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)(an1 + an2b + an3 b2 + ... + ab2 + bn1) = 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)(an1 an2b + an3 b2 — ... — abn2 + bn1) = 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+93p)+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+93p))+18p)
= (p+3) *2*(
p2+93p+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(x2y2), 15(x22xy+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 (a2b2) whose factors are (a+b)
and (ab) with a=x and b= y
The factors of first term are 10, (x+y) and
(xy)
10(x2y2)=10(x+y)(xy)
2. The second term has an expression of
the form (a22ab+b2) whose factors are (ab) and (ab) with a=x and b= y
The factors of second term are 15, (xy) and
(xy)
15(x22xy+y2)= 15(xy) (xy)
3. The third term has an expression of
the form (x3y3) whose factors are (xy) and (x2 +y2 +xy) with a=x and b=y
The factors of third term are 20, (xy) and (x2 +y2 +xy)
4. The fourth term can be rewritten as
5*3(xy)
The factors of fourth term are 15, (xy)
5*3(xy) =
5*(3)(xy)=15, (xy)
Step 2: Follow the division method to
find HCF and LCM
The common
factors are 5 and (xy) so let us start dividing terms by these two together
5 (xy) 
10(x+y) (xy), 15(xy) (xy), 20(xy)(x2 +y2 +xy), 15(xy)
2(x+y), 3(xy), 4(x2 +y2 +xy), 3
We stop
further division as there are no more common factors for all the terms
Therefore
HCF = 5(xy)
To find LCM,
we start division with 5(xy)
5(xy) 
10(x+y) (xy), 15(xy) (xy), 20(xy)(x2 +y2 +xy), 15(xy)
2
2(x+y), 3(xy), 4(x2 +y2 +xy), 3 (We continue division as some terms
have common factors)
3
(x+y), 3(xy), 2(x2 +y2 +xy), 3
(x+y), (xy), 2(x2 +y2 +xy) 1
We stop
further division as there are no common factors among any 2 terms
Therefore
LCM =5(xy)* 2*3*(x+y)*(xy)*2(x2 +y2 +xy)
= 60*(xy)(x+y)*(xy)(x2 +y2 +xy) (Note that
(xy)(x2 +y2 +xy) is of the form (ab)( (a2 +b2 +ab) with a=x and b= y)
=
60*(x2y2)* (x3y3)
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(xy) = 5*(32) = 5
and
LCM= 60*(x2
y2)* (x3y3) = 60*(94)*)(278)=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(x2y2),
15(x22xy+y2) 20(x3 y3),5(3x +3y)
Therefore
the terms are 10(3222), 15(322*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. 96+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. 17+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+2x3) = (x+1) (x+2) (x+3) (x1) { (x2+2x3) = (x+3)(x1)}
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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