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66 questions

Unit 10 · Highest Common Factor and Lowest Common Multiple

Exercise 10.1

Question 1

a.

What is meant by the HCF of algebraic expressions?

Answer

The HCF (highest common factor) of two or more algebraic expressions is the product of the factors common to all of them. It is the expression of highest degree that divides each of them exactly. For example, the HCF of x2−4 x^2 - 4 and x3−8 x^3 - 8 is x − 2.

b.

When is the HCF of algebraic expressions 1?

Answer

When the expressions have no common factor (other than 1). For example, p3−q3=(p−q)(p2+pq+q2) p^3 - q^3 = (p - q)(p^2 + pq + q^2) and p3+q3=(p+q)(p2−pq+q2) p^3 + q^3 = (p + q)(p^2 - pq + q^2) share no factor, so their HCF is 1.

c.

What is the product of the common factors of 4x3y 4x^3y and 3z3 3z^3 ?

Answer

4x3y=2×2×x×x×x×y 4x^3y = 2 \times 2 \times x \times x \times x \times y and 3z3=3×z×z×z 3z^3 = 3 \times z \times z \times z . They have no factor in common, so the product of the common factors, their HCF, is 1.

Question 2

a.

(x+y)2 (x + y)^2 and (x + y)(x − y)

Answer

(x + y)(x + y) and (x + y)(x − y): (x + y) is in both, once in the second. HCF = x + y.

b.

(x−y)(x2+xy+y2) (x - y)(x^2 + xy + y^2) and (x2+xy+y2)(x2−xy+y2) (x^2 + xy + y^2)(x^2 - xy + y^2)

Answer

The factor in both: HCF =x2+xy+y2 = x^2 + xy + y^2 .

c.

(2x−3y)(4x2+6xy+9y2) (2x - 3y)(4x^2 + 6xy + 9y^2) and 4xy(2x − 3y)(2x + 3y)

Answer

Only (2x − 3y) is in both: HCF = 2x − 3y.
(2x + 3y) is not a factor of 4x2+6xy+9y2 4x^2 + 6xy + 9y^2 : (2x+3y)2=4x2+12xy+9y2 (2x + 3y)^2 = 4x^2 + 12xy + 9y^2 . The first expression is 8x3−27y3 8x^3 - 27y^3 , which has no (2x + 3y) in it.

d.

4a3b(a−b)(a+b−1) 4a^3b(a - b)(a + b - 1) and 16a2b2(a+2b)(a+b−1) 16a^2b^2(a + 2b)(a + b - 1)

Answer

The HCF of 4 and 16 is 4; the lowest powers are a2 a^2 and b; and (a + b − 1) is in both. HCF =4a2b(a+b−1) = 4a^2b(a + b - 1) .

Question 3

a.

a2−ab a^2 - ab and a3b−a2b2 a^3b - a^2b^2

Answer

a2−ab=a(a−b) a^2 - ab = a(a - b) and a3b−a2b2=a2b(a−b) a^3b - a^2b^2 = a^2b(a - b) . HCF = a(a − b).

b.

3x2+9x 3x^2 + 9x and 7x + 21

Answer

3x2+9x=3x(x+3) 3x^2 + 9x = 3x(x + 3) and 7x + 21 = 7(x + 3). HCF = x + 3.

c.

a3+1 a^3 + 1 and a4+a2+1 a^4 + a^2 + 1

Answer

a3+1=(a+1)(a2−a+1) a^3 + 1 = (a + 1)(a^2 - a + 1) .
a4+a2+1=(a2+1)2−a2=(a2+a+1)(a2−a+1) \begin{aligned} a^4 + a^2 + 1 &= (a^2 + 1)^2 - a^2 \\ &= (a^2 + a + 1)(a^2 - a + 1) \end{aligned}
HCF =a2−a+1 = a^2 - a + 1 .

d.

8x3+27y3 8x^3 + 27y^3 and 16x4+36x2y2+81y4 16x^4 + 36x^2y^2 + 81y^4

Answer

8x3+27y3=(2x)3+(3y)3=(2x+3y)(4x2−6xy+9y2) 8x^3 + 27y^3 = (2x)^3 + (3y)^3 = (2x + 3y)(4x^2 - 6xy + 9y^2) .
16x4+36x2y2+81y4=(4x2+9y2)2−72x2y2+36x2y2=(4x2+9y2)2−(6xy)2=(4x2+6xy+9y2)(4x2−6xy+9y2) \begin{aligned} &16x^4 + 36x^2y^2 + 81y^4 \\ &= (4x^2 + 9y^2)^2 - 72x^2y^2 + 36x^2y^2 \\ &= (4x^2 + 9y^2)^2 - (6xy)^2 \\ &= (4x^2 + 6xy + 9y^2)(4x^2 - 6xy + 9y^2) \end{aligned}
HCF =4x2−6xy+9y2 = 4x^2 - 6xy + 9y^2 .

