Table of Contents

Last modified on October 30th, 2024

Polynomial comes from ‘poly-’ (meaning ‘many’) and ‘-nomial’ (meaning ‘terms’). A polynomial is a mathematical expression consisting of two main parts, variables and constants, joined together by mathematical operators like addition, subtraction, and multiplication.

The exponent of each of the variables in a polynomial is always a whole number (non-negative integer).

Here are a few examples of polynomials:

- 7xy
- 2x
^{2}+ 4x + 7 - ${-x^{3}-\dfrac{7}{2}x+1}$
- 11

However, the expressions ${\sqrt{x}}$, ${\dfrac{1}{x^{3}}}$, and 5x^{-1}y^{2}z^{-2} are **NOT** polynomials.

The polynomial function is denoted by P(x), where x represents the variable. It is written in the descending power of the variable.

In standard form, it is written as:

**P(x) = a**_{n}**x**^{n}** + a**_{n – 1}**x**^{n – 1}** + … + a**_{1}**x + a**_{0}

Here,

- a
_{n}, a_{n-1}, …, a_{1}are coefficients of x^{n}, x^{n-1}, …, x - a
_{n}(≠ 0) is the leading coefficient - a
_{0}is constant and ∀ a_{n}∈ ℝ

The polynomial equation in general form is written as:

**a**_{n}**x**^{n}** + a**_{n – 1}**x**^{n – 1}** + … + a**_{1}**x + a**_{0}** = 0**^{ }

In an expression, terms are each part of a polynomial that is separated by a ‘+’ or ‘-‘ sign.

For example, the polynomial 7xy has 1 term (i.e., 7xy), while 2x^{2} + 4x + 7 has 3 terms (i.e., 2x^{2}, 4x, and 7).

Based on the number of terms present in the expression, a polynomial is of 3 types:

A polynomial consisting of only one term is called a monomial expression.

**Examples**: 3x, 4xyz, and ${2x^{2}}$.

A polynomial with two terms is known as a binomial expression.

**Examples**: 2x + y, 4z + 7, and 10x^{2} + 5x^{3}.

A polynomial with three terms is called a trinomial expression.

**Examples**: 7x^{2} + 3x + 5, -6x + 3 + 3y, and 5x + 7y + 3z.

The degree of a polynomial is the highest exponent of the variable in a polynomial. It is used to determine the maximum number of solutions of a polynomial equation.

For example, the degree of the polynomial 2x^{2} + 4x + 7 is 2

A few polynomials based on their degrees are classified here.

Polynomial | Degree | Standard Form | Example |
---|---|---|---|

Constant | 0 | P(x) = a_{0} | P(x) = 7 |

Linear polynomial | 1 | P(x) = a_{1}x + a_{0} | P(x) = x + 7 |

Quadratic polynomial | 2 | P(x) = a_{2}x^{2} + a_{1}x + a_{0} | P(x) = x^{2} + 7x + 4 |

Cubic polynomial | 3 | P(x) = a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0} | P(x) = 3x^{3} + x^{2} + 7x + 4 |

Quartic polynomial | 4 | P(x) = a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0} | P(x) = 5x^{4} + 3x^{3} + x^{2} + 7x + 4 |

Quintic polynomial | 5 | P(x) = a_{5}x^{5} + a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0} | P(x) = x^{5} – 5x^{4} + 3x^{3} + x^{2} + 7x + 4 |

Here, a_{0}, a_{1}, a_{2}, a_{3}, and a_{4} are constants. However, the leading coefficient in the polynomial will be non-zero (coefficient of the variable of the highest power).

Certain properties are followed while solving polynomial expressions.

If a polynomial P(x) is divided by a polynomial G(x), the result is a quotient Q(x) with remainder R(x). According to the division algorithm, it can be expressed as:

P(x) = G(x) • Q(x) + R(x)

Here,

R(x) = 0 or the degree of R(x) < the degree of G(x)

It states that a polynomial P(x) is divisible by a binomial (x – a) if and only if P(a) = 0.

It states that if a polynomial P(x) is divided by (x – a) with remainder *r*, then P(a) = r.

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It states that dividing a polynomial P(x) by Q(x) results in R(x) with zero remainders if and only if Q(x) is a factor of P(x).

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It states that if P(x) is a polynomial, and P(x) ≠ P(y) for (x < y), then P(x) takes every value from P(x) to P(y) in the closed interval [x, y].

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The addition, subtraction, and multiplication of polynomials P(x) and Q(x) result in a polynomial where,

- Degree(P ± Q) ≤ Degree(P or Q)
- Degree(P × Q) = Degree(P) + Degree(Q)

If a polynomial P(x) is divisible by a polynomial Q(x), then every zero of Q(x) is also a zero of P(x).

