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CBSE NOTES CLASS 10 MATHEMATICS CHATER 2

Polynomials

Some Important Results and Formulae

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

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

3x -2 = 0 x = 23.

Hence 23 is zero of 3x -2.

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.

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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.

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

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).

Algebraic identities for factorization

(x + y)2 = x2 + 2xy + y2x2 + y2 = (x + y)2 - 2xy

(x – y)2 = x22xy + y2x2 + 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 – y33xy(x – y)

= x3 – 3x2y + 3xy2y3

x3 + y3 + z33xyz = (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