Polynomial

A polynomial is an algebraic expression that includes constants, variables, and exponents. It is the expression in which the variables have only positive integral powers.

Example

1. 4x³ + 3x² + x +3 is a polynomial in variable x.

2. 4x² + 3x^-1 – 4 is not a polynomial as it has negative power.

3. 3x^3/2 + 2x – 3 is not a polynomial, since it power is not an integer.

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• Polynomials are denoted by p (x), q (x), etc.
• In the above polynomials, 2x², 3y, and 2 are the terms of the polynomial.
• 2 and 3 are the coefficients of x² and y, respectively.
• x and y are the variables.
• 2 is the constant term, which has no variable.

Polynomials in One Variable

If there is only one variable in the expression, then this is called the polynomial in one variable.

Example

• x³ + x – 4 is polynomial in variable x and is denoted by p(x).
• r² + 2 is polynomial in variable r and is denoted by p(r).

Types of polynomials on the basis of the number of terms

Types of polynomials on the basis of the number of degrees

The highest value of the power of the variable in the polynomial is the degree of the polynomial.

Zeros of a Polynomial

If p(x) is a polynomial, then the number ‘a’ will be the zero of the polynomial with p(a) = 0. We can find the zero of the polynomial by equating it to zero.

Example: 1

The given polynomial is p(x) = x – 4

To find the zero of the polynomial, we will equate it to zero.

x – 4 = 0

x = 4

p(4) = x – 4 = 4 – 4 = 0

This shows that if we place 4 in place of x, we get the value of the polynomial as zero. So 4 is the zero of this polynomial. And also, we are getting the value 4 by equating the polynomial with 0.

So 4 is the zero of the polynomial or the root of the polynomial.

The root of the polynomial is basically the x-intercept of the polynomial.

If the polynomial has one root, it will intersect the x-axis at one point only, and if it has two roots, it will intersect at two points, and so on.

Example: 2

Find p (1) for the polynomial p (t) = t² – t + 1

p (1) = (1)² – 1 + 1

= 1 – 1 + 1

= 1

Remainder Theorem

We know the property of division which follows in the basic division, i.e.

Dividend = (Divisor × Quotient) + Remainder

This follows the division of polynomials.

If p(x) and g(x) are two polynomials in which the degrees of p(x) ≥ degree of g(x) and g(x) ≠ 0 are given, then we can get the q(x) and r(x) so that:

P(x) = g(x) q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x).

It says that p(x) divided by g(x), gives q(x) as a quotient and r(x) as a remainder.

Let’s understand it with an example

Division of a Polynomial with a Monomial

We can see that ‘x’ is common in the above polynomial, so we can write it as

Hence, 3x² + x + 1 and x the factors of 3x³ + x² + x.

Steps of the Division of a Polynomial with a Non –Zero Polynomial

Divide x² – 3x -10 by 2 + x

Step 1:  Write the dividend and divisor in descending order, i.e., in the standard form. x² – 3x -10 and x + 2

Divide the first term of the dividend with the first term of the divisor.

x²/x = x this will be the first term of the quotient.

Step 2: Now multiply the divisor by this term of the quotient and subtract it from the dividend.

Step 3: Now the remainder is our new dividend, so we will repeat the process again by dividing the dividend by the divisor.

Step 4: – (5x/x) = – 5 

Step 5: 

The remainder is zero.

Hence x² – 3x – 10 = (x + 2)(x – 5) + 0

Dividend = (Divisor × Quotient) + Remainder

The Remainder Theorem says that if p(x) is any polynomial of degree greater than or equal to one and let ‘t’ be any real number and p (x) is divided by the linear polynomial x – t, then the remainder is p(t).

As we know,

P(x) = g(x) q(x) + r(x)

If p(x) is divided by (x-t) then

If x = t

P (t) = (t – t).  q (t) + r = 0

To find the remainder or to check the multiple of the polynomial, we can use the remainder theorem.

Example:

What is the remainder if a⁴ + a³ – 2a² + a + 1 is divided by a – 1.

Solution:

P(x) = a⁴ + a³ – 2a² + a + 1

To find the zero of (a – 1), we need to equate it to zero.

a -1 = 0

a = 1

p (1) = (1)⁴ + (1)³ – 2(1)² + (1) + 1

= 1 + 1 – 2 + 1 + 1

= 2

So by using the remainder theorem, we can easily find the remainder after the division of the polynomial.

Factor Theorem

The factor theorem says that if p(y) is a polynomial with degree n≥1 and t is a real number, then

1. (y – t) is a factor of p(y), if p(t) = 0, and
2. P (t) = 0 if (y – t) is a factor of p (y).

Example: 1

Check whether g(x) = x – 3 is the factor of p(x) = x³ – 4x² + x + 6 using the factor theorem.

Solution:

According to the factor theorem, if x – 3 is the factor of p(x), then p(3) = 0, as the root of x – 3 is 3.

P (3) = (3)³ – 4(3)² + (3) + 6

= 27 – 36 + 3 + 6 = 0

Hence, g (x) is the factor of p (x).

Example: 2

Find the value of k, if x – 1 is a factor of p(x) = kx² – √2x + 1

Solution:

As x -1 is the factor, p(1) = 0

Factorization of Polynomials

Factorization can be done by three methods

1. By taking out the common factor

If we have to factorise x² –x then we can do it by taking x common.

x(x – 1) so that x and x-1 are the factors of x² – x.

2. By grouping

ab + bc + ax + cx = (ab + bc) + (ax + cx)

= b(a + c) + x(a + c)

= (a + c)(b + x)

3. By splitting the middle term

x² + bx + c = x² + (p + q) + pq

= (x + p)(x + q)

This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.

Example: 1

Factorize 6x² + 17x + 5 by splitting the middle term.

Solution:

If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.

Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.

6x² + 17x + 5 =6 x² + (2 + 15) x + 5

= 6 x² + 2x + 15x + 5

= 2 x (3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Example: 2

Factorize 8x³ + 27y³ + 36x²y + 54xy²

Solution:

The given expression can be written as

= (2x)³ + (3y)³ + 3(4x²) (3y) + 3(2x) (9y²)

= (2x)³ + (3y)³ + 3(2x)²(3y) + 3(2x)(3y)²

= (2x + 3y)³ (Using Identity VI)

= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.

Example: 3

Factorize 4x² + y² + z² – 4xy – 2yz + 4xz.

Solution:

4x² + y² + z² – 4xy – 2yz + 4xz = (2x)² + (–y)² + (z)² + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)

= [2x + (- y) + z]² (Using Identity V)

= (2x – y + z)² = (2x – y + z) (2x – y + z)