e.

m3+1+2m2+2m m^3 + 1 + 2m^2 + 2m and m3−1 m^3 - 1

Answer

m3+1+2m2+2m=(m+1)(m2−m+1)+2m(m+1)=(m+1)(m2−m+1+2m)=(m+1)(m2+m+1) \begin{aligned} &m^3 + 1 + 2m^2 + 2m \\ &= (m + 1)(m^2 - m + 1) + 2m(m + 1) \\ &= (m + 1)(m^2 - m + 1 + 2m) \\ &= (m + 1)(m^2 + m + 1) \end{aligned}
m3−1=(m−1)(m2+m+1) m^3 - 1 = (m - 1)(m^2 + m + 1) . HCF =m2+m+1 = m^2 + m + 1 .

f.

8(6x4−x3−2x2) 8(6x^4 - x^3 - 2x^2) and 12(2x6+3x5+x4) 12(2x^6 + 3x^5 + x^4)

Answer

8(6x4−x3−2x2)=8x2(6x2−4x+3x−2)=8x2{2x(3x−2)+(3x−2)}=8x2(3x−2)(2x+1) \begin{aligned} &8(6x^4 - x^3 - 2x^2) \\ &= 8x^2(6x^2 - 4x + 3x - 2) \\ &= 8x^2\{2x(3x - 2) + (3x - 2)\} \\ &= 8x^2(3x - 2)(2x + 1) \end{aligned}
12(2x6+3x5+x4)=12x4(2x2+2x+x+1)=12x4(2x+1)(x+1) \begin{aligned} &12(2x^6 + 3x^5 + x^4) \\ &= 12x^4(2x^2 + 2x + x + 1) \\ &= 12x^4(2x + 1)(x + 1) \end{aligned}
The HCF of 8 and 12 is 4, the lower power of x is x2 x^2 , and (2x + 1) is in both. HCF =4x2(2x+1) = 4x^2(2x + 1) .

g.

a2+4ab+4b2−c2 a^2 + 4ab + 4b^2 - c^2 and a2−4b2+ac−2bc a^2 - 4b^2 + ac - 2bc

Answer

a2+4ab+4b2−c2=(a+2b)2−c2=(a+2b+c)(a+2b−c) \begin{aligned} &a^2 + 4ab + 4b^2 - c^2 \\ &= (a + 2b)^2 - c^2 \\ &= (a + 2b + c)(a + 2b - c) \end{aligned}
a2−4b2+ac−2bc=(a+2b)(a−2b)+c(a−2b)=(a−2b)(a+2b+c) \begin{aligned} &a^2 - 4b^2 + ac - 2bc \\ &= (a + 2b)(a - 2b) + c(a - 2b) \\ &= (a - 2b)(a + 2b + c) \end{aligned}
HCF = a + 2b + c.

h.

x4+4y4 x^4 + 4y^4 and 2x3y+4xy3+4x2y2 2x^3y + 4xy^3 + 4x^2y^2

Answer

x4+4y4=(x2+2y2)2−4x2y2=(x2+2y2)2−(2xy)2=(x2+2xy+2y2)×(x2−2xy+2y2) \begin{aligned} x^4 + 4y^4 &= (x^2 + 2y^2)^2 - 4x^2y^2 \\ &= (x^2 + 2y^2)^2 - (2xy)^2 \\ &= (x^2 + 2xy + 2y^2) \\ &\quad \times (x^2 - 2xy + 2y^2) \end{aligned}
2x3y+4xy3+4x2y2=2xy(x2+2xy+2y2) 2x^3y + 4xy^3 + 4x^2y^2 = 2xy(x^2 + 2xy + 2y^2) . HCF =x2+2xy+2y2 = x^2 + 2xy + 2y^2 .

i.

x2−10x+24+6y−9y2 x^2 - 10x + 24 + 6y - 9y^2 and x2+3xy−6x x^2 + 3xy - 6x

Answer

x2−10x+24+6y−9y2=(x−5)2−1+6y−9y2=(x−5)2−(3y−1)2=(x−5+3y−1)(x−5−3y+1)=(x+3y−6)(x−3y−4) \begin{aligned} &x^2 - 10x + 24 + 6y - 9y^2 \\ &= (x - 5)^2 - 1 + 6y - 9y^2 \\ &= (x - 5)^2 - (3y - 1)^2 \\ &= (x - 5 + 3y - 1)(x - 5 - 3y + 1) \\ &= (x + 3y - 6)(x - 3y - 4) \end{aligned}
x2+3xy−6x=x(x+3y−6) x^2 + 3xy - 6x = x(x + 3y - 6) . HCF = x + 3y − 6.

j.

(1−x2)(1−y2)+4xy (1 - x^2)(1 - y^2) + 4xy and 1−2x+y−x2y+x2 1 - 2x + y - x^2y + x^2

Hint

Multiply out the first, then group it as (1+xy)2−(x−y)2 (1 + xy)^2 - (x - y)^2 .

