Polynomial, GCD, HCF, Algebra Mathematics, Math Assignment Help

mathematics - Polynomial, GCD, HCF, Algebra Mathematics, Math

Polynomial: An expression of the form a0xn + a1xn-1 + a2xn-2 + ... + an-1 x+an, where a0, a1, a2, ....., an are real numbers, n is a non-negative integer and a0 ¹ 0, is called a polynomial of degree n.   

Example:

(i)      7x - 2 is a polynomial of degree one.

(ii)     3x2 - 2x +  is a polynomial of degree two.

(iii)    x3 - 3x2 +  x - 9 is a polyno­mial of degree three.

In the above definition a0 , a1, a2, ......... an are known as the coefficients of the polynomial. If a0, a1, a2,...... an are integers. then the polynomial p(x) = a0xn + a1xn-1 + ... + an-1 x + an is known as the polynomial with integral coefficients.

Example:

(i)      2x3 - 3x2 + 4x - 9 is a polynomial of degree 3 with integral coefficients.

(ii)     21/2x2-3x+5 is a polynomial of degree 2. The coefficients are not all integers. So, it is a polynomial with non-integral coefficients.

 

Divisor: A polynomial d(x) is called a divisor of a polynomial p(x), if d(x) is a factor of p(x).

In other words, a polynomial d(x) is a divisor of a polynomial p(x) if there exists a polynomial q(x) such that p(x) = d(x) q(x). 

Example:

Let p(x) = X2 - 3x + 2.

Then, p(x) = (x - 1) (x - 2). Clearly, (x - 1) and (x - 2) are factors of p(x). Therefore (x - 1) and (x - 2) are divisors of p(x).

Greatest common divisor:

Consider two polynomials p(x) = (x - 1)2 (x + 2) (x + 3)2 and q(x) = (x - 1)3 (x + 2)2

(x + 5). Clearly, (x - 1), - (x - 1),

(x + 2), - (x + 2), (x - 1) (x + 2), -

(x -1) (x + 2): (x - 1)2 (x + 2), -

(x - 1)2 (x + 2) are common divisors of p(x) and q(x). We observe that among all the common divisors, (x-1)2 (x+2) and - (x-1)2 (x+2) are two common divisors of highest degree. The divisor - (x - 1)2 (x + 2) has its highest degree term coefficient as negative whereas the divisor (x - 1)2 (x + 2) has its highest degree term coefficient positive. Such a divisor is called the highest common divisor or greatest common divisor as defined below.

Greatest Common Divisor (GCD) or Highest Common Factor (HCF):

The greatest common divisor (g.c.d.) of two polynomials p(x) and q(x) is that common divisor which has highest degree among all common divisors and in which the coefficient of highest degree term is positive.

The g.c.d. of two polynomials can be obtained by using the following algorithm. 

ALGORITHM FOR FINDING G.C.D. BY FACTORISATION METHOD

Step I.        Obtain two polynomials. Let the polyn0mials be p(x) and q(x).

Step II.      Factories the polynomials p(x) and q(x).

Step III.     Express p(x) and q(x) in the form of product of powers of different factors.

Step IV.      Determine the maximum power of each factor which is common to the given polynomials.

Step V.       Obtain the product of all factors obtained in step IV.

The product obtained in step V is the g.c.d. of the given polynomials.

The following formulae wm be helpful in factorizing polynomials.

(i)      (x + y)2 = x2 + 2xy + y2

(ii)     (x - y)2 = x2 - 2xy + y2

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

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

(v)     (x - y)3 = x3 - 3xy (x - y) -y3

(vi)    x2 - y2 = (x - y) (x + y)

(vii)   x3 + y3 = (x + y) (x2 - xy + y2)

(viii)   x3 - y3 = (x - y) (x2 + xy + y2)

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

(x)     x4 - y4 = (x2- y2) (x2 + y2) = (x -y) (x + y) (x2 + y2)

(xi)    x8 - y8 = (x4 - y4) (x4 + y4) = (x - y) (x + y) (x2 + y2) (x4 + y4)

 

Example:

Find the g.c.d. (x + 2)2 (x - 3)2 (x + 1)2 and (x + 1)3 (x + 2)3 (x -3).

 

Solution:

Let p(x) = (x + 2)2 (x - 3)2 (x + 1)2 and q(x) = (x + 1)3 (x +2)3(x- 3).

The highest degree common divisors of two polynomials are

(x + 2)2, (x - 3) and (x + 1)2.

Therefore, g.c.d. of p(x) and q(x) is (x + 2)2 (x - 3) (x + 1)2.

