Saturday, 3 March 2012

HCF AND LCM OF POLYNOMIAL




HCF AND LCM OF POLYNOMIAL

HCF of Polynomials

The HCF of two or more given polynomials is the polynomial of highest degree and largest numerical coefficients which is a factor of all the given polynomials.



Method to find HCF of polynomial



The following method is suggested for finding the HCF of several polynomials



(a)    Write each polynomial as a product of prime factors.



(b)    The HCF is the product obtained by taking each factor to the lowest power to which it occurs in both the polynomials. For example,



The HCF of 23 32 (x - y)3(x + 2y)2; 2233 (x - y)2 (x + 2y)3 and 32(x - y)2 (x + 2y) is 32(x - y)2 (x + 2y).



Note-  Two or more polynomials are relatively prime if their HCF is 1.



LCM of Polynomials

The LCM of two or more given polynomials is the polynomial of lowest degree and smallest numerical coefficients for which each of the given polynomials will be its factor.



Method to find LCM of polynomial



The following procedure is suggested for determining the LCM of several polynomials.



(a)   Write each polynomial as a product of prime factors.



(b)   The LCM is the product obtained by taking each factor to the highest power to which it occurs. For example, The LCM of 2332 (x — y)3(x + 2y)2; 2233(x — y)2 (x + 2y)3; 32(x — y)2(x + 2y) is 2333 (X — y)3 (x + 2y)3.



Example1. Find the HCF and LCM of (I) 9X4y2 and 12X3y3



(ii) 6x — 6y and 4x2 — 4y2.

Sol. (i) 9x4y2 = 32 x x4 x y2 and

12X3y3 = 22x 3 x X3 x y3.

HCF = 3x3y2 and

LCM = 32 x 22 X x4 X y3 =36X4y3

So, HCF = 3x3y2, LCM = 36x4y3



(ii) 6x — 6y = 2 x 3 x (x — y) and

4x2 — 4y2 = 22(x2— y2) = 22 x (x + y) (x — y) HCF = 2(x — y) and

LCM = 22x 3 x (x — y)(x + y) = 12(x — y) (x + y). So, HCF = 2(x — y), LCM = 12(x + y) (x — y).



Special Products

The following are some of the products which occur frequently in mathematics and the student should become familiar with them as soon as possible.



Product of a monomial and a binomial a (c + d) = ac + ad.



Product of the sum and the difference of two terms (a + b)(a — b) = a2 — b2.



Square of a binomial                                          

(a + b)2 = a2 + 2ab + b2.

(a — b)2 = a2 — 2ab + b2



Product of two binomials



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



(ax + b)(cx + d) = acx2 + (ad + bc)x + bd. ,(a+b)(c+d)=ac+bc+ad+bd.



Cube of a binomial

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

(a — b)3 = a3 — 3a2b + 3ab2 — b3.



Square of a trinomial



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



Products that give answers of the form a n ± bn.



It may be verified by multiplication that (a — b)(a2 + ab + b2) = a3 — b3



(a — b)(a3 + a2b + ab2 + b3) = a4 — b4



(a — b)(a4 + a3b + a2b2 + ab3 + b4) = a5 — b5



(a — b) (a 5 + a 4 b + a 3  b 2 + a 2 b 3 + a b 4 + b 5 ) = a 6 — b 6  and so on These may be summarized as (a — b)(an-1 + an-2b + an-3 b2 + ... + ab2 + bn-1) = an — bn where n is any positive integer (1, 2, 3, 4, ...). Similarly, it may be verified that (a + b)(a2 - ab + b2) = a3 + b3



(a + b)(a4 — a3b + a2b2 — ab3 + b4) = a5 + b5



(a + b)(a6 — a5b + a4b2 — a3 b3 + a2 b4 — abs + b6) = a7 + b7 and so on.



These may be summarized as



(a + b)(an-1- an-2b + an-3 b2 — ... — abn-2 + bn-1) = an + bn  Where n is any positive odd integer (1, 3, 5, 7, ...).



Example2 . Simplify (3xy + 1)(2x2 — 3y).



Sol. (3xy)(2x2) + (3xy)(-3y) + (1)(2x2) + (1)(-3y) = 6x3y — 9xy2 + 2x2 — 3y.



Example3 .Simplify (x + y + z + 1)2.



Sol. [(x + y) + (z + 1)]2



= (x + y)2 + 2(x + y)(z + 1) + (z + 1)2



= x2 + 2xy + y2 + 2x2 + 2x + 2yz + 2y + z2 + 2z + 1.