Answer

(1−x2)(1−y2)+4xy=1−x2−y2+x2y2+4xy=(1+2xy+x2y2)−(x2−2xy+y2)=(1+xy)2−(x−y)2=(1+xy+x−y)(1+xy−x+y) \begin{aligned} &(1 - x^2)(1 - y^2) + 4xy \\ &= 1 - x^2 - y^2 + x^2y^2 + 4xy \\ &= (1 + 2xy + x^2y^2) \\ &\quad - (x^2 - 2xy + y^2) \\ &= (1 + xy)^2 - (x - y)^2 \\ &= (1 + xy + x - y)(1 + xy - x + y) \end{aligned}
1−2x+y−x2y+x2=(1−2x+x2)+y(1−x2)=(1−x)2+y(1−x)(1+x)=(1−x)(1−x+y+xy) \begin{aligned} &1 - 2x + y - x^2y + x^2 \\ &= (1 - 2x + x^2) + y(1 - x^2) \\ &= (1 - x)^2 + y(1 - x)(1 + x) \\ &= (1 - x)(1 - x + y + xy) \end{aligned}
HCF = 1 − x + y + xy.

k.

8a3+1 8a^3 + 1 and 16a4−4a2+4a−1 16a^4 - 4a^2 + 4a - 1

Answer

8a3+1=(2a+1)(4a2−2a+1) 8a^3 + 1 = (2a + 1)(4a^2 - 2a + 1) .
16a4−4a2+4a−1=16a4−(4a2−4a+1)=(4a2)2−(2a−1)2=(4a2+2a−1)(4a2−2a+1) \begin{aligned} &16a^4 - 4a^2 + 4a - 1 \\ &= 16a^4 - (4a^2 - 4a + 1) \\ &= (4a^2)^2 - (2a - 1)^2 \\ &= (4a^2 + 2a - 1)(4a^2 - 2a + 1) \end{aligned}
HCF =4a2−2a+1 = 4a^2 - 2a + 1 .

l.

2a3−a2+a−2 2a^3 - a^2 + a - 2 and a3−a2+a−1 a^3 - a^2 + a - 1

Answer

2a3−a2+a−2=2(a3−1)−a(a−1)=2(a−1)(a2+a+1)−a(a−1)=(a−1)(2a2+2a+2−a)=(a−1)(2a2+a+2) \begin{aligned} &2a^3 - a^2 + a - 2 \\ &= 2(a^3 - 1) - a(a - 1) \\ &= 2(a - 1)(a^2 + a + 1) - a(a - 1) \\ &= (a - 1)(2a^2 + 2a + 2 - a) \\ &= (a - 1)(2a^2 + a + 2) \end{aligned}
a3−a2+a−1=a2(a−1)+(a−1)=(a−1)(a2+1) a^3 - a^2 + a - 1 = a^2(a - 1) + (a - 1) = (a - 1)(a^2 + 1) . HCF = a − 1.

Question 4

a.

a3+b3 a^3 + b^3 , a3−a2b+ab2 a^3 - a^2b + ab^2 and a4+a2b2+b4 a^4 + a^2b^2 + b^4

Answer

a3+b3=(a+b)(a2−ab+b2) a^3 + b^3 = (a + b)(a^2 - ab + b^2) ; a3−a2b+ab2=a(a2−ab+b2) a^3 - a^2b + ab^2 = a(a^2 - ab + b^2) ; a4+a2b2+b4=(a2+ab+b2)(a2−ab+b2) a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2) .
HCF =a2−ab+b2 = a^2 - ab + b^2 .

b.

x3+2x2+2x+1 x^3 + 2x^2 + 2x + 1 , x3−1 x^3 - 1 and x4+x2+1 x^4 + x^2 + 1

Answer

x3+2x2+2x+1=(x3+1)+2x(x+1)=(x+1)(x2−x+1+2x)=(x+1)(x2+x+1) \begin{aligned} &x^3 + 2x^2 + 2x + 1 \\ &= (x^3 + 1) + 2x(x + 1) \\ &= (x + 1)(x^2 - x + 1 + 2x) \\ &= (x + 1)(x^2 + x + 1) \end{aligned}
x3−1=(x−1)(x2+x+1) x^3 - 1 = (x - 1)(x^2 + x + 1) and x4+x2+1=(x2+x+1)(x2−x+1) x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1) .
HCF =x2+x+1 = x^2 + x + 1 .

c.

x3−4x x^3 - 4x , 4x3−10x2+4x 4x^3 - 10x^2 + 4x and 3x4−8x3+4x2 3x^4 - 8x^3 + 4x^2

