H.C.F And L.C.M Of Algebraic Expressions
We had some coloured ribbons of different lengths. We had a green ribbon of length 32 metre and a yekkow ribbon of length 104 metre and a yellow ribbon of length 56 metre.
We have decided that we cut off the gretedt length of same side from the ribbons in such away that no portion of the ribbon is left cut.
Question 1. Let us calculate what will be the G.C.D of 32,104,56
Solution.
Given
32,104,56
G.C.D of 32, 104, 56
32=2×2×2×2×2
104=2×2×2×13
56= 2×2×2×7
∴ G.C.D of 32,104,and56 2×2×2×8
∴ The greatest length of ribbon we have to cut is metre, i.e., we factorise the three numbers into prime factors and consider the highest common factor, G.C.D of the three numbers.
Question 2. If we had a green ribbon of length 2a2b metre, a yellow ribbon of length 4ab2metre and a blue ribbon of length 6a2b2, let’s calculate the greatest length of same size we have cut from the ribbons in such a way that no portion of the ribbon is left.
How will I get the G.C.D of 2a2b,4ab2 and 6a2b2?
Solution.
Given
If we had a green ribbon of length 2a2b metre, a yellow ribbon of length 4ab2metre and a blue ribbon of length 6a2b2
G.C.D of 2a2b,4ab2 and 6a2b2
At first we resolve 2a2b,4ab2 and 6a2b2 into prime factors.
2a2b= 2×a×a×b
4ab2= 2×2×a×b×b
6a2b2= 3×2×a×a×b×b
Alternately, G.C.D of 2,4 and 6=2 Lowest power of a is a, the lowest power of is bis b in the expressions 2a2b,4ab2 and 6a2b2.
∴ The required G.C.D = 2a1b1=2ab
∴ G.C.D of 2a2,4ab2 and 6a2b2 is 2ab
Read and Learn More WBBSE Solutions For Class 8 Maths
Question 3. Let’s find the G.C.D of 6x2yz3,10x3y3z3 and 8x2yz4.
Solution:
Given
6x2yz³,10x3y3z3 and 8x²yz4.
At first we factorize 6x2yz3, 10x3y3z3and 8x²yz4into prime factors.
6x2yza = 3×2×x×x×y×z×z×z
10x3y3z3 = 5×2×x×x×x×y×y×y×z×z×z
8x2yz4 = 2×2×2×x×x×y×z×z×z×z
∴ G.C.D of 6x2yz3, 10x3y3z3and 8x2yz4 is = 2x2z3
Alternately, G.C.D of 6,10 and 8 is = 2
In 6x2yz3, 10x3y3z3and 8x2yz4 the lowest power of x is 2, the lowest power of y is 1 and the lowest power of z is 3;
∴ G.C.D of 6x2yz3, 10x3y3z3and 8x2yz4 is = 2x2z3
Question 4. Let’s find the G.C.D of (x3 + 2×2) and (x3+11×2+18x).
Solution.
Given,
First expression, x3+2x2 =x2(x + 2)
Second expression, x3 + 11 x2 + 18x = x (x2 + 11 x + 18).
= x (x2 + 9x + 2x + 18) = x (x+9) (x+2)
∴ Required G.C.D is = x (x+2)
Algebraic Expressions Exercise
Question 1. Let’s find the G.C.D of the algebraic expressions x (x2-9), x2-x-12
Solution:
Given algebraic expressions x (x2-9), x2-x-12
First expression = x (x2-9)
= x {(x)2-(3)2}
= x (x+3) (x-3)
Second expression = x2 – x – 12
= x2-(4-3) x – 12
= x2 – 4x + 3x – 12
= x (x-4) +3 (x-4)
= (x-4) (x+3)
Required G.C.D is = (x+3)
Question 2. However; if we stich 4xy2z decimeter red ribbon, 6yz2x decimeter blue ribbon, and 10zx2y decimeter green ribbon on the curtain step by step then let’s calculate the least amount of curtain is required to stich the ribbons completely.
Solution:
Given
If we stich 4xy2z decimeter red ribbon, 6yz2x decimeter blue ribbon, and 10zx2y decimeter green ribbon on the curtain step by step
Let’s try to find the L.C.M of 4xy2z, 6yz2z and 10zx2y.