If a polynomial P(x) is divisible by two co-prime polynomials, Q(x) and R(x), then it is divisible by (Q • R).

It states that if P(x) = a_{0} + a_{1}x + a_{2}x^{2} + … + a_{n}x^{n} is a polynomial such that degree(P) = n (≥ 0), then P(x) has at most *n* distinct roots.

The number of positive real zeros in a polynomial function P(x) is less than or equal to the number of sign changes in the coefficients by an even number. So, if there are* k* sign changes, the number of positive roots will be k or (k – a), where *a* is some even number.

It states that every non-constant, single-variable polynomial with complex coefficients contains at least one complex zero.

If a polynomial P(x) with real coefficients contains one complex zero, then there exists another complex zero (called complex conjugate)

Mathematically, if x = a – bi is a zero of P(x), then x = a + bi will also be a zero of P(x).

When simplifying polynomials, first, we combine the like terms and then perform the basic mathematical operations following the rule of PEMDAS.

To add and subtract polynomials, we add or subtract all the like terms (the terms with the same variable and power). The addition or subtraction of two or more polynomials always results in a polynomial of the same degree or a lower degree.

Let us simplify (x^{2} + 5x + 4) + (x^{3} – 2x^{2} – 3x + 1) – (3x^{3} – 2x^{2} + 3x – 6)

Using the rules of basic mathematical operations,

x^{2} + 5x + 4 + x^{3} – 2x^{2} – 3x + 1 – 3x^{3} + 2x^{2} – 9x + 6

Grouping all the like terms,

(x^{3} – 3x^{3}) + (x^{2} – 2x^{2} + 2x^{2}) + (5x – 3x – 9x) + (4 + 1 + 6)

Now, adding and subtracting all like terms,

= -2x^{3} + x^{2} – 7x + 11

Thus, (x^{2} + 5x + 4) + (x^{3} – 2x^{2} – 3x + 1) – (3x^{3} – 2x^{2} + 3x – 6) = -2x^{3} + x^{2} – 7x + 11

When multiplying polynomials, every term of the first expression is multiplied by every term of the second. The product of two or more polynomials always results in a polynomial of a higher degree (unless one of them is a constant polynomial).

Whereas the division of two polynomials may or may not result in a polynomial. However, to divide polynomials, we use the long division or synthetic division method.

Let us multiply (x – 2)(x^{2} + 2x – 3)

Here, the first polynomial is (x – 2) and the second one is (x^{2} + 2x – 3)

Now, multiplying them,

= x(x^{2} + 2x – 3) – 2(x^{2} + 2x – 3)

= x^{3} + 2x^{2} – 3x – 2x^{2} – 4x + 6

Combining the like terms,

= x^{3} + (2x^{2} – 2x^{2}) + (-3x – 4x) + 6

Now, adding and subtracting all like terms,

= x^{3} – 7x + 6

Thus, (x – 2)(x^{2} + 2x – 3) = x^{3} – 7x + 6

Now, let us divide (x^{2} + 2x – 3) by (x – 2)

Since ${\dfrac{x^{2}+2x-3}{x-2}}$ can not be simplified, the final answer (including the remainder) will be in the fraction form.

Thus, the division of these polynomials does not result in a Polynomial.

However, using long division, the quotient of the polynomials is

Factoring polynomials is expressing a polynomial as a product of its simpler polynomial factors. It helps in solving polynomial equations, simplifying expressions, and understanding polynomial behavior.

A polynomial can be factored using different methods:

- Method of Common Factors
- Grouping Method
- Factoring by Splitting Terms
- Factoring Using Algebraic Identities

Polynomials with one variable are easy to graph as they have smooth and continuous lines, as shown.

In the above graph, we observe that the polynomial function f(x) = x^{4} – 4x^{2} + 4x – 1 has a smooth curve.

Following is a list of general polynomial identities used to solve polynomial expressions:

(a + b)^{2} | a^{2} + 2ab + b^{2} |

(a – b)^{2} | a^{2} – 2ab + b^{2} |

(a^{2} – b^{2}) | (a + b)(a – b) |

(x + a)(x + b) | x^{2} + x(a + b) + ab |

(a + b)^{3} | a^{3} + 3a^{2}b + 3ab^{2} + b^{3} |

(a – b)^{3} | a^{3} – 3a^{2}b + 3ab^{2} – b^{3} |

a^{3} + b^{3} | (a + b)(a^{2} – ab + b^{2}) |

a^{3} – b^{3} | (a – b)(a^{2} + ab + b^{2}) |

a^{3} + b^{3} + c^{3} – 3abc | (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca) |

