynomial expression equal to zero to form a complete polynomial equation.

What is known as Polynomial?

Polynomials are algebraic expressions that are composed of two or more algebraic terms. The algebraic terms are Constant, Variables and Exponents. The word poly means “many “and nominal means term in the polynomial.

What is Polynomial Function?

Polynomial Function is a type of polynomial equation which retains only one single variable in which a variable can take place in the given polynomial equation much time which varies the degree of the exponent. Polynomial Function graph can be drawn using various elements such as intercepts, end behavior, turning points, and the intermediate value theorem.

Example of Polynomial Function:-

F(x) =4x²+6x+21

Degree of a Polynomial

The exponent of the highest degree term in a polynomial is known as its degree.

For example: – f(x) =4x³-2x²+8x-21 and g(x) =7x²-3x+12 are polynomials of degree 3 and degree 2 respectively.

Based on the degree of a polynomial, there are 5 standard names for Polynomial Equations.

1.  Constant Polynomial-A polynomial of degree 0 is called a constant polynomial.   

For example-f(x) =2, g(x) = -14, h(y) =5/2 etc are constant polynomials. Constant polynomial 0 or f(x) =o is called the zero polynomial.

2.  Linear Polynomial-A polynomial of degree 1 is called a linear polynomial.           

For example- f(x)=x-12 ,g(x)=12x,h(x)=-7x+8 are linear polynomials

3.  Quadratic Polynomial-A polynomial of degree 2 is known as quadratic polynomial. 

For Example-f(x) =2x²-3x+15, g(x) =3/2y²-4y+11/3 etc are quadratic polynomials.

4.  Cubic Polynomial-A polynomial of degree 3 is called a cubic polynomial.            

For Example-f(x) =12x³-4x²+7x-6,g(x)=7x³+4x-12 are cubic polynomials.

5.  BI-Quadratic Polynomial-A polynomial of degree 4 is known as a biquadratic polynomial. For Example-f(x)=12x⁴-7x³+8x²-12c-20 is biquadratic polynomial

How to solve Polynomial Equations?

Polynomial Equations can be solved with the usage of some general algebraic and factorization rules. The foremost step to solving any of the polynomial equations is to fix 0 on the right side of the polynomial equation.

Method to solve linear polynomials

The foremost step to solving linear polynomials is to make the polynomial equation equals to zero. Then apply some algebraic concepts to solve your equation.

For Example:-

Solve 4x-8

Solution

The first step to make the above equation equals to 0

4X-8=0

4X=8

X=8/4

X=2

Thus the solution of 4x-8 is x=2

Method to solve Quadratic Polynomials

The foremost step to solving a quadratic polynomial is to write the polynomial expression in ascending form. Then place the 0 to the right side of the equation.

For Example:-

Solve 4x²-5x+x³-6

The first step is to arrange the above equation in ascending form and place the zero on the right side of the equation.

=x³+4x²-5x-20

Now take the common terms

=X²(x+4)-5(x+4)

=(x²-5)(x+4)

So the required solution will be considered as

X²=5 and x=-4

And x is also equal to √5

Solved Polynomial Example

1.     Find the value of the f (2) and f (-3) in the given Polynomial equation f(x) =2x³-13x²+17x+12.

Solution: – We have f(x) =2x³-13x²+17x+12

f (2)=2*(2)³-13*(2)²*17*2+12

=2*8-13*4+34+12=16-52+34+12=10

 f(-3)=2*(3)³-13*(-3)²+17*(-3)+12

=2*27-13*9+17*-3+12

= 54-117-51+12=-210

2.     Find the value of polynomial 5x-4x²+3 when x=0.

Solution- Let f(x) = 6x-5x²+2

Putting 0 in place of x, we will get

f(0) =6x (0)-5*(0)²+2

    =0-0+2

    = 2

Quiz Time

What are the variables in polynomials known as?

a.   Symbols

b.  Terms

c.   Coefficients

d.  Degrees

What will be the expanded form of (x+8) (x-10)

a.       x²-8x-80

b.      x²-2x-80

c.       x²+2x+80

d.      x²-2x+80

Fun Facts

A linear polynomial may be binomial or monomial. For example- f(x) =7x-15 is a binomial whereas g(x) =3x is a monomial.

A quadratic polynomial may be a binomial or monomial or trinomial. For Example-f(x) =7x² is a monomial, g(x) =2x²+3 is a binomial and h(x) =3x²-2x+4 is a trinomial.

Polynomials can be dealt with more than one variable. For Example-x²+y²+2xyz (where x, y, z are variable. So it is a polynomial with three variables.

FAQs (Frequently Asked Questions)

Define Monomial, Binomial and Trinomial?

Monomial-A monomial is considered as a polynomial with one term. For Example-4x is a polynomial as it has only one term.