Answer

x3−4x=x(x+2)(x−2) x^3 - 4x = x(x + 2)(x - 2) .
4x3−10x2+4x=2x(2x2−5x+2)=2x(2x−1)(x−2) 4x^3 - 10x^2 + 4x = 2x(2x^2 - 5x + 2) = 2x(2x - 1)(x - 2) .
3x4−8x3+4x2=x2(3x2−8x+4)=x2(3x−2)(x−2) 3x^4 - 8x^3 + 4x^2 = x^2(3x^2 - 8x + 4) = x^2(3x - 2)(x - 2) .
HCF = x(x − 2).

d.

y2+2y−8 y^2 + 2y - 8 , y2−5y+6 y^2 - 5y + 6 and y2+5y−14 y^2 + 5y - 14

Answer

(y + 4)(y − 2), (y − 2)(y − 3) and (y + 7)(y − 2). HCF = y − 2.

e.

x2+2x+1 x^2 + 2x + 1 , x2+5x+6 x^2 + 5x + 6 and 2x2−5x+2 2x^2 - 5x + 2

Answer

(x+1)2 (x + 1)^2 , (x + 2)(x + 3) and (2x − 1)(x − 2): no factor is in all three. HCF = 1.

f.

x2+2x−8 x^2 + 2x - 8 , x2−2x−24 x^2 - 2x - 24 and x2+5x+4 x^2 + 5x + 4

Answer

(x + 4)(x − 2), (x − 6)(x + 4) and (x + 4)(x + 1). HCF = x + 4.

g.

x3+2x2+4x x^3 + 2x^2 + 4x , x4+4x2+16 x^4 + 4x^2 + 16 and x3−8 x^3 - 8

Answer

x3+2x2+4x=x(x2+2x+4) x^3 + 2x^2 + 4x = x(x^2 + 2x + 4) .
x4+4x2+16=(x2+4)2−4x2=(x2+2x+4)(x2−2x+4) \begin{aligned} x^4 + 4x^2 + 16 &= (x^2 + 4)^2 - 4x^2 \\ &= (x^2 + 2x + 4)(x^2 - 2x + 4) \end{aligned}
x3−8=(x−2)(x2+2x+4) x^3 - 8 = (x - 2)(x^2 + 2x + 4) . HCF =x2+2x+4 = x^2 + 2x + 4 .

h.

8x3+27y3 8x^3 + 27y^3 , 16x4+36x2y2+81y4 16x^4 + 36x^2y^2 + 81y^4 and 4x3−6x2y+9xy2 4x^3 - 6x^2y + 9xy^2

Answer

From Question 3d, 8x3+27y3=(2x+3y)(4x2−6xy+9y2) 8x^3 + 27y^3 = (2x + 3y)(4x^2 - 6xy + 9y^2) and 16x4+36x2y2+81y4=(4x2+6xy+9y2)(4x2−6xy+9y2) 16x^4 + 36x^2y^2 + 81y^4 = (4x^2 + 6xy + 9y^2)(4x^2 - 6xy + 9y^2) .
4x3−6x2y+9xy2=x(4x2−6xy+9y2) 4x^3 - 6x^2y + 9xy^2 = x(4x^2 - 6xy + 9y^2) . HCF =4x2−6xy+9y2 = 4x^2 - 6xy + 9y^2 .

i.

2x3−54 2x^3 - 54 , 24x4+18x2+162 24x^4 + 18x^2 + 162 and 2x2+6x+18 2x^2 + 6x + 18

Answer

2x3−54=2(x3−27)=2(x−3)(x2+3x+9) 2x^3 - 54 = 2(x^3 - 27) = 2(x - 3)(x^2 + 3x + 9) .
24x4+18x2+162=6(4x4+3x2+27) 24x^4 + 18x^2 + 162 = 6(4x^4 + 3x^2 + 27) .
2x2+6x+18=2(x2+3x+9) 2x^2 + 6x + 18 = 2(x^2 + 3x + 9) .
4x4+3x2+27 4x^4 + 3x^2 + 27 has neither x − 3 nor x2+3x+9 x^2 + 3x + 9 as a factor: put x = 1 and it is 34, which 13 (the value of x2+3x+9 x^2 + 3x + 9 ) doesn’t divide; at x = 3 it isn’t 0. So only the number is common: the HCF of 2, 6 and 2. HCF = 2.

j.

