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2.9
HCF and LCM of Polynomials/Algebraic expressions:
We use the division method to find HCF and LCM of
monomials or binomials or trinomials or any algebraic expressions
2.9
Problem 1: Find
HCF and LCM of (p+3)^{3},
2p^{3}+54+18p(p+3), (p^{2}+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 2^{nd} term:
2p^{3}+54+18p(p+3)
= 2(p^{3}+27)+18p(p+3)
= 2*(p+3)( p^{2}+9-3p)+18p(p+3), (p^{3}+27 is of the form a^{3}+b^{3}
with a=p and b=3_{}we can apply the formula a^{3}+b^{3} =(a+b) (a^{2} +b^{2} -ab)
=(p+3)*((2*(p^{2}+9-3p))+18p)
= (p+3) *2*( p^{2}+9-3p+9p)
=2(p+3)( p^{2}+9+6p) ( _{}(p^{2}+9+6p) is of the form ( a^{2}+ b^{2}+2ab)
with a=p and b=3, but ( a^{2}+ b^{2}+2ab)= (a+b)^{2}
= 2(p+3)(p+3)^{2}
_{} The factors of 2p^{3}+54+18p(p+3) are 2, (p+3),(p+3),(p+3)
3. we have already seen above
that (p^{2}+6p+9) =(p+3)^{2}
_{} The factors of (p^{2}+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*2^{3}+54+18*2(2+3)),
(2^{2}+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.
2.9
Problem 2: Find
HCF and LCM of 10(x^{2}-y^{2}),
15(x^{2}-2xy+y^{2}) 20(x^{3}- y^{3}),5(-3x +3y)
Solution:
Step 1: Find factors of all the
expressions first.
1. The first term has an expression of the form (a^{2}-b^{2})
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(x^{2}-y^{2})=10(x+y)(x-y)
2. The second term has an expression of the form (a^{2}-2ab+b^{2})
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(x^{2}-2xy+y^{2})= 15(x-y)
(x-y)
3. The third term has an expression of the form (x^{3}-y^{3})
whose factors are (x-y) and (x^{2} +y^{2} +xy)
with a=x and b=y
_{} The factors of third
term are 20, (x-y) and (x^{2} +y^{2}
+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)(x^{2} +y^{2} +xy),
-15(x-y)
2(x+y), 3(x-y), 4(x^{2} +y^{2} +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)(x^{2} +y^{2} +xy), -15(x-y)
2| 2(x+y), 3(x-y), 4(x^{2} +y^{2} +xy), -3
(We continue division as some terms have common factors)
3| (x+y), 3(x-y),
2(x^{2} +y^{2}
+xy), -3
(x+y), (x-y), 2(x^{2} +y^{2} +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(x^{2} +y^{2}
+xy)
= 60*(x-y)(x+y)*(x-y)(x^{2} +y^{2} +xy)
(Note that (x-y)(x^{2} +y^{2} +xy) is
of the form (a-b)( (a^{2} +b^{2} +ab) with a=x and b= y)
= 60*(x^{2}-y^{2})*
(x^{3}-y^{3})
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*(x^{2}- y^{2})* (x^{3}-y^{3})
= 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(x^{2}-y^{2}), 15(x^{2}-2xy+y^{2})
20(x^{3}- y^{3}),5(-3x +3y)
Therefore the terms are 10(3^{2}-2^{2}),
15(3^{2}-2*3*2+2^{2}), 20(3^{3}- 2^{3}),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.
2.9
Problem 3 : For what value of a
and b the polynomials
p(x) = (x^{2}+3x+2) (x^{2}+2x+a)
and
q(x) = (x^{2}+7x+12) (x^{2}+7x+b)
have (x+1)(x+3) as their HCF
Solution:
(x^{2}+3x+2) = (x+1)(x+2)
(x^{2}+7x+12) = (x+4)(x+3)
_{} p(x)
= (x+1)(x+2)(x^{2}+2x+a)
q(x) = (x+4)(x+3) (x^{2}+7x+b)
Since it is given that (x+1)(x+3)
is HCF of p(x),we conclude that
(x+3) is factor of (x^{2}+2x+a)
This implies x=-3 satisfies the equation (x^{2}+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 (x^{2}+7x+b)
This implies x=-1 satisfies the equation (x^{2}+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) = (x^{2}+3x+2) (x^{2}+2x-3)
= (x+1) (x+2) (x+3) (x-1) { _{}(x^{2}+2x-3) = (x+3)(x-1)}
q(x) =(x^{2}+7x+12) (x^{2}+7x+6)
= (x+4) (x+3)
(x+1) (x+6) { _{}(x^{2}+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)
2.9 Summary of learning
No |
Points studied |
1 |
Division
method to find HCF and LCM of algebraic expressions |