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CBSE NOTES CLASS 10 MATHEMATICS CHATER 2

Polynomials

Some Important Results and Formulae

• 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 real constants and an ≠ 0.

• The powers of x in a polynomial can only be positive integers.
• The constants a0, a1, a2, . . ., an are respectively the coefficients of x0, x, x2, . . ., xn. A coefficient can be any real number.
• Each of anxn, an–1xn–1, …, a2x2, a1x, a0, with an ≠ 0, is called a term of the polynomial p(x).
• Polynomials are denoted by p(x), q(x), g(x), r(x) etc.

For example in p(x) = 2x2 + 3x - 1

Here the terms are 2x2, 3x and - 1

The coefficient of x2 is 2, of x if 3 and constant term is-1

• Variable names can be x, y, t u, v etc.
• A polynomial of one term is called a monomial.

A polynomial of two terms is called a binomial.

A polynomial of three terms is called a trinomial.

• The highest power of the variable x (that is n) is called degree of the polynomial.

A polynomial with degree 0 has only a non-zero constant term and it is called a constant polynomial, i.e. p(x) = 2.

Only a constant term with value 0 is called zero polynomial. The degree of zero polynomial is not defined.

A polynomial of degree one is called a linear polynomial, i.e.,

p(x) = 3x + 1

A polynomial of degree two is called a quadratic polynomial, i.e.,

p(x) = 2x2 + 3x - 1.

A polynomial of degree three is called a cubic polynomial, i.e.,

p(x) = 4x3+2x2+3x-1

• In standard form of a polynomial, the terms are arranged from highest power of variable to lowest power.

For a quadratic polynomial the standard form is ax2+bx+c, a ≠ 0

Value of a polynomial

By putting x = a, we get the value of polynomial at x = a.

For example, if p(x) = 5x3 - 2x2 + 3x – 4, then,

p(0) = value of polynomial at (x=0)

= 5×03 - 2×02 + 3×0 - 4 = -4

p(1) = value of polynomial at (x=1)

= 5×13 - 2×12 + 3×1 - 4

= 5 – 2 + 3 - 4 = 2

And so on.

Zeroes of a polynomial

• A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0. ‘a’ is also called a root of the equation p(x) = 0.
• Every linear polynomial in one variable has a unique zero
• A non-zero constant polynomial has no zero.
• Every real number is a zero of the zero polynomial.
• For any polynomial, number of zeroes ≤ degree of polynomial
• For finding zeroes of a polynomial, put p(x) = 0 and solve for x.

For example, to find zero of 3x – 2, we put,

3x -2 = 0 x = $\frac{2}{3}$.

Hence $\frac{2}{3}$ is zero of 3x -2.

• Geometrically the zeroes of a polynomial p(x) are precisely the x-coordinates of the points, where the graph of y = p(x) intersects the x -axis.
• A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have at most 3 zeroes.
• Number of zeroes is equal to the number of times the graph of the polynomial intersects the x-axis.
• If α and β are the zeroes of the quadratic polynomial ax2 + bx + c, then

• If α, β, γ are the zeroes of the cubic polynomial ax3 + bx2 + cx + d, then

• If α and β are the zeroes of a quadratic polynomial, the polynomial can be written as (x- α)(x- β)
• Find all the zeroes of 2x4 – 3x3 – 3x2 + 6x – 2, if you know that two of its zeroes are $\sqrt{2}$ and $-\sqrt{2}$ .

Since two zeroes are $\sqrt{2}$ and −$\sqrt{2}$ , (x$\sqrt{2}$)(x+ $\sqrt{2}$ ) = x2–2 is a factor of the given polynomial. Now, we divide the given polynomial by x2 – 2 and the quotient will be a factor of the polynomial.

Factorize it and find the other zeroes.

Dividing one polynomial by another

Example

Divide p(x) = x + 3x21 by g(x) =1 + x

Step 1: Write the dividend x + 3x2 – 1 and the divisor 1 + x in the standard form, i.e., after arranging the terms in the descending order of their degrees. So, the dividend is 3x2 + x –1 and divisor is x + 1.

Step 2: Divide the first term of the dividend by the first term of the divisor, i.e., we divide

3x2 by x, and get 3x. This gives the first term of the quotient.

Step 3: Mmultiply the divisor by the first term of the quotient, and subtract this product from the dividend, i.e., we multiply x + 1 by 3x and subtract the product 3x2 + 3x from the dividend 3x2 + x – 1. This gives us the remainder as –2x – 1. Step 4: Treat the remainder –2x – 1 as the new dividend. The divisor remains the same. Repeat Step 2 to get the next term of the quotient, i.e., we divide the first term – 2x of the (new) dividend by the first term x of the divisor and obtain – 2. Thus, – 2 is the second term in the quotient.

Step 5: Multiply the divisor by the second term of the quotient and subtract the product from the dividend. That is, we multiply x + 1 by – 2 and subtract the product – 2x – 2 from the dividend – 2x – 1. This gives us 1 as the remainder.

• This process continues till the remainder is 0 or the degree of the new dividend is less than the degree of the divisor. At this stage, this new dividend becomes the remainder and the sum of the quotients gives us the whole quotient.

Step 6: Thus, the quotient in full is 3x – 2 and the remainder is 1.

The division algorithm states that given any polynomial p(x) and any non-zero polynomial g(x), there are polynomials q(x) and r(x) such that

dividend = divisor × quotient + remainder

p(x) = g(x) × q(x) + r(x),

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

Factor

If r(x) = 0, then p(x) = g(x) × q(x) ⇒ q(x) and g(x) are called factors of p(x).

If g(x) is a factor of p(x) then, we put g(x) = 0 and solve for zeroes of g(x). If we put the value of this zero in p(x) we should get r(x) = 0

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).

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

• We can verify the result by dividing p(x) by x-a by long division method. We see that remainder = p(a).

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.

Proof: By the Remainder Theorem,

p(x) = (xa) q(x) + p(a).

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

p(x) = (xa) g(x) for same polynomial g(x).

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

• Factorization by mid-term splitting

Find a and b such that

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

is satisfied.

• Factorization by factor theorem
• Factorization by mixed combination

Algebraic identities for factorization

 (x + y)2 = x2 + 2xy + y2 ⇒ x2 + y2 = (x + y)2 - 2xy (x – y)2 = x2 – 2xy + y2 ⇒ x2 + y2 = (x - y)2 + 2xy (x + y)2 = (x - y)2 + 4xy ⇒ (x - y)2 = (x + y)2 - 2xy x2 – y2 = (x + y) (x – y) (x + a) (x + b) = x2 + (a + b)x + ab (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx (x + y)3 = x3 + y3 + 3xy(x + y) = x3 + 3x2y + 3xy2 + y3 (x – y)3 = x3 – y3 – 3xy(x – y) = x3 – 3x2y + 3xy2 – y3 x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) x3 + y3 = (x + y) (x2 + y2 - xy) x3 - y3 = (x - y) (x2 + y2 + xy) If x + y + z = 0 then x3 + y3 + z3 = 3xyz