Let’s factorize 4xy2z, 6yz2x and 10zx2y.
4xy2Z = 2×2× x ×y ×y ×z
6yz2x = 3×2× x × y × z ×z
10zx2y=5×2×x×x×y×z
∴ The common factor of 4xy2z, 6yz2x and 10zx2y is 2xyz, and other factors are 2,3,5,x,y,z.
∴ L.C.M. of 4xy2z, 6yz2x and 10zx2y is 2xyz×2×3×5×x×y×z = 60x2y2z2
In 4xy2z, 6yz2x and 10zx2y the highest power of x is 2, the highest power of y is 2 and the highest power of z is 2.
∴ L.C.M. of 4xy2z, 6yz2x and 10zx2y = 60x2y2z2
Algebraic Expressions Exercise
Question 1. Let’s find the L.C.M of 2(x-4) and (x2-3x+2). And Let’s find the G.C.D L.C.M of and (x3-8), (x2+3x-10) and (x3+2x2+8x).
Solution:
Given expressions (x3-8), (x2+3x-10) and (x3+2x2+8x)
First expression, x3-8
= x3-(2)3
= (x-2) (x2+2x+4)
Second expression, x2+3x-10
= x2+5x-2x-10
= x(x+5)-2 (x+5)
= (x+5) (x-2)
= x(x2+2x-8)
Third expression, x3+2x2-8x
= x (x2+2x-8)
= x {x2+4x-2x-8}
= x {x(x+4)-2 (x+4)}
= x(x+4) x (x-2)
Required G.C.D. = x – 2
Required L.C.M. = x(x-2)(x2+2x+4)(x-5)(x+4)
Algebraic Expressions Exercise
Question 1. Let’s find the G.C.D and the L.C.M of (y3-8), (y3-4y2+4y) and (y2+y-6).
Solution:
Given (y3-8), (y3-4y2+4y) and (y2+y-6).
First expression, y3-8
= (y)3-(2)3
= (y-2)(y2+2y+4)
Second expression, y3-4y2+4y
= y (y2-4y+4)
= y{(y)2-2.y.2 + (2)2}
= y (y-2)2 = y (y-2) (y-2)
Third expression, y2+y-6
= y2+(3-2) y-6
= y2+3y-2y-6
= y(y+3)-2 (y+3)
= (y+3) (y-2)
Required G.C.D. = (y-2)
Required L.C.M = y(y+3)(y-2)2(y2+2y+4)
Algebraic Expressions Exercise
Question 1. Let’s find the G.C.D of the following algebraic expressions :
1. 4a2b2, 20ab2
Solution:
Given 4a2b2, 20ab2
4a2b2 = 2×2×a×a×b×b
20ab2 =2×2×5×a×b×b
Required G.C.D. = 2×2×a×b×b
= 4ab2
G.C.D = 4ab2
2. 5p2q2,10p2q2, 25p4q3
Solution:
Given 5p2q2, 10p2q2, 25p4q3
5p2q2= 5×p×p×q×q
10p2q2 = 2×5×p×p×q×q
25p4q3= 5×5×p×p×p×p×q×q×q
Required G.C.D. = 5×p×p×q×q = 5p2q2
G.C.D. = 5p2q2
3. 7y3z6, 21 y2,14z2
Solution:
Given 7y3z6, 21 y2,14z2
7y3z6= 7×y×y×y×z×z×z×z×z×z
21y2 =3×7×y×y
14z2 = 2×7 × z × z
Required G.C.D. = 7
4. 3a2b2c, 12a2b4c2, 9a5b4
Solution:
Given 3a2b2c, 12a2b4c2, 9a5b4
3a2b2c= 3×a×a×b×b×c .