Division Method



We use the division method to find HCF and LCM of monomials or binomials or trinomials or any algebraic expressions

 Problem 1: Find HCF and LCM of  (p+3)3, 2p3+54+18p(p+3), (p2+6p+9)

Solution:

Step 1: Find factors of all the expressions first.

1.The factors of (p+3)3 are (p+3),(p+3) and (p+3)

2. Let us simplify the 2nd term:

2p3+54+18p(p+3)

= 2(p3+27)+18p(p+3)

= 2*(p+3)( p2+9-3p)+18p(p+3),  (p3+27 is of the form a3+b3 with a=p and b=3we can apply the formula a3+b3 =(a+b) (a2 +b2 -ab)

=(p+3)*((2*(p2+9-3p))+18p)

= (p+3) *2*( p2+9-3p+9p)

=2(p+3)( p2+9+6p) ( (p2+9+6p) is of the form ( a2+ b2+2ab) with a=p and b=3, but ( a2+ b2+2ab)= (a+b)2

= 2(p+3)(p+3)2

 The factors of 2p3+54+18p(p+3) are 2, (p+3),(p+3),(p+3)

3. we have already seen above that (p2+6p+9) =(p+3)2

 The factors of (p2+6p+9) are p+3, p+3

Follow the division method to find HCF and LCM

The given terms can be rewritten as ( p+3)(p+3)(p+3),  2(p+3)(p+3)(p+3), (p+3)(p+3)

The first common factor is p+3 and let us start dividing terms by this term

(p+3) | ( p+3)(p+3)(p+3),  2(p+3)(p+3)(p+3), (p+3)(p+3)

(p+3) | (p+3)(p+3),            2(p+3)(p+3),          (p+3)

           (p+3),                     2(p+3)                            1

We stop further division as there are no more common factors for all the terms

Therefore HCF = (p+3)(p+3)= (p+3)2  and

(p+3) | ( p+3)(p+3)(p+3),  2(p+3)(p+3)(p+3), (p+3)(p+3)

(p+3) | (p+3)(p+3),            2(p+3)(p+3),          (p+3)

(p+3) | (p+3),                     2(p+3)                            1

              1,                          2,                                    1            

We stop further division as there are no more common factors for any 2 terms

Therefore LCM = (p+3)(p+3)(p+3)*1*2*1 = 2(p+3)3

Verification:

Let us cross check the solution by substituting p=2 in the above problem.

From the solution the HCF is (p+3)2 = (2+3)2 =25 and LCM= 2(p+3)3= 2(2+3)3= 2*125=250

Let us find the HCF and LCM of the given terms after converting them to numbers by substituting p=2.

Therefore the terms are  (2+3)3, (2*23+54+18*2(2+3)), (22+6*2+9)

= {125, 250,25}

By close observation we notice that HCF=25 and LCM=250

Since both the methods give same HCF and LCM our solution is correct.



 Problem 2: Find HCF and LCM of  10(x2-y2), 15(x2-2xy+y2) 20(x3- y3),5(-3x +3y)

Solution:

Step 1: Find factors of all the expressions first.

1. The first term has an expression of the form  (a2-b2) whose factors are (a+b) and (a-b) with a=x and b= y

 The factors of first term are 10, (x+y) and (x-y)

 10(x2-y2)=10(x+y)(x-y)

2. The second term has an expression of the form (a2-2ab+b2) whose factors are (a-b) and (a-b) with a=x and b= y

 The factors of second term are 15, (x-y) and (x-y)

 15(x2-2xy+y2)= 15(x-y) (x-y)

3. The third term has an expression of the form (x3-y3) whose factors are (x-y) and (x2 +y2 +xy) with a=x and b=y

 The factors of third term are  20, (x-y) and (x2 +y2 +xy)

4. The fourth term can be rewritten as 5*-3(x-y)

 The factors of fourth term are -15, (x-y)

5*-3(x-y) = 5*(-3)(x-y)=-15, (x-y) 

Step 2: Follow the division method to find HCF and LCM

The common factors are 5 and (x-y) so let us start dividing terms by these two together

5 (x-y) | 10(x+y) (x-y), 15(x-y) (x-y), 20(x-y)(x2 +y2 +xy), -15(x-y)

             2(x+y),             3(x-y),             4(x2 +y2 +xy),        -3

We stop further division as there are no more common factors for all the terms

Therefore HCF = 5(x-y)

To find LCM, we start division with 5(x-y)