9x2−4y2−8yz−4z2 9x^2 - 4y^2 - 8yz - 4z^2 , 4z2−4y2−9x2−12xy 4z^2 - 4y^2 - 9x^2 - 12xy and 9x2+12xz+4z2−4y2 9x^2 + 12xz + 4z^2 - 4y^2

Answer

9x2−4y2−8yz−4z2=(3x)2−(2y+2z)2=(3x+2y+2z)(3x−2y−2z) \begin{aligned} &9x^2 - 4y^2 - 8yz - 4z^2 \\ &= (3x)^2 - (2y + 2z)^2 \\ &= (3x + 2y + 2z)(3x - 2y - 2z) \end{aligned}
4z2−4y2−9x2−12xy=(2z)2−(3x+2y)2=(2z+3x+2y)(2z−3x−2y) \begin{aligned} &4z^2 - 4y^2 - 9x^2 - 12xy \\ &= (2z)^2 - (3x + 2y)^2 \\ &= (2z + 3x + 2y)(2z - 3x - 2y) \end{aligned}
9x2+12xz+4z2−4y2=(3x+2z)2−(2y)2=(3x+2z+2y)(3x+2z−2y) \begin{aligned} &9x^2 + 12xz + 4z^2 - 4y^2 \\ &= (3x + 2z)^2 - (2y)^2 \\ &= (3x + 2z + 2y)(3x + 2z - 2y) \end{aligned}
HCF = 3x + 2y + 2z.

k.

2ax2+2ax−12a 2ax^2 + 2ax - 12a , 3a2x2−7a2x−6a2 3a^2x^2 - 7a^2x - 6a^2 and a3x2+4a3x−21a3 a^3x^2 + 4a^3x - 21a^3

Answer

2ax2+2ax−12a=2a(x2+x−6)=2a(x+3)(x−2) 2ax^2 + 2ax - 12a = 2a(x^2 + x - 6) = 2a(x + 3)(x - 2) .
3a2x2−7a2x−6a2=a2(3x2−7x−6)=a2(3x+2)(x−3) 3a^2x^2 - 7a^2x - 6a^2 = a^2(3x^2 - 7x - 6) = a^2(3x + 2)(x - 3) .
a3x2+4a3x−21a3=a3(x2+4x−21)=a3(x+7)(x−3) a^3x^2 + 4a^3x - 21a^3 = a^3(x^2 + 4x - 21) = a^3(x + 7)(x - 3) .
(x − 3) is in the second and third only. The only factor in all three is a: HCF = a.
If the first expression were 2ax2−2ax−12a=2a(x−3)(x+2) 2ax^2 - 2ax - 12a = 2a(x - 3)(x + 2) , all three would share a(x − 3), and that would be the HCF.

l.

x3+64y3 x^3 + 64y^3 , x4+16x2y2+256y4 x^4 + 16x^2y^2 + 256y^4 and 4x3−16x2y+64xy2 4x^3 - 16x^2y + 64xy^2

Answer

x3+64y3=(x+4y)(x2−4xy+16y2) x^3 + 64y^3 = (x + 4y)(x^2 - 4xy + 16y^2) .
x4+16x2y2+256y4=(x2+16y2)2−16x2y2=(x2+4xy+16y2)×(x2−4xy+16y2) \begin{aligned} &x^4 + 16x^2y^2 + 256y^4 \\ &= (x^2 + 16y^2)^2 - 16x^2y^2 \\ &= (x^2 + 4xy + 16y^2) \\ &\quad \times (x^2 - 4xy + 16y^2) \end{aligned}
4x3−16x2y+64xy2=4x(x2−4xy+16y2) 4x^3 - 16x^2y + 64xy^2 = 4x(x^2 - 4xy + 16y^2) . HCF =x2−4xy+16y2 = x^2 - 4xy + 16y^2 .

m.

(a+b)3−3ab(a+b) (a + b)^3 - 3ab(a + b) , a4+a2b2+b4 a^4 + a^2b^2 + b^4 and a4−2a3b+a2b2−b4 a^4 - 2a^3b + a^2b^2 - b^4

Answer

(a+b)3−3ab(a+b)=(a+b)(a2−ab+b2) (a + b)^3 - 3ab(a + b) = (a + b)(a^2 - ab + b^2) , as in Example 3b.
a4+a2b2+b4=(a2+ab+b2)(a2−ab+b2) a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2) .
a4−2a3b+a2b2−b4=(a2−ab)2−(b2)2=(a2−ab+b2)(a2−ab−b2) \begin{aligned} &a^4 - 2a^3b + a^2b^2 - b^4 \\ &= (a^2 - ab)^2 - (b^2)^2 \\ &= (a^2 - ab + b^2)(a^2 - ab - b^2) \end{aligned}
HCF =a2−ab+b2 = a^2 - ab + b^2 .

Question 5

5.

The areas of three rectangular rooms are (x + 3)(x + 6), x2+8x+15 x^2 + 8x + 15 and x2+7x+12 x^2 + 7x + 12 square units. The rooms all have the same breadth. Find it.

Answer

The breadth is a factor of every area, so it is their common factor. x2+8x+15=(x+3)(x+5) x^2 + 8x + 15 = (x + 3)(x + 5) and x2+7x+12=(x+3)(x+4) x^2 + 7x + 12 = (x + 3)(x + 4) , and the first is (x + 3)(x + 6).
The breadth is x + 3 units, and the lengths are x + 6, x + 5 and x + 4 units.
What it means: if the rooms stand side by side along one wall, x + 3 is the depth of that row; with x = 2, they are 5 units deep and 8, 7 and 6 units long.