12a2b4c2= 2×2×3×a×a×b×b×b×b×c×c
9a5b4= 3×3×a×a×a×a×a×b×b×b×b
Required G.C.D. = 3a2b2
Question 2. Let’s find the L.C.M. of the following expressions :
1. 2x2y3,10x3y
Solution:
Given 2x2y3, 10x3y
2x2y3= 2×X×x×y×y×y
10x3y= 2×5×X×X×X×y
2. 7p2q3, 35p3q, 42pq4
Solution:
Given 7p2q3, 35p3q, 42pq4
7p2q3 = 7 x p x p x q x q x q
35p3q = 5×7×p×p×p×q
42pq4= 2×3×7×p×q×q×q×q
Required L.C.M. = 2×3×5×7×p3q3
= 210 p3q4
3. 5a5b, 15ab2c, 25a2b2c2
Solution:
Given 5a5b, 15ab2c, 25a2b2c2
5a5b = 5 × a × a × a × a × a × b
15ab2c = 3×5×a×b×b×c
25a2b2 c2= 5×5×a×a×b×b×c×c
Required L.C.M. = 75a5b2c2
4. 11a2bc2, 33a2b2c, 55a2bc2
Solution:
Given 11 a2bc2, 33a2b2c, 55a2bc2
11 a2bc2 = 11×a×a×b×c×c
33a2b2c = 3×11×a×a×b×b×c
55a2bc2= 5×11×a×a×b×c×c
Required L.C.M. =165a2b2c2
Question 3. Let’s find G.C.D of the following algebraic expressions :
1. 5x (x+y), x3-xy2
Solution:
Given 5x (x+y), x3-xy2
First expression, = 5x (x+y)
Second expression, x3-xy2
= x (x2-y2) .
= x (x+y) (x-y)
Required G.C.D. = x (x+y)
2. x3-3x2y, x2-9y2
Solution:
Given x3-3x2y, x2-9y2
First expression, x3-3x2y
= x2(x-3y)
Second expression, x2-9y2
=(x)2-(3y)2
= (x+3y) (x-3y)
Required G.C.D. = (x-3y)
3. 2ax(a-x)2, 4a2x (a-x)3
Solution:
Given 2ax(a-x)2, 4a2x (a-x)3
First expression, 2ax(a-x)2
=2ax (a-x) (a-x)
Second expression, 4a2x (a-x)3
= 2×2×a×a×x (a-x) (a-x) (a-x)
Required G.C.D. = 2ax (a-x)2
4. x2-1, x2-2x+1, x3+x2-2x
Solution:
Given x2-1, x2-2x+1, x3+x2-2x
First expression, x2-1
= (x)2-(1)2
= (x+1) (x-1)
Second expression, x2-2x+1
= (x)2-2.x.1 + (1)2
= (x-1)2
= (x-1) (x-1)
Third expression, x3+x2-2x
= x(x2+x-2)
= x (x2+ 2x – x-2)
= x {x(x +2) -1 (x+2)}
= x (x+2) (x-1)
Required G.C.D. = (x-1)
5. a2-1, a3-1, a2+a-2
Solution:
Given a2-1, a3-1, a2+a-2
First expression, a2-1
= (a)2-(1)
= (a+1) (a-1)
Second expression, a3-1
= (a)3– (1 )3
= (a-1) (a2+a-2)
Third expression, a2+a-2
= a2+(2-1) a-2
= a2+2a-a-2
= a (a+2) -1 (a+2)
= (a+2) (a-1)
Required G.C.D. = (a-1)
6. x2+3x+2, x2+4x+3, x2+5x+6
Solution:
Given x2+3x+2, x2+4x+3, x2+5x+6
First expression, x2+3x+2
= x2+(2+1)x+2
= x2+2x+x+2
= x (x+2)+1 (x+2)
= (x+2) (x+1)
Second expression, x2+4x+3
= x2 + (3+1)x + 3
= x2 + 3x + x + 3
= x (x+3) +1 (x+3)
= (x+3) (x+1)
Third expression, x2+5x+6
= x2 + (3+2) x+6
= x2+3x+2x+6
= x(x+3) + 2 (x+3)
= (x+3) (x+2)
Required G.C.D. = 1
7. x2+xy, xz + yz, x2+ 2xy+y2
Solution:
Given x2+xy, xz + yz, x2+ 2xy+y2
First expression, x2+xy
= x (x+y)
Second expression, xz + yz
= z (x+y)
Third expression, x2+ 2xy+y2
= (x+y)2
= (x+y) (x+y)
Required G.C.D. = (x+y)
8. 8 (x2-4), 12 (x3+8), 36 (x2-3x-10)
Solution:
Given 8 (x2-4), 12 (x3+8), 36 (x2-3x-10)
First expression, 8 (x2-4)
= 8{(x)2-(2)2}
= 2×2×2 (x+2) (x-2)
Second expression, 12 (x3+8)
= 12{(x)3+(2)3}.