**Add the polynomials:****a) (3x ^{2} + 5x – 4) + (2x^{2} – 3x + 6)**

Solution:

a) Here, (3x^{2} + 5x – 4) + (2x^{2} – 3x + 6)

= 3x^{2} + 5x – 4 + 2x^{2} – 3x + 6

= (3x^{2} + 2x^{2}) + (5x – 3x) + (-4 + 6)

= 5x^{2} + 2x + 2

Thus, the sum is 5x^{2} + 2x + 2

b) Here, (x^{3} – 2x^{2} + 3x) + (4x^{3} + 5x – 7)

= x^{3} – 2x^{2} + 3x + 4x^{3} + 5x – 7

= (x^{3} + 4x^{3}) – 2x^{2} + (3x + 5x) – 7

= 5x^{3} – 2x^{2} + 8x – 7

Thus, the sum is 5x^{3} – 2x^{2} + 8x – 7

**Subtract: (x ^{4} – 3x^{2} + x) – (2x^{4} + x^{2} – 4)**

Solution:

Here, (x^{4} – 3x^{2} + x) – (2x^{4} + x^{2} – 4)

= x^{4} – 3x^{2} + x – 2x^{4} – x^{2} + 4

= (x^{4} – 2x^{4}) – (3x^{2} + x^{2}) + x + 4

= -x^{4} – 4x^{2} + x + 4

Thus, the difference is -x^{4} – 4x^{2} + x + 4

**Multiply the polynomials: (x + 3)(x – 4)**

Solution:

Here, (x + 3)(x – 4)

= x(x – 4) + 3(x – 4)

= x^{2} – 4x + 3x – 12

= x^{2} + (-4x + 3x) – 12

= x^{2} – x – 12

Thus, the product is x^{2} – x – 12

**Divide: ${\dfrac{x^{2}+4x-12}{x-2}}$**

Solution:

Here, ${\dfrac{x^{2}+4x-12}{x-2}}$

= ${\dfrac{x^{2}+6x-2x-12}{x-2}}$

= ${\dfrac{\left( x+6\right) \left( x-2\right) }{x-2}}$

= ${x+6}$

Thus, the quotient is x + 6

**Problem: **Solving for **LINEAR POLYNOMIALS**

**Solve 4x + 48**

Solution:

Here, the linear polynomial is 4x + 48

Setting the equation as 0,

4x + 48 = 0

Now, solving for x,

⇒ 4x = -48

⇒ x = ${\dfrac{-48}{4}}$

⇒ x = -12

Thus, x = -12

**Problem: **Solving **QUADRATIC POLYNOMIALS **

**Solve 2x ^{2} – 3x – 5**

Solution:

Here, the quadratic polynomial is 2x^{2} – 3x – 5

Setting the equation as 0,

2x^{2} – 3x – 5 = 0 …..(i)

As we know, the standard form of a quadratic polynomial is ax^{2} + bx + c, where a (a ≠ 0), b, and c are constants.

Comparing the equation (i) with the standard form, we get

a = 2, b = -3, and c = -5

Now, using the quadratic formula,

x = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

⇒ x = ${\dfrac{-\left( -3\right) \pm \sqrt{\left( -3\right) ^{2}-4\times 2\times \left( -5\right) }}{2\times 2}}$

⇒ x = ${\dfrac{3\pm \sqrt{9+40}}{4}}$

⇒ x = ${\dfrac{3\pm \sqrt{49}}{4}}$

⇒ x = ${\dfrac{3\pm 7}{4}}$

⇒ x = ${\dfrac{5}{2}}$ and x = ${-1}$

Thus, x = ${\dfrac{5}{2}}$ and x = ${-1}$

**Problem – **Solving **CUBIC POLYNOMIALS**

**Solve x ^{3} – 6x^{2} + 11x – 6**

Solution:

Here, x^{3} – 6x^{2} + 11x – 6

Setting the equation as 0,

x^{3} – 6x^{2} + 11x – 6 = 0

Now, solving for x by the factoring method,

⇒ x^{3} – x^{2} – 5x^{2} + 5x + 6x – 6 = 0

⇒ x^{2}(x – 1) – 5x(x – 1) + 6(x – 1) = 0

⇒ (x – 1)(x^{2} – 5x + 6) = 0

Again, by the factoring method,

⇒ (x – 1)(x^{2} – 3x – 2x + 6) = 0

⇒ (x – 1)[x(x – 3) – 2(x – 3)] = 0

⇒ (x – 1)(x – 2)(x – 3) = 0

⇒ x = 1, x = 2, and x = 3Thus, x = 1, 2, and 3 are the solutions (roots) of the cubic polynomial x^{3} – 6x^{2} + 11x – 6

Last modified on October 30th, 2024