5(x-y) | 10(x+y) (x-y), 15(x-y) (x-y), 20(x-y)(x2 +y2 +xy), -15(x-y)

        2|  2(x+y),             3(x-y),             4(x2 +y2 +xy),        -3 (We continue division as some terms have common factors)

        3|   (x+y),             3(x-y),              2(x2 +y2 +xy),        -3

   (x+y),                (x-y),              2(x2 +y2 +xy)         -1

We stop further division as there are no common factors among any 2 terms

Therefore LCM =5(x-y)* 2*3*(x+y)*(x-y)*2(x2 +y2 +xy)

=  60*(x-y)(x+y)*(x-y)(x2 +y2 +xy) (Note that (x-y)(x2 +y2 +xy) is of the form (a-b)( (a2 +b2 +ab)   with a=x and b= y)

= 60*(x2-y2)* (x3-y3)

Verification:

Since it is very difficult to cross verify easily, we will cross check the solution for at least  for one value of x and y by substituting x=3 and y=2 in the above problem

From the solution the HCF is 5(x-y) = 5*(3-2) = 5  and

LCM= 60*(x2- y2)* (x3-y3) = 60*(9-4)*)(27-8)=60*5*19=5700

Let us find the HCF and LCM of the given terms after converting them to numbers by substituting x=3 and y=2 in

10(x2-y2), 15(x2-2xy+y2) 20(x3- y3),5(-3x +3y)

Therefore the terms are 10(32-22), 15(32-2*3*2+22), 20(33- 23),5(-3*3 +3*2)

= {50, 15, 380, -15}

By observation we note that HCF=5

Let us use the division method to find LCM

5 | 50,15,380,-15

 2 | 10,3,76,-3

  3 | 5,3,38,-3

     |  5,1,38,-1

LCM = 5*2*3*5*38=5700

Since both the methods give same HCF and LCM our solution is not incorrect.



 Problem 3 : For what value of  a and b the polynomials

p(x) = (x2+3x+2) (x2+2x+a) and

q(x) = (x2+7x+12) (x2+7x+b)

have (x+1)(x+3) as their HCF

Solution:

(x2+3x+2) = (x+1)(x+2)

(x2+7x+12) = (x+4)(x+3)

 p(x) = (x+1)(x+2)(x2+2x+a)

q(x) = (x+4)(x+3) (x2+7x+b)

Since it is given that (x+1)(x+3) is HCF of  p(x),we conclude that

(x+3) is factor of (x2+2x+a)

This implies x=-3 satisfies the equation (x2+2x+a) =0

 (-3)2+2(-3)+a =0

I.e. 9-6+a =0

a =-3

Since it is given that (x+1)(x+3) is HCF of  q(x),we conclude that

(x+1) is factor of (x2+7x+b)

This implies x=-1 satisfies the equation (x2+7x+b) =0

 (-1)2+7(-1)+b =0

I.e. 1-7+b =0

b =6

Verfication :  By substituting value for a an b in p(x) and q(x) we get

p(x) = (x2+3x+2) (x2+2x-3) = (x+1) (x+2) (x+3) (x-1) { (x2+2x-3) = (x+3)(x-1)}

q(x) =(x2+7x+12) (x2+7x+6) = (x+4) (x+3) (x+1) (x+6) { (x2+7x+6)= (x+1)(x+6)}

By looking at factors of p(x) and q(x) we conclude that HCF of p(x) and q(x) is (x+1) (x+3)



Relationship between H.C.F. and L.C.M. of Two Polynomials



If f(x) and g(x) are two polynomials, then

           f(x). g(x) = {HCF of f(x) and g(x)} x {L.C.M of f(x) and g(x)}.               

Example: Find the L.C.M of the following pairs of polynomials with the help of their H.C.F.:

3 (x2 - 7x + 12) and 24(x2 - 9x + 20)

Solution:

Given, p(x) = 3 (x2 - 7x + 12)

= 3(x - 3)(x - 4)

and, q(x)  = 24(x2 - 9x + 20)

= 23 x 3 x (x - 4)(x - 5)

Clearly, H.C.F. of p(x) and q(x) is 3 (x - 4).