If p(x) arid q(x) are two polynomials then we define

I.c.m. of pix) and q(x)

p(x). q(x) / g.c.d.p(x) and q(x)

It follows from this definition that the I.c.m. of two polynomials pix) and q(x) is the polynomial of lowest degree which has p(x) and q(x) as divisors and whose coefficient of highest degree term has same sign as the sign of the coefficient of highest degree term in the product p(ix). q(x).

Example:

Find the I.c.m.

3x2 + 5x - 2 and 3x2 - 7x + 2

Solution:

Let p(x) = 3x2 + 5x - 2 and q(x) = 3x2 - 7x + 2. Then,

p(x) = 3x2 + 5x - 2 = 3x2 + 6x - x - 2 = 3x (x + 2) - (x + 2) = (3x - 1) (x + 2) and q(x) = 3x2 - 7x + 2 = 3x2 - 6x - x + 2 = 3x (x - 2) - (x - 2) = (ax - 1) (x - 2).

So, g.c.d. of p(x) and q(x) is (3x - 1).

Hence, I.c.m. of p(x) and q(x)

p(x). q(x) / g.c.d of p(x) and q(x)

= (3x-1)(x+2).(3x-1)(x-2) / 3x-1

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

Aliter: The I.c.m. of two polynomials can also be determined by taking each factor of pix) and q(x) and if a factor is common we take that factor which has the highest degree in p(x), q(x).

REMAINDER AND FACTOR THEOREMS

Remainder theorem: Let fix) be a polynomial of degree greater than or equal to one and a be any real number. If f(x) is divisible by (x - a), then the remainder is equal to f(a). 

Example:

Determine the remainder when the polynomial f(x) = x4 - 3x2 + 2x + 1 is divided by (x - 1).

 

Solution: By remainder theorem, the required remainder is equal to f(1).

Now, f(x) = x, - 3x2 + 2x + 1

Þ f(1) = 1 - 3 + 2 + 1 = 1.

Hence, the required remainder is equal to 1.

Factor theorem: Let fix) be a polynomial of degree greater than or equal to one and a be a real number Such that f(a) = 0, then (x - a) is a factor of f(x). Conversely, if (x - a) is a factor of f(x), then f(a) = O.

 

Remark:

(i)      (x + a) is a factor of a polynomial f(x) iff f(-a) = 0

(ii)     (ax - b) is a factor of a polynomial f(x) iff fib/a) = 0

(iii)    (ax + b) is a factor of a polynomial f(x) iff f(-b/a) = 0

(iv)    (x - a) (x - b) is a factor of a polynomial f(x) iff f(a) = 0 and f(b) = 0.

Example: (x - 3) is a factor of the polynomial x3 - 3x2 + 4x - 12.

Solution: Let pix) = x3 - 3x2 + 4x - 12 be the given polynomial. By factor theorem, (x - a) is a factor of a polynomial p(x) if p(a) = 0. Therefore, in order to prove that x - 3 is a factor of p(x), it is sufficient to show that p(3) = 0.

Now, p(x) = x3 - 3x2 + 4x - 12

p(3) = 33 - 3 × 32 + 4 × 3 - 12

= 27 - 27 + 12 - 12 = 0

Hence, (x - 3) is a factor of pix) = x3 - 3x2 + 4x - 12.

FACTORISATION TO OPEN PRODUCT INTO FACTORS

Type I:       Taking out the common factors: When each term of an expression has a common factor, we divide each term of the expression by this factor and take it out as a multiple.

Type II:      Grouping: The terms of the expression may be grouped so as to have a common factor. It can now be factorised as in type I.

Type III:    a2 ± 2ab + b2 = (a ± b)2.

Type IV:     Difference of two squares: a2 - b2 = (a - b) (a + b).

Type V:       Sum and difference of two cubes: a3 + b3 = (a + b)

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

Type VI:     x2 + (a + b) x + ab = (x + a) (x + b)

Type VII:   ax2 + bx + c.

Factorisation of polynomals using remainder theorem: The following algorithm is very helpful in the factorization of polynomials Over integers.

 

Algorithm:

Step I. Obtain the given polynomial f(x) (say).

Step II. Obtain the constant term in f(x) and find its all Possible factors.

For example, in the polynomial f(x) = x3 - 6x2 + 11x - 6 the constant term is - 6 and its factors are ± 1, ± 2, ± 3, ± 6.

Step III. Take one of the factors, say a1 and replace x by it in the given polynomial. If the polynomial reduces to zero, then x - a1 is a factor of the polynomial. Otherwise, take another factor and continue this procedure till you get as many as factors as the degree of the polynomial.

Let x-a1, x - a2, x - a3,.........a3, ...... be the factors obtained by applying the above procedure.