Exercise 10.2

Question 1

a.

Define LCM.

Answer

The LCM (lowest common multiple) of two or more algebraic expressions is the product of their common factors and their remaining factors, each factor taken as many times as it appears in the expression that has it most. It is the expression of lowest degree that each of them divides exactly.

b.

For two expressions, how are their HCF and LCM related to the expressions?

Answer

HCF × LCM = first expression × second expression. For example, x2−4 x^2 - 4 and x3−8 x^3 - 8 have HCF x − 2 and LCM (x+2)(x−2)(x2+2x+4) (x + 2)(x - 2)(x^2 + 2x + 4) , and the HCF times the LCM is (x2−4)(x3−8) (x^2 - 4)(x^3 - 8) .

Question 2

a.

2x(x + 2)(x − 2) and 4x2(3x+7)(x−2) 4x^2(3x + 7)(x - 2)

Answer

The LCM of 2 and 4 is 4; the higher power is x2 x^2 ; the brackets (x − 2), (x + 2), (3x + 7) once each.
LCM =4x2(x−2)(x+2)(3x+7) = 4x^2(x - 2)(x + 2)(3x + 7) .

b.

3x2y(x−y)(x2+3xy+9y2) 3x^2y(x - y)(x^2 + 3xy + 9y^2) and 10x(x2+y2)(x+y)(x−y) 10x(x^2 + y^2)(x + y)(x - y)

Answer

The LCM of 3 and 10 is 30; then x2 x^2 , y, and each bracket once:
LCM =30x2y(x−y)(x+y) = 30x^2y(x - y)(x + y) (x2+y2)(x2+3xy+9y2) (x^2 + y^2)(x^2 + 3xy + 9y^2) .

c.

(2x−y)(x2+xy+y2)(x2−xy+y2) (2x - y)(x^2 + xy + y^2)(x^2 - xy + y^2) and 25xy(2x − y)

Answer

LCM = 25xy(2x − y)(x2+xy+y2)(x2−xy+y2) (x^2 + xy + y^2)(x^2 - xy + y^2) .

d.

8x3y2(a+b+1)(a−b+1) 8x^3y^2(a + b + 1)(a - b + 1) and 5x3y(a+b+1)(a+b+2) 5x^3y(a + b + 1)(a + b + 2)

Answer

The LCM of 8 and 5 is 40; the higher powers are x3 x^3 and y2 y^2 :
LCM =40x3y2(a+b+1) = 40x^3y^2(a + b + 1) (a − b + 1)(a + b + 2).

Question 3

3.

The HCF and LCM of two expressions are (a+b)2 (a + b)^2 and 3a(a+b)2(2a+b) 3a(a + b)^2(2a + b) , and the second expression is 3a(a+b)2 3a(a + b)^2 . Find the first expression.
The textbook gives the LCM as 3a2(a+b)(2a+b) 3a^2(a + b)(2a + b) , but its numbers don’t work: an LCM always contains the HCF and each expression, and that one has (a + b) only once. With the LCM 3a(a+b)2(2a+b) 3a(a + b)^2(2a + b) they fit.

Answer

For two expressions, first × second = HCF × LCM, so
first=HCF×LCMsecond=(a+b)2×3a(a+b)2(2a+b)3a(a+b)2=(a+b)2(2a+b) \begin{aligned} \text{first} &= \frac{\text{HCF} \times \text{LCM}}{\text{second}} \\ &= \frac{(a + b)^2 \times 3a(a + b)^2(2a + b)}{3a(a + b)^2} \\ &= (a + b)^2(2a + b) \end{aligned}
The first expression is (a+b)2(2a+b) (a + b)^2(2a + b) .
Check: (a+b)2(2a+b) (a + b)^2(2a + b) and 3a(a+b)2 3a(a + b)^2 share exactly (a+b)2 (a + b)^2 ✓, and together they need 3a(a+b)2(2a+b) 3a(a + b)^2(2a + b) ✓.
(With the book’s printed LCM, the same division gives a(a + b)(2a + b); but that and 3a(a+b)2 3a(a + b)^2 have HCF a(a + b), not (a+b)2 (a + b)^2 , so it doesn’t pass the check.)

Question 4

4.

Two expressions are (x + 5) and (x+5)(x2−5x+25) (x + 5)(x^2 - 5x + 25) , and their HCF is (x + 5). Find their LCM.

Answer

LCM=first×secondHCF=(x+5)(x+5)(x2−5x+25)x+5=(x+5)(x2−5x+25) \begin{aligned} \text{LCM} &= \frac{\text{first} \times \text{second}}{\text{HCF}} \\ &= \frac{(x + 5)(x + 5)(x^2 - 5x + 25)}{x + 5} \\ &= (x + 5)(x^2 - 5x + 25) \end{aligned}
LCM =(x+5)(x2−5x+25) = (x + 5)(x^2 - 5x + 25) , which is x3+125 x^3 + 125 . It is the second expression itself: the first divides the second, so the second is already a multiple of both.

