Polynomials

In class VIII you have learnt about algebraic expressions and polynomials in one variable. In this chapter, we strengthen our knowledge of operations like Addition, Subtraction, Multiplication and Division.

In addition to that, we also study how to factorise some algebraic expressions with the help of some Identities.

We shall also study about the Remainder Theorem and Factor Theorem and their use in the factorisation of Polynomials.

Constant: A constant is a number or an alphabet which remains constantly with the variable.

Variable: A variable is an alphabet which changes its value when the equation changes.

Monomial: A polynomial which is having one term is called a monomial. Ex: 2xyz,6bc2,…

Binomial: A polynomial which is having two terms is called a binomial. Ex: 2x+5y, 4xyz-3abc,…

Trinomial: A polynomial which is having three terms is called a trinomial. Ex: 12ab+13ca- 10abc,…

A polynomial of degree one is called Linear polynomial.

A polynomial of degree two is called Quadratic polynomial.

A polynomial of degree three is called Cubic polynomial.

We use distributive laws for multiplication of polynomials.

We can solve the multiplication of polynomials in two methods. We can solve them by directly or by using column method.

Note: The degree of the product = Sum of the degrees of the multiplicand and the multiplier.

We have observed the multiplication of polynomials. In case of division of polynomials, the degree of the quotient is equal to the degree of the dividend minus (-), the degree of the divisor.

The remainder may be zero or its degree is at least one less than that of the divisor.

(x+y)2 = x2 + 2xy + y2

(x+y)2 = x2 + 2xy + y2

(x+a) (x+b) = x2 + (a+b) x + ab

x2 – y2 = (x + y) (x – y)

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

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

(x-y)3 = x3 –y3 – 3xy(x-y)

X3 + y3 + z3 -3xyz = (x+y+z) (x2 + y2+ z2–xy–yz-zx

(a+b) (a2-ab+b2) = a3+b3

(a-b) (a2+ab-b2) = a3-b3

(a+b)3 = a3+3a2b+3ab2+b3

(a-b)3 = a3-3a2b+3ab2-b3

(a+b+c)2 = a2+b2+c2+2ab+2bc+2ac

(a+b+c) (a2+b2+c2-ab-bc-ca) = a3+b3+c3 – 3abc

(ax+b) (cx+b) = ax (cx+d) +b(cx+d)

(x+a) (x+b) (x+c) = (x+a) [x2 +x(b+c)+bc

Let p(x) be any polynomial with 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).

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.

Therefore, p(x) = (x – a) q(x) + r(x)

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

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

which proves the theorem.b

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) is 0, and

(ii) p(a) is 0, if x – a is a factor of p(x).

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.