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, 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 2^nd 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.

2.9 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.

2.9 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)

2.9 Summary of learning

No	Points studied
1	Division method to find HCF and LCM of algebraic expressions