= 2x2x3 (x+2) (x2-2x+4)
Third expression, 36 (x2-3x-10)
= 36 {x2-(5-2) x-10}
= 36 (x2-5x+2x-10)
= 36 {x (x-5) +2 (x-5)}
= 2×2×3×3 (x-5) (x+2)
Required G.C.D. = 4(x+2)
9. a2-b2-c2+2bc, b2-c2-a2+2ac, c2-a2-b2+2ab
Solution:
Given a2-b2-c2+2bc, b2-c2-a2+2ac, c2-a2-b2+2ab
First expression, a2-b2-c2+2bc
= a2-(b2-2bc + c2)
= (a)2– (b-c)2
= (a+b-c) (a-b+c)
Second expression, b2-c2-a2+2ac
= b2– (a2-2ac+c2)
= (b)2 – (a-c)2
= (b+a-c) (b-a+c)
= (a+b-c) (b-a+c)
Third expression, c2-a2-b2+2ab
= c2-(a2-2ab+b2)
= (c)2 – (a-b)2 = (c+a-b) (c-a+b)
= (a-b+c) (c-a+b)
Required G.C.D. = 1
10. x3-16x, 2x3+9×2+4x, 2x3+x2-28x
Solution:
Given x3-16x, 2x3+9x2+4x, 2x3+x2-28x
First expression, x3-16x
= x (x2-16)
= x {(x)2– (4)2}
x (x+4) (x-4)
= x (2x2+8x+x+4)
= x {2x (x+4) +1 (x+4)}
= x (x+4) (2x +1)
Third expression, 2x3+x2-28x = x (2x2+x-28)
= x (2x2 + 8x – 7x – 28)
= x {2x (x + 4) – 7 (x + 4)}
= x (x + 4) (2x – 7)
Required G.C.D. = x (x + 4)
11. 4x2-1, 8x3-1, 4x2-4x+1
Solution:
Given 4x2-1,8x3-1,4x2-4x+1
First expression, 4x2-1
= (2x)2– (1)2
= (2x +1) (2x-1)
Second expression, 8x3-1
= (2x)3 – (1)3
= (2x-1) {(2x)2+2x.1 + (1)2}
= (2x-1) (4×2+2x+1)
Third expression, 4x2-4x+1
= (2x)2– 2.2x.1 + (1)2
= (2x-1)2= (2x-1) (2x-1)
Required G.C.D. = (2x-1)
12. x3-3x2-10x, x3+6x2+8x, x4-5x3-14x2
Solution:
Given x3-3x2-10x, x3+6x2+8x, x4-5x3-14x2
First expression, x3-3x2-10x
= x (x2-3x-10)
= x {x2– (5-2) x -10}
= x (x2-5x+2x-10)
= x {x (x-5) +2 (x-5)}
= x (x-5) (x+2)
= x (x+4) (x+2)
Third expression, x4-5×3-14x2
= x2 (x2-5x-14)
= x2 (x2-7x+2x-14)
= x2{x (x-7)+2(x-7)}
= x.x(x-7)(x+2)
Required G.C.D. = x(x+2)
13. 6x2-13xa+6a22, 6x2+11xa-10a2, 6x2+2xa-4a2
Solution:
Given 6x2-13xa+6a2, 6x2+11xa-10a2, 6x2+2xa-4a2
First expression, 6x2-13xa+6a2
= 6x2 – (9+4) xa + 6a2
= 6x2 – 9ax – 4ax + 6a2
= 3x (2x-3a)-2a(2x-3a)
= (2x – 3a) (3x – 2a)
Second expression, 6x2+11xa-10a2
=6x2+(15-4) xa -10a2
= 6x2+ 15ax-4ax-10a2
= 3x (2x+5a) -2a (2x+5a)
= (2x+5a) (3x-2a)
Third expression, 6x2+2xa-4a2
= 2 (3x2+xa-2a2)
= 2 (3x2+3ax-2ax-2a2)
= 2 (3x (x+a) -2a (x+a)}