Binomial-A binomial is considered as a polynomial with 2 terms. For Example-4x²+5y² is a binomial as it has two terms

Trinomial-A trinomial is considered as a polynomial with 3 terms. For Example-2a²+5a+7 are a trinomial because it has 3 terms.

What are Terms and their Coefficients?

If (f) x) = anxn + an-1xn-1 + an-2xn-2 + a1x +a0  is a polynomial equation with variable x then anxn + an-1xn-1 + an-2xn-2 …..a₁, a₀ and x will be known as terms of polynomial whereas an, an-1 and an-2 will be considered as coefficients in polynomial equations.

The coefficient of an of the highest degree term is known as the leading coefficient whereas a₀ will be known as the constant term.

For Example-f(x) = 2x₂-7x+8

2x₂.-7x and 8 are its terms and 2,-7 and 8 are coefficients of x², x, and constant terms respectively.

Sample Questions

Ques. What are the factors of the given polynomial formula x² – 10x + 25? (3 Marks)

Ans. x² – 10x + 25

= x² – 2(5x) + 52

= x² – 2(5)(x) + 52

= (x – 5)²

As a result, the factors are (x – 5) and (x – 5) respectively.

Ques. What is the difference between polynomials and equations? (1 Mark)

Ans. The primary distinction between polynomials and equations is that an equation is a mathematical statement stating the equivalence of two expressions, whereas a polynomial is a mathematical expression containing variables and constants.

Ques. Is this statement true or false? The equation g(x) = 0 will always have three real zeros since the polynomial function g(x) has a degree of three. (5 Marks)

Ans. There are only three possible real zeros in the equation g(x) = 0. This means it could have three genuine zeros or none at all. When g(x) = x³ + x, one function can be used to show that the statement is false.

Let’s solve the equation for x and see what happens. We’ll factor x out and equate x² + 1 to 0 because the expression is still factorable.

x3 + x = 0

x(x² + 1) = 0

x = 0

x² + 1 = 0

x² = -1

Only when x = -i or x = I will this be true. Despite having a degree of three, this plainly proves that g(x) can’t have three genuine zeros. As a result, the statement is false.

Ques. If one of the polynomial’s zeroes is 2, get the value of “p” from the polynomial x² + 3x + p. (3 Marks)

Ans. Because the polynomial’s zero is 2.

We know that if the polynomial p(x) has a zero, then p() = 0.

x = 2 is substituted for x² + 3x + p 

⇒ 22 + 3(2) + p = 0

⇒ 4 + 6 + p = 0

⇒ 10 + p = 0

⇒ p = -10

Ques. Verify the division algorithm by dividing the polynomial f(x)=3x²-x³-3x+5 by the polynomial g(x)=x-1-x². (5 Marks)

Ans. Given, f(x)=3x² – x³ – 3x + 5 g(x)=x – 1 – x²

Dividing f(x)=3x² – x³ – 3x+5 by g(x) = x – 1 – x²

Here,

Quotient = q(x) = x – 2

Remainder = r(x) = 3

By using a polynomial division algorithm,

(Quotient *Divisor) + Remainder = Dividend

So,

[q(x) × g(x)] + r(x) = (x – 2)(x – 1 – x²) + 3

= x² – x – x3 – 2x + 2 + 2x² + 3

= 3x² – x³ – 3x + 5

= f(x)

As a result, the division algorithm has been verified.

Ques. Solve the equation: x² +16x + 64 = 0 (3 Marks)

Ans. Factors of x² +16x + 64

The 64 factors are 2, 4, 8, 16, 32, and 64. The total for the middle term is eight.

Now, x² + (8 + 8) x + 8 ? 8

=x² +8x +8x + 8 ? 8

= x(x + 8) + 8 (x + 8)

= (x + 8) (x + 8)

Ques. Solve x³ – 7x2 + 12x = 0 (3 Marks)

Ans. To factor x³−7x²+12x

= x(x²−5x+6)

Now, we factorise x² − 7x + 12

factors of the 12 = 3, 4

Now, x² − (3 + 4) x + 3 ? 4

=x² -3x -4x + 3 ? 4

= x(x – 3) – 4 (x – 3) = (x−3) (x−4)

Factors of x³−7x²+12x are : x (x−3) (x−4)

Ques. Find a quadratic polynomial whose zeros are reciprocals of the zeros of f(x) = ax² + bx + c, a ≠ 0, c ≠ 0. (5 Marks)

Ans. Let the polynomial f(x) = ax² + bx + c’s zeroes be α and β.

So, α + β = -b/a

αβ = c/a

The zeroes of the needed quadratic polynomial, according to the provided, are 1/α and 1/β.

Now, the sum of zeroes = (1/α) + (1/β) 

= (α + β)/αβ

= (-b/a)/ (c/a)

= -b/c

Product of two zeroes = (1/α) (1/β)

= 1/αβ

= 1/(c/a)

= a/c

k[x² – (sum of zeroes)x + (product of zeroes)] is the needed quadratic polynomial.

