# Chapter 2.Polynomials

1.Important Identities of Polynomials

1. (x + y) ² = x² + 2xy + y²

2 .(x y) ² = x² – 2xy + y²

3. x² – y² = (x + y) (x y)

2.Use of Polynomial Identities

1) factorisation.

2.find out  Square of Big Numbers

3.Remainder Theorem

4. Factor Theorem

Example:- ax+b+c=0   is monomial , degree of x is 1

ax²+bx+c=0 is Binomial (or Quadratic Polynomial) , degree of x is 2

ax3+ax²+bx+c=0 is cubic Polynomial, degree of x is 3.

3.Remainder Theorem : Let p(x) be any polynomial of degree greater than or

equal to one and let a be any real number. If p(x) is divided by the linear

polynomial x – a, then the remainder is p(a).

Proof : Let p(x) be any polynomial with degree greater than or equal to 1. Suppose

that when p(x) is divided by x a, the quotient is q(x) and the remainder is r(x), i.e.,

p(x) = (x a) q(x) + r(x)

Since the degree of x a is 1 and the degree of r(x) is less than the degree of x a,

the degree of r(x) = 0. This means that r(x) is a constant, say r.

So, for every value of x, r(x) = r.

Therefore, p(x) = (x a) q(x) + r

In particular, if x = a, this equation gives us

p(a) = (a a) q(a) + r

= r,

which proves the theorem.

4. Factor Theorem : If p(x) is a polynomial of degree n > 1 and a is any real number,

then (i) x a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x a is a factor of p(x).

Proof: By the Remainder Theorem, p(x)=(x a) q(x) + p(a).

(i) If p(a) = 0, then p(x) = (x a) q(x), which shows that x a is a factor of p(x).

(ii) Since x – a is a factor of p(x), p(x) = (x a) g(x) for same polynomial g(x).

In this case, p(a) = (a a) g(a) = 0.

5. Factorisation of the polynomial ax2 + bx + c by splitting the middle term is as

follows:

Let its factors be (px + q) and (rx + s). Then

ax2 + bx + c = (px + q) (rx + s) = pr x2 + (ps + qr) x + qs

Comparing the coefficients of x2, we get a = pr.

Similarly, comparing the coefficients of x, we get b = ps + qr.

And, on comparing the constant terms, we get c = qs.

This shows us that b is the sum of two numbers ps and qr, whose product is

(ps)(qr) = (pr)(qs) = ac.

Therefore, to factorise ax2 + bx + c, we have to write b as the sum of two

numbers whose product is ac.

6. Algebraic Identities

Identity I: (a + b)2 = a2 + 2ab + b2

Identity II: (a – b)2 = a2 – 2ab + b2

Identity III: a2 – b2= (a + b)(a – b)

Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab

Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)

Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)

Identity VIII: a3 + b3 + c– 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Summary of this chapter

1. A polynomial p(x) in one variable x is an algebraic expression in x of the form

p(x) = anxn + an–1xn – 1 + . . . + a2x2 + a1x + a0,

where a0, a1, a2, . . ., an are constants and an ¹ 0.

a0, a1, a2, . . ., an are respectively the coefficients of x0, x, x2, . . ., xn, and n is called the degree

of the polynomial. Each of anxn, an–1 xn–1, …, a0, with an ¹ 0, is called a term of the polynomial

p(x).

2. A polynomial of one term is called a monomial.

3. A polynomial of two terms is called a binomial.

4. A polynomial of three terms is called a trinomial.

5. A polynomial of degree one is called a linear polynomial.

6. A polynomial of degree two is called a quadratic polynomial.

7. A polynomial of degree three is called a cubic polynomial.

8. A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0. In this case, a is also called a

root of the equation p(x) = 0.

9. Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial

has no zero, and every real number is a zero of the zero polynomial.

10. Remainder Theorem : If p(x) is any polynomial of degree greater than or equal to 1 and p(x)

is divided by the linear polynomial x a, then the remainder is p(a).

11. Factor Theorem : x a is a factor of the polynomial p(x), if p(a) = 0. Also, if x a is a factor

of p(x), then p(a) = 0.

12. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

13. (x + y)3 = x3 + y3 + 3xy(x + y)

14. (x y)3 = x3 – y3 – 3xy(x y)

15. x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy yz zx)