= 2 (x+a) (3x-2a)
Required G.C.D. = (3x – 2a)
Question 4. Let’s find the L.C.M of the following algebraic expressions :
1. p2-q2, (p+q)2
Solution:
Given p2-q2, (p+q)2
First expression, p2-q2
= (p+q) (p-q)
Second expression, (p+q)2
= (p+q) (p+q)
Required L.C.M. .= (p+q)2 (p-q)
2. (x2y2-x2), (xy2-2xy+x)
Solution:
Given (x2y2-x2), (xy2-2xy+x)
First expression, (x2y2-x2)
= x2(y2-1)
= x2(y+1) (y-1)
Second expression, (xy2-2xy+x)
= x(y2-2y+1)
= x (y-1 )2
= X (y-1) (y-1)
Required L.C.M. = x2 (y-1)2 (y+1)
3. (p+q) (p+r), (q+r) (r+p), (r+p) (p+q)
Solution:
Given (p+q) (p+r), (q+r) (r+p), (r+p) (p+q)
Required L.C.M. = (p+q) (q+r) (r+p)
4. ab4-8ab, a2b4+8a2b, ab4-4ab2
Solution:
Given ab4-8ab, a2b4+8a2b, ab44-4ab2
First expression, ab4-8ab
= ab(b3-8)
= ab {(b)3-(2)3}
= ab (b-2) {(b)2+b2 + (2)2}
= ab (b-2) (b2+2b+4)
Second expression, a2b4+8a2b
= a2b (b3+8)
= a2b {(b)3+(2)3}
= a2b (b+2) {(b)2-b.2. + (2)2}
= a2b (b+2) (b2-2b+4)
Third expression, ab4-4ab2–
= ab2(b2-4)
= ab2 {(b)2 – (2)2}
= ab2 (b+2) (b-2)
Required L.C.M. = a2b2 (b-2) (b+2) (b2+ 2b+4) (b2– 2b+4)
5. x4+x2y2+y4, x3y+y4, (x2-xy)3
Solution:
Given x4+x2y2+y4, x3+y4, (x2-xy)3
First expression, x4+x2y2+y4
= (x2)2 + 2.x2.y2+ (y2)2 – x2y2
= (x2+y2)2 – (xy)2
= (x2+y2+xy) (x2+y2-xy)
= (x2+xy+y2) (x2-xy+y2)
Second expression, x3y+y4 = y (x3+y3)
= y (x+y) (x2-xy+y2)
Third expression, (x2-xy)3
= {x (x-y)}3
= x3 (x-y)3 .
Required L.C.M. = x3y (x+y) (x-y)3 (x2+ xy + y2) (x2-xy+y2)
6. p2 + 2p, 2p4+3p3-2p2, 2p3-3p2-14p.
Solution:
Given p2 + 2p, 2p4+3p3-2p2, 2p3-3p2-14p
First expression, p2+ 2p
= P (P+2)
Second expression, 2p4+3p3-2p2
= p2 (2p2+3p-2)
= p2 (2p2+4p-p-2)
= p2 (2p (p+2) -1 (p+2)}
= p2 (p+2) (2p-1)
Third expression, 2p3-3p2-14p
= p (2p2-3p-14)
= P (2.P2– (7-4) p-14}
= p (2p2-7p+4p-14)
= P (P (2p-7) +2 (2p-7)}
= P (2p-7) (p+2)
Required L.C.M. = p2 (p+2) (2p-1) (2p-7)
7. x2-y2+z2-2xz, x2-y2-z2+2yz, xy+zx+y2-z2
Solution:
Given x2-y2+z2-2xz, x2-y2-z2+2yz, xy+zx+y2-z2
First expression, x2-y2+z2-2xz
= x2-2xz+z2-y2
= (x-z)2– (y)2
= (x-z+y) (x-z-y)
= (x+y-z) (x-y-z) .