= k[x² – (-b/c)x + (a/c)]

= k[x² + (b/c) + (a/c)]

Ques. Is there a real zero in the polynomial a⁴ + 4a² + 5? (3 Marks)

Ans. Let a² = x in the aforementioned polynomial.

The polynomial now has the form,

x² + 4x + 5

When compared to ax² + bx + c,

Here, b² – 4ac = 42 – 4(1)(5) = 16 – 20 = -4

So, D = b² – 4ac < 0

The above polynomial has no actual roots or zeroes because the discriminant (D) is negative.

Ques. Find the values of a and b if the polynomial x³ – 3x² + x + 1’s zeroes are a – b, a, a + b. (5 Marks)

Ans. Assume the following polynomial:

p(x) = x³ – 3x² + x + 1

Given,

The p(x) zeroes are a – b, a, and a + b.

Comparing the specified polynomial equation to the generic expression is the next step.

px³ + qx² + rx + s = x³ – 3x² + x + 1

Here, p = 1, q = -3, r = 1 and s = 1

To find the sum of zeroes, use the following formula:

The total number of zeros will be = a – b + a + a + b.

-q/p = 3a

Substitute q and p for their respective values.

-(-3)/1 = 3a

a = 1

As a result, the zeros are 1 – b, 1, and 1 + b.

For the product of zeroes, use the following formula:

Product of zeroes = 1(1 – b)(1 + b)

-s/p = 1 – ????²

=> -1/1 = 1 – ????²

Or, ????² = 1 + 1 =2

So, b = √2

Thus, 1 – √2, 1, 1 + √2 are the x³, 3x² + x + 1 equation’s zeros.

Example 1

Given that f(x) = -2x³ + 4x² – 7x – 6, how many sign changes are there in f(x) and f(-x)? Interpret the results.

Solution

We can immediately inspect f(x) for its sign changes. We have two sign changes: -2x³ and 4x² and as +4x² and -7x.

As for f(-x), let’s go ahead and find the expression for f(-x) first.

f(-x) = 2x³ + 4x² + 7x -6

From this, we can see that f(-x) has only one sign change: between 7x and -6. Using Descartes’ Rule of Sign, we can conclude that:

• When f(x) is equated to 0, the resulting equation may have two or zero positive real zeros.
• Similarly, the resulting equation may have one negative real zero.

Example 2

True or False? Given that the polynomial function, g(x), has a degree of 3, the equation g(x) = 0 will always have three real zeros.

Solution

The equation g(x) = 0 will have at most three possible real zeros. This means that it may or may not have three real zeros exactly. One function to further show that the statement is not true is when g(x) = x³ + x.

Let’s solve the equation and observe the results for x. Since the expression is still factorable, we’ll factor x out and equate x² + 1 to 0.

x³ + x = 0

x(x² + 1) = 0

x = 0

x² + 1 = 0

x² = -1

This will only be true when x = -i or x = i. This clearly shows that it is possible for g(x) to not have three real zeros despite having a degree of 3. Hence, the statement is not true.

Example 3

Find the values of x that satisfies the given equation: 4x⁵ – 4x⁴ + 73x² = -18(x -1)+ 73x³.

The equation is still not in its standard form, so let’s go ahead and isolate all terms on the left-hand side.

4x⁵ – 4x⁴ – 73x³ + 73x²+ 18x – 18 = 0

Using the rational zeros theorem, let’s list down the possible rational zeros for the polynomial equation.

Factors of p	±1, ±2, ±4
Factors of q	±1, ±2, ±3, ±6, ±9, ±18
Possible zeros (p/q)	±1/18, ±1/9, ±1/6, ±1/3, ±2/9, ± 1/2, ±2/3, ±4/9, ±1, ±4/3, ±2, ±4

Let’s check if x = 1 is a root of f(x) using synthetic division.

Since the remainder is 0, (x – 1) is a factor of the expression and x = 1 is a solution. Let’s go ahead and try x = 1/2 and x = -1/2 to see if they are zeros of the equation too.

We can see that x = -1/2 and x = 1/2 are both zeros of the polynomial equation from these two consecutive synthetic divisions.

We now have (x – 1)(x – 1/2)(x + 1/2)(4x² – 72) = 0. Since the remaining expression is a quadratic expression, we can equate it to 0 and solve the polynomial equation’s remaining zeros.

4x² – 72 = 0

4(x² – 18) = 0

x² – 18 = 0

x² = 18

x = ±√18

x = ±2√2

We now have five zeros for the polynomial equation (this is already the maximum number of zeros possible for a polynomial equation with a degree of 5).

Hence, the equation has a solution set of: {-2√2, -1/2, 1/2, 1, 2√2}.

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