Question 5

a.

x2−x+1 x^2 - x + 1 and x4+x x^4 + x

Answer

x4+x=x(x3+1)=x(x+1)(x2−x+1) x^4 + x = x(x^3 + 1) = x(x + 1)(x^2 - x + 1) , which already contains x2−x+1 x^2 - x + 1 .
LCM =x(x+1)(x2−x+1) = x(x + 1)(x^2 - x + 1) .

b.

4x + 16 and 5x + 20

Answer

4(x + 4) and 5(x + 4). LCM = 20(x + 4).

c.

3x + 27 and 8x3+72x2 8x^3 + 72x^2

Answer

3(x + 9) and 8x2(x+9) 8x^2(x + 9) . LCM =24x2(x+9) = 24x^2(x + 9) .

d.

(x−y)3 (x - y)^3 and x3−y3 x^3 - y^3

Answer

(x−y)3 (x - y)^3 and (x−y)(x2+xy+y2) (x - y)(x^2 + xy + y^2) : (x − y) at most three times.
LCM =(x−y)3(x2+xy+y2) = (x - y)^3(x^2 + xy + y^2) .

e.

(a+b)3 (a + b)^3 and a3+b3 a^3 + b^3

Answer

(a+b)3 (a + b)^3 and (a+b)(a2−ab+b2) (a + b)(a^2 - ab + b^2) . LCM =(a+b)3(a2−ab+b2) = (a + b)^3(a^2 - ab + b^2) .

f.

x4+4 x^4 + 4 and 2x3−4x2+4x 2x^3 - 4x^2 + 4x

Answer

x4+4=(x2+2)2−(2x)2=(x2+2x+2)(x2−2x+2) \begin{aligned} x^4 + 4 &= (x^2 + 2)^2 - (2x)^2 \\ &= (x^2 + 2x + 2)(x^2 - 2x + 2) \end{aligned}
2x3−4x2+4x=2x(x2−2x+2) 2x^3 - 4x^2 + 4x = 2x(x^2 - 2x + 2) .
LCM =2x(x2+2x+2)(x2−2x+2) = 2x(x^2 + 2x + 2)(x^2 - 2x + 2) .

g.

a4+a2+1 a^4 + a^2 + 1 and a2−a+1 a^2 - a + 1

Answer

a4+a2+1=(a2+a+1)(a2−a+1) a^4 + a^2 + 1 = (a^2 + a + 1)(a^2 - a + 1) , which contains the second.
LCM =(a2+a+1)(a2−a+1) = (a^2 + a + 1)(a^2 - a + 1) , the first expression itself.

h.

x4+x2y2+y4 x^4 + x^2y^2 + y^4 and x3−y3 x^3 - y^3

Answer

(x2+xy+y2)(x2−xy+y2) (x^2 + xy + y^2)(x^2 - xy + y^2) and (x−y)(x2+xy+y2) (x - y)(x^2 + xy + y^2) .
LCM =(x−y)(x2+xy+y2) = (x - y)(x^2 + xy + y^2) (x2−xy+y2) (x^2 - xy + y^2) .

i.

1+4p+4p2−16p4 1 + 4p + 4p^2 - 16p^4 and 1+2p−8p3−16p4 1 + 2p - 8p^3 - 16p^4

Answer

1+4p+4p2−16p4=(1+2p)2−(4p2)2=(1+2p+4p2)(1+2p−4p2) \begin{aligned} &1 + 4p + 4p^2 - 16p^4 \\ &= (1 + 2p)^2 - (4p^2)^2 \\ &= (1 + 2p + 4p^2)(1 + 2p - 4p^2) \end{aligned}
1+2p−8p3−16p4=(1+2p)−8p3(1+2p)=(1+2p)(1−8p3)=(1+2p)(1−2p)(1+2p+4p2) \begin{aligned} &1 + 2p - 8p^3 - 16p^4 \\ &= (1 + 2p) - 8p^3(1 + 2p) \\ &= (1 + 2p)(1 - 8p^3) \\ &= (1 + 2p)(1 - 2p)(1 + 2p + 4p^2) \end{aligned}
LCM=(1+2p)(1−2p)×(1+2p+4p2)(1+2p−4p2) \begin{aligned} \text{LCM} &= (1 + 2p)(1 - 2p) \\ &\quad \times (1 + 2p + 4p^2)(1 + 2p - 4p^2) \end{aligned}

j.