Step IV. Put f(x) equal to k(x - a1) (x - a2) (x - a3) ... i.e., write f(x) = k(x - a1) (x - a2) (x - a3) ..... ,where k is a constant.

Step V. Substitute any value of x other than a1, a2, a3, ........ in the equation obtained in step IV and get the value of k.

Step VI. Substitute the value of k in f(x) = k(x - a1) (x - a2) (x - a3) Following examples will illustrate the above procedure

Example Using factor theorem, factorize the polynomial

x3 - 6x2 + 11x - 6.

Solution: let f(x) = x3 - 6x2 + 11x - 6.

The constant term in f(x) is equal to -6 are ± 1, ± 2, ± 3, ± 6.

Putting x = 1 in f(x), we have

f(1) = 13 - 6 × 12 + 11 × 1 - 6 = 1 - 6 + 11 - 6 = 0

\       (x - 1) is a factor of f(x). Similarly, x - 2 and x - 3 are factors of f(x).

Since f(x) is a polynomial of degree 3. So, it cannot have more than three linear factors.

Let f(x) = k(x - 1) (x - 2) (x - 3).

Then,

x3 - 6x2 + 11x - 6 = k(x - 1) (x - 2) (x - 3)

Putting x = 0 on both sides, we get

- 6 = k(0 - 1) (0 - 2) (0 - 3) Þ 6 = - 6k Þ k = 1

Putting k = 1 in f(x) = k(x - 1) (x - 2) (x - 3), we get f(x) = (x - 1)

(x - 2) (x - 3)

Hence, x3 - 6x2 + 11x - 6 = (x - 1) (x - 2) (x - 3).

 

Note: Sometimes it is not convenient to get all factors of a given polynomial by using the above procedure. In such cases if the given polynomial is a cubic, then we find one factor by using factor theorem and then divide the polynomial by it to get a quadratic polynomial as the quotient. The quadratic polynomial is then factorized by splitting the middle term. If the given polynomial is a fourth degree polynomial, then we find two linear factors by using factor theorem and then divide the polynomial by their product to get a quadratic polynomial as quotient. Now, we factorize the quadratic polynomial to get all possible factors of the given polynomial

Symmetric expressions: An algebraic expression involving two or more letters is said to be symmetric with respect to a pair of letters if it remains unchanged when these letters are interchanged.

Example: The expression a2 + 5ab + b2 is symmetric with respect to a and b since when a and b are interchanged, the resulting expression b2 + 5ba + a2 is same as the original one.

Homogeneous expression: An expression in two or more variables is said to be a homogeneous expression of degree n, if the sum of the powers of all the variables in each term is equal to n. 

Example: Clearly, the expression a + b + c is a homogeneous expression of degree one.

Factorization of cyclic expressions using factor theorem: We shall factorize cyclic expressions by using the same procedure as in factorization of polynomials.

Principle of indeterminate coefficients:

If two polynomials of the same degree in one variable are identically equal, then the coefficients of the like powers of the variable in the two polynomials are separately equal i.e.,

a0 + a1x + a2x2 + ......... + anxn = b0 + b1 + b2x2 + ....... + bnxn

Þ 0 a0 = b0, a1 = b1, a2, = b2, ......., a0 = bn.

Example: Using factor theorem we prove that

a2 (b - c) + b2 (c - a) + c2 (a - b) = - (a - b) (b - c) (c - a)

Solution: We have:

LHS = a2 (b - c) + b2 (c - a) + c2 (a - b) = a2b - a2c + b2c - b2a + c2a - c2b.

Clearly, it is a homogeneous expression of degree 3.

Putting a = b in the given expression it becomes

= b2. b - b2 c + b2 c - b2 . b + c2 b - c2 b = 0

So, (a - b) is a factor of the given expression.

Similarly, by symmetry each of (b - c) and (c - a) is a factor of the expression.

Now, since the given expression is a symmetrical and 'homogeneous expression of degree 3. So, it cannot have more than 3 linear factors. Therefore, let

a2 (b - c) + a2 (c - a) + c2 (a - b) = k (a - b) (b - c) (c - a) where k is a constant.

Now, putting any convenient values of a, b, c in the above identity such that neither side of (i) becomes zero, we may obtain the value of k.

putting a = 0, b = 1 and c = - 1 in

(i), we get

0 + 1 (- 1 - 0) + (- 1)2 (0 - 1) = k

(0 -1) (1 + 1) (- 1 - 0)

Þ 0 - 2 = 2k Þ 0 k = - 1

Substituting k = - 1 in (i), we get

a2 (b _ c) + b2 (c - a) + c2 (a - b) = - (a - b) (b - c) (c - a). 

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