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Revision Notes on Polynomials
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. 3x3/2 + 2x – 3 is not a polynomial, since it power is not an integer.
.
- 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
- (y – t) is a factor of p(y), if p(t) = 0, and
- 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)
| Algebraic Identities |
| 1. (x + y)2 = x2 + 2xy + y2 |
| 2. (x – y)2 = x2 – 2xy + y2 |
| 3. (x + y) (x – y) = x2 – y2 |
| 4. (x + a) (x + b) = x2 + (a + b)x + ab |
| 5. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx |
| 6. (x + y)3 = x3 + y3 + 3xy(x + y) = x3+ y3 + 3x2y + 3xy2 |
| 7. (x – y)3 = x3– y3 – 3xy(x – y) = x3 – y3 – 3x2y + 3xy2 |
| 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) x3 + y3 + z3 = 3xyz if x + y + z = 0 |
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)
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