Second expression, x2-y2-z2+2yz
=x2 – (y2-2yz+z2)
= (x)2 – (y-z)2
= (x+y-z) (x-y+z)
Third expression, xy+zx+y2-z2
= x (y+z) + (y+z) (y-z)
= (y+z) (x+y-z)
Required LC.M. = (y+z) (x+y-z) (x-y+z) (x-y-z)
7. x2-xy-2y2, 2x2-5xy+2y2, 2x2+xy-y2
Solution:
Given x2-xy-2y2,2x2-5xy+2y2,2x2+xy-y2
First expression, x2-xy-2y2
= x2-(2-1 )xy-2y2
= x2-2xy + xy – 2y2
= x (x-2y) + y (x-2y)
= (x-2y) (x+y)
Second expression, 2x2-5xy+2y2
= 2x2-(4+1) xy + 2y2
= 2x2 – 4xy – xy + 2y2
= 2x (x-2y) -y (x-2y)
= (x-2y) (2x-y)
Third expression, 2x2+xy-y2
= 2x2+(2-1) xy -y2
= 2x2+2xy-xy-y2
= 2x (x+y) – y (x+y)
= (x+y) (2x-y)
Required L.C.M. = (x+y) (x-2y) (2x-y)
8. 3x2-15x+18, 2x2+2x-24, 4x2+36x+80
Solution:
3x2-15x+18, 2x2+2x-24, 4x2+36x+80
First expression, 3x2-15x+18
= 3(x2-5x+6)
= 3 (x2-(3+2) x+6)
= 3 (x2-3x-2x+6)
= 3 (x (x-3) -2 (x-3)}
= 3 (x-3) (x-2)
= 2 {x (x+4) -3 (x+4)}
= 2 (x+4) (x-3)
Third expression, 4x2+36x+80
= 4 (x2+9x+20)
= 4 {x2 + (5+4) x+20}
= 4 (x2+5x+4x+20)
= 4{x (x+5) +4 (x+5)}
= 2x2 (x+5 (x+4) .
Required L.C.M. =2×2×3 (x-2) (x-3) (x+4) (x+5)
= 12 (x-2) (x-3) (x+4) (x+5)
L.C.M. = 12 (x-2) (x-3) (x+4) (x+5)
9. (a2+2a)2, 2a3+3a3-2a, 2a4-3a3-14a2
Solution:
Given (a2+2a)2, 2a3+3a3-2a, 2a4-3a3-14a2
First expression, (a2+2a)2
= {a(a+2)}2
= a2 (a+2)2
= a.a (a+2) (a+2)
Second expression, 2a3+3a3-2a
= a (2a2+3a-2)
= a {2a2+(4-1) a-2}
= a (2a2+4a-a-2)
= a {2a (a+2) -1 (a+2)
= a (a+2) (2a-1)
Third expression, 2a4-3a3-14a2
= a2 (2a2-3a-14)
= a2{2a2-(7-4) a-14}
= a2 (2a2-7a+4a-14)
= a2 {a (2a-7) +2 (2a-7)}
= a.a (2a-7) (a+2)
Required L.C.M. = a2 (a+2)2 (2a-1) (2a-7)
10. 3a2-5ab-12b2, a5-27a2b3, 9a2+24ab+16b2
Solution:
Given 3a2-5ab-12b2, a5-27a2b3, 9a2+24ab+16b2
First expression, 3a2-5ab-12b2
= 3a2-(9-4)ab-12b2
= 3a2-9ab+4ab-12b2
= 3a (a-3b) + 4b (a-3b)
= (a-3b) (3a+4b)
Second expression, a5-27a2b3
= a2 (a3-27b3)
= a2 {(a)3-(3b)3}
= a2 (a-3b) {(a)2+a.3b+(3b)2}
= a2 (a-3b) (a2+3ab+9b2)
Third expression, 9a2+24ab+16b2
= (3a)2+2.3a.4b + (4b)2
= (3a+4b)2= (3a+4b) (3a+4b)
Required L.C.M. = a2 (a-3b) (3a+4b)2(a2+3ab+9b2)
Question 5. Let’s find the G.C.D and the L.C.M of the following expressions :
