## Long Division of Polynomials

➢ Write the polynomials in standard form i.e., arrange the terms in decreasing order of their degrees.

➢ Put them in the division bracket.

➢ Divide the first term of the dividend by the first term of the divisor, the result is the first term of the quotient.

➢ Multiply the divisor by the quotient, subtract the product from the dividend.

➢ The remainder found is the new dividend. Divide it’s first term by first term of the divisor to get the second term of the quotient.

➢ Multiply the divisor by the second term of the quotient, subtract the product from the dividend.

➢ 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.

## Division Algorithm for Polynomials

If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x) and g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that:

p(x) = g(x)q(x) + r(x),

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

Link between the remainder and value of the dividend at zero of the divisor when divided by a linear polynomial.

↬ Zero of the divisor, g(x) = 0, x=?.

↬ Value of dividend at zero of the divisor.

↬ Remainder = Value of dividend at zero of the divisor

Let’s find the remainder obtained on dividing

p(x) = x2 − 1 by g(x) = x + 1

Zero of x + 1 is −1

p(−1) = (−1)2 − 1 = 0

∴ Remainder = 0

## 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).

Q. Find the remainder when x3 – ax2 + 6x – a is divided by x – a

Ans.

x – a = 0

⇒ x = a

∴ Zero of x – a is a

p(a) = a3 – aa2 + 6a – a

= a3a3 + 5a

= 5a

∴ Remainder = 5a

## Factorisation of Polynomials

### Factor Theorem

If p(x) is a polynomial of degree n ≥ 1 and a is any real number, then

(i) xa 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.

Q.Find the value of k,

if x – 1 is a factor of p(x) in

p(x) =
x2
+ x + k

Ans

x – 1 is a factor of p(x)

p(1) = 0

⇒ 12 + 1 + k = 0

⇒ 2 + k = 0

⇒ k = −2

### Factorisation of Polynomials by splitting the middle term

Polynomials of second degree can be factorised by splitting the middle term

ax2 + bx + c

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

ax2 + bx + c

= (px + q) (rx + s)

= prx2 + (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.

Factorise: 6x2 + 17x + 5

We have to write 17 as sum of two numbers whose product is 6×5(=30)

{2+15=17 and 2×15=30}

∴ 6x2 + 17x + 5

= 6x2 + (2+15)x + 5

= 6x2 + 2x + 15x + 5

= 2x(3x + 1) + 5(3x + 1)

= (3x + 1)(2x + 5)

### Factorisation of Polynomials using Factor Theorem

Look for all the factors of the constant term in p(x). let them be a, b, c etc

Check if any of these factor reduces the p(x) to zero,

If p(a) = 0, then (x − a) is the factor of p(x),

Find the other factor by dividing p(x) by (x−a),

Second factor can be factorised by splitting the middle term.

Or,

Find the factors of constant term equal in number to the degree of p(x)

(x − a), (x −b), (x−c)…

Write p(x) = k(x − a), (x −b), (x−c)…where k is a constant

Put any value of x (other than a, b, c…) in p(x) to find the value of k.

Factorise:
x3 − 2x2x + 2

Let p(x) = x3 − 2x2x + 2

Factors of 2 are ±1, ±2

p(−1) = (−1)3 − 2⋅(−1)2 − (−1) + 2

= −1 − 2⋅1 + 1 + 2 = 0

∴ (x+1) is a factor of p(x)

p(1) = (1)3 − 2⋅(1)2 − (1) + 2 = 0

∴ (x−1) is a factor of p(x)

p(−2) = (−2)3 − 2⋅(−2)2 − (−2) + 2

= −8 − 8+ 2 + 2 = −12

p(2) = (2)3 − 2⋅(2)2 − (2) + 2

= 8 − 8 −2 + 2 = 0

∴ (x−2) is a factor of p(x)

∵ Degree of p(x) is 3

p(x) = k(x+1)(x−1)(x−2)

p(3) = k(3+1)(3−1)(3−2)

⇒ 33 − 2⋅32 − 3 + 2 = k(3+1)(3−1)(3−2)

⇒ 27 − 18 − 1 = k(4)(2)(1)

⇒ 8 = 8k

k = 1

p(x) = (x+1)(x−1)(x−2)

A cubic polynomial has at most three zeroes.

If we are given only one zero, we can find the other two.

First we divide the polynomial by the known zero, then the quotient so obtained is factorised by splitting it’s middle term to get the other two zeroes.

Maximum Number of Unique factors (or Roots) for a Polynomial = Degree of the Polynomial.

## Algebraic Identities

### Standard Identities   (full)⏩

I. (x + y)2 = x2 + y2 + 2xy
II. (x − y)2 = x2 + y2 − 2xy
III. (x + y)(x − y) = x2 − y2
IV. (x + a)(x + b) = x2 + (a + b)x + ab

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

Proof:

Let, x + y = u

LHS
= (u + z)2
= u2 + z2 + 2uz

= (x + y)2 + z2 + 2(x + y)z

= x2 + y2 + 2xy+ z2 + 2xz + 2yz

= x2 + y2 + z2 + 2xy + 2yz + 2zx

= RHS

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

Proof:

(x + y)3

= (x + y) (x + y)2

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

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

= x3 + 2x2y + xy2 + x2y + 2xy2 + y3

= x3 + 3x2y + 3xy2 + y3

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

= RHS

#### (x + y)3 = x3 − y3 − 3xy (x − y)

Proof:

(xy)3 = (x y) (xy)2

= (x y)(x2 − 2xy + y2)

= x(x2 − 2xy + y2) − y(x2 − 2xy + y2)

= x3 − 2x2y + xy2x2y + 2xy2y3

= x3 − 3x2y + 3xy2y3
= x3y3 − 3xy(xy)

= RHS

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

Proof:

LHS

= x(x2 + y2 + z2 xy yz zx)

+ y(x2 + y2 + z2xyyzzx)

+ z( x2 + y2 + z2xyyzzx)

= x3 + xy2 + xz2x2yxyzzx2

+ x2y + y3 + yz2xy2y2zxyz

+ x2z + y2z + z2xyzyz2 − xz2

= x3+ y3+ z3

+ xy2xy2x2y+ x2y + xz2xz2

zx2 + x2z + 2yz2y2z + y2z

– 3xyz

= x2+ y2+ z2– 3xyz

= RHS

Factorise:

(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz

(ii) 8a3 + b3 + 12a2b + 6ab2

Ans

(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz

= (−2x)2 + (−3y)2 + (4z)2 + 2(−2x)(−3y )+ 2(−3y )(4z) + 2(−2x)(4z)

= (−2x −3y + 4z)2

(ii) 8a3 + b3 + 12a2b + 6ab2

= (2a)3 + b3 + 3(2a)2b + 3(2a)b2

= (2a + b)3