Therefore,

L.C.M. of p(x) and q(x)

= [p(x). q(x)] / H.C.F. of p(x) and q(x)

= { [3 (x - 3) (x - 4)1 x [23 x 3 x (x - 4) (x- 5)]}/ 3 (x - 4)

= (x - 3) x 23 x 3 x (x - 4) (x - 5)

= 24(x - 3)(x - 4)(x - 5)



 FREE HAND EXERCISE QUESTIONS



1. Find the HCF of the following:

(a) 2x4 - 2y4 and 3x3 + 6x2y - 3xy2 - 6y3

(b) 12(x3 + x2 + x + 1) and 18(x4 - 1)

(c) x3 + 2x2 - 3x and 2x3 + 5x2 - 3x

(d) 2(x4 - y4) and 3(x3 + 2x2y - xy2 - 2y3)

(e) 18(x3 - x2 + x - 1) and 12(x4 - 1)

(f) 45(x4 - x3 - x2) and 75(8x5 + x2)

(g) 36(3x4 + 5x3 - 2x2) and 54(27x4 - x)

(h) 42(2x3 - 5x2 - 3x) and 60(8x4 + x)

(i) 4(x4 - 1) and 6(x3 - x2 - x + 1)



2. Find the LCM of the following polynomials:

(a) 35(x4 - 27x) and 40(2x3 - 5x2 - 3x)

(b) 20(2x3 + 3x2 - 2x) and 45(x4 + 8x)

(c) 15(4x3 - 4x2 + x) and 35(2x2 - 7x + 3)

(d) 25(x2 + 7x + 12) and 15x(x2 - 16)

(e) x(8x3 + 27) and 2x2 (2x2 + 9x + 9)



3. Find the GCD and LCM of the polynomials P(x) and Q(x), where

P(x) = (x3 - 27) (x2 - 3x + 2) and

Q(x) =( x2 + 3x + 9) (x2 - 5x + 6)



4. The LCM and GCD of two polynomials, P(x) and Q(x) are 56(x4 + x) and 4(x2 - x + 1) respectively. If P(x) = 28(x3 + 1), find Q(x) .

5. For what value of k, the g.c.d. of x2 + x - (2k + 2) and 2x2 + kx - 12 is x + 4?

6. Find the value of K for which the g.c.d. of x2 - 2x - 24 and x2 - kx - 6 is x - 6.

7. (x2 - x - 6) is the GCD of the expression (x + 2) (2x2 + ax + 3) and (x - 3) (3x2 + bx + 8). Find the value of a and b.

8. (x + 1) ( x - 4) is the g.c.d. of the polynomials ( x - 4) (2x2 + x - a) and ( x + 1) (2x2 + bx - 12) find a and b .

9. (x - 3) is the g.c.d. of (x3 -2x2 + px + 6) and ( x2 - 5x + q) . Find (6p + 5q) .

10. Find the value of a and b so that the polynomials P(x) and Q(x) have (x - 1) (x + 4) as their HCF:

P(x) = (x2 - 3x + 2) (x2 + 7x + a)

Q(x) = (x2 + 5x + 4) (x2 - 5x + b)

11. Find the value of a and b so that the polynomial x3 + ax2 + bx + 15 is divisible by x2 + 2x - 15.

12. Find the value of p and q so that the polynomial f(x) = px3 + 2x2 - 19x + 9 is divisible by x2 + x - 6.

13. Determine the value of k such that x + 3 is a factor of the polynomial

f(x) = kx3 + x2 - 22x - 21

14. If x - 2 is a factor of x2 + ax - 6 = 0 and x2 - 9x + b = 0, find the value of a and b.Answers



ANSWERS

1. (a) x2 - y2                          (b) 6(x + 1) (x2 + 1)

    (c) x(x + 3)                         (d) x2 - y2

    (e) 6(x - 1) (x2 + 1)           (f) 15x2(2x + 1)

    (g) 18x(3x - 1)                   (h) 6x(2x + 1)

    (i) 2(x2 - 1)

2. (a) 280x(x - 3) (2x + 1) (x2 + 3x + 9)

    (b) 180x(x + 2) (2x - 1) (x2 - 2x + 4)

    (c) 105x((x - 3) (2x - 1)2

    (d) 75x(x2 - 16) (x + 3)

    (e) 2x2(8x2 + 27) (x + 3)

3. G. C. D. = (x3 - 27) (x - 2)

     LCM = (x - 1) (x - 2) (x - 3) (x2 + 3x + 9)

4. 8(x3 - x2 + x)

5. K = 5

6. K = 5

7. a = -7, b = 10

8. a = 1, b= -5

9. 0

10. a = 12, b = 4

11. a = 1, b = -17

12. p = 3, q = 6

13. K = 2              
14. a = 1, b = 14

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