1. x3-8, x2+3x-10, x3+2x2-8x
Solution:
Given x3-8, x2+3x-10, x3+2x2-8x
First expression, x3-8
= (x)3 – (2)3
= (x-2) (x2+2x+4)
Second expression, x2+3x-10
= x2+(5-2) x-10
= x2 + 5x- 2x-10
= x (x+5) -2 (x+5)
= (x+5) (x-2)
Third expression, x3+2x2-8x
= x (x2+2x-8)
= x {x2+(4-2)x-8}
= x (x2+4x-2x-8)
= x (x(x+4)-2(x+4)}
= x (x+4) (x-2)
Required G.C.D. = (x-2)
Required L.C.M. = x (x-2) (x+4) (x+5) (x2+2x+4)
2. 3y2-15y+18, 2y2+2y-24, 4y2+36y+80
Solution:
Given 3y2-15y+18, 2y2+2y-24, 4y2+36y+80
First expression, 3y2-15y+18
= 3 (y2-5y+6)
= 3 (y2-3y-2y+6)
= 3 {y (y-3) -2 (y-3)}
= 3 (y-3) (y-2)
Second expression, 2y2+2y-24 = 2(y2+y-12)
= 2 (y2+4y-3y-12)
= 2 {y (y+4) -3 (y+4)}
= 2 (y+4) (y-3)
Third expression, 4y2+36y+80
= 4 (y2+9y+20)
= 4 (y2+5y+4y+20)
= 4 {y (y+5) +4 (y+5)}
= 2x2 (y+5) (y+4)
Required G.C.D. = 1
Required L.C.M. = 12 (y-2) (y-3) (y+4) (y+5)
3. a3-4a2+4a, a2+a-6, a3-8
Solution:
Given a3-4a2+4a, a2+a-6, a3-8
First expression, a3-4a2+4a = a (a2-4a+4)
= a {(a)2-2.a.2 + (2)2}
= a (a-2)2 = a (a-2) (a-2)
Second expression, a2+a-6
= a2 + 3a – 2a – 6
= a (a+3) -2 (a+3)
= (a+3) (a-2)
Third expression, a3-8 = (a)3-(2)3
= (a-2) {(a)2 + a.2 + (2)2}
= (a-2) (a2+2a+4)
Required G.C.D. = (a-2)
Required L.C.M. = a(a-2)2(a+3) (a2+2a+4)
4. a2+b2-c2+2ab, c2+a2-b2+2ca, b2+c2-a2+2bc
Solution:
Given a2+b2-c2+2ab, c2+a2-b2+2ca, b2+c2-a2+2bc
First expression, a2+b2-c2+2ab
= a2+2ab+b2-c2
= (a+b)2-(c)2 = (a+b+c) (a+b-c)
Second expression, c2+a2-b2+2ca
= a2+2ca+c2-b2
= (a+c)2-(b)2
= (a+c+b) (a+c-b)
= (a+b+c) (a-b+c)
Third expression, b2+c2-a2+2bc
= b2+2bc+c2-a2
= (b+c)2-(a)2
= (b+c+a) (b+c-a)
= (a+b+c) (b+c-a)
Required G.C.D. = (a+b+c)
Required L.C.M. = (a+b+c) (a-b+c) (a+b-c) (b+c-a)
5. x3-4x, 4(x2-5x+6), (x2-4x+4)
Solution:
Given x3-4x, 4(x2-5x+6), (x2-4x+4)
First expression, x3-4x
= x (x2-4)
= x {(x)2-(2)2}
= x (x+2) (x-2)
Second expression, 4(x2-5x+6)
= 4 {x2– (3+2) x+6}
= 4 (x2-3x-2x+6)
= 4 {x (x-3) -2 (x-3)}
= 2x2 (x-3) (x-2) .
Third expression, (x2-4x+4)
= (x)2-2.x.2 + (2)2
= (x-2)2
= (x-2) (x-2)
Required G.C.D. = (x-2)
Required L.C.M. = 4x (x+2) (x-2) (x-3)