x3+x2+x+1 x^3 + x^2 + x + 1 and x3−x2+x−1 x^3 - x^2 + x - 1

Answer

x3+x2+x+1=x2(x+1)+(x+1)=(x+1)(x2+1) x^3 + x^2 + x + 1 = x^2(x + 1) + (x + 1) = (x + 1)(x^2 + 1) .
x3−x2+x−1=x2(x−1)+(x−1)=(x−1)(x2+1) x^3 - x^2 + x - 1 = x^2(x - 1) + (x - 1) = (x - 1)(x^2 + 1) .
LCM =(x+1)(x−1)(x2+1) = (x + 1)(x - 1)(x^2 + 1) , which is x4−1 x^4 - 1 .

k.

y4+(2b2−a2)y2+b4 y^4 + (2b^2 - a^2)y^2 + b^4 and y3−ay2+b2y y^3 - ay^2 + b^2y

Answer

y4+(2b2−a2)y2+b4=y4+2b2y2+b4−a2y2=(y2+b2)2−(ay)2=(y2+ay+b2)(y2−ay+b2) \begin{aligned} &y^4 + (2b^2 - a^2)y^2 + b^4 \\ &= y^4 + 2b^2y^2 + b^4 - a^2y^2 \\ &= (y^2 + b^2)^2 - (ay)^2 \\ &= (y^2 + ay + b^2)(y^2 - ay + b^2) \end{aligned}
y3−ay2+b2y=y(y2−ay+b2) y^3 - ay^2 + b^2y = y(y^2 - ay + b^2) .
LCM =y(y2+ay+b2)(y2−ay+b2) = y(y^2 + ay + b^2)(y^2 - ay + b^2) .

l.

x4y4+y4x4+1 \dfrac{x^4}{y^4} + \dfrac{y^4}{x^4} + 1 and x3y3+y3x3 \dfrac{x^3}{y^3} + \dfrac{y^3}{x^3}

Hint

xy×yx=1 \dfrac{x}{y} \times \dfrac{y}{x} = 1 , so these are a4+a2b2+b4 a^4 + a^2b^2 + b^4 and a3+b3 a^3 + b^3 with a=xy a = \dfrac{x}{y} , b=yx b = \dfrac{y}{x} and ab = 1.

Answer

x4y4+y4x4+1=(x2y2+y2x2)2−2+1=(x2y2+1+y2x2)×(x2y2−1+y2x2) \begin{aligned} &\frac{x^4}{y^4} + \frac{y^4}{x^4} + 1 \\ &= \left(\frac{x^2}{y^2} + \frac{y^2}{x^2}\right)^2 - 2 + 1 \\ &= \left(\frac{x^2}{y^2} + 1 + \frac{y^2}{x^2}\right) \\ &\quad \times \left(\frac{x^2}{y^2} - 1 + \frac{y^2}{x^2}\right) \end{aligned}
x3y3+y3x3=(xy+yx)(x2y2−1+y2x2) \begin{aligned} &\frac{x^3}{y^3} + \frac{y^3}{x^3} \\ &= \left(\frac{x}{y} + \frac{y}{x}\right)\left(\frac{x^2}{y^2} - 1 + \frac{y^2}{x^2}\right) \end{aligned}
LCM=(xy+yx)(x2y2+1+y2x2)×(x2y2−1+y2x2) \begin{aligned} \text{LCM} &= \left(\frac{x}{y} + \frac{y}{x}\right)\left(\frac{x^2}{y^2} + 1 + \frac{y^2}{x^2}\right) \\ &\quad \times \left(\frac{x^2}{y^2} - 1 + \frac{y^2}{x^2}\right) \end{aligned}

Question 6

a.

x3+1 x^3 + 1 , x4−x3+x2 x^4 - x^3 + x^2 and x3−x2+x x^3 - x^2 + x

Answer

(x+1)(x2−x+1) (x + 1)(x^2 - x + 1) , x2(x2−x+1) x^2(x^2 - x + 1) and x(x2−x+1) x(x^2 - x + 1) . The highest power of x is x2 x^2 .
LCM =x2(x+1)(x2−x+1) = x^2(x + 1)(x^2 - x + 1) .

b.

a3+1 a^3 + 1 , a4+a2+1 a^4 + a^2 + 1 and a2+a+1 a^2 + a + 1

Answer

(a+1)(a2−a+1) (a + 1)(a^2 - a + 1) , (a2+a+1)(a2−a+1) (a^2 + a + 1)(a^2 - a + 1) and (a2+a+1) (a^2 + a + 1) .
LCM =(a+1)(a2−a+1)(a2+a+1) = (a + 1)(a^2 - a + 1)(a^2 + a + 1) .

c.

x2−3x+2 x^2 - 3x + 2 , x2−5x+6 x^2 - 5x + 6 and x2−8x+12 x^2 - 8x + 12

Answer

(x − 1)(x − 2), (x − 2)(x − 3) and (x − 2)(x − 6).
LCM = (x − 1)(x − 2)(x − 3)(x − 6).

d.

2x3+16 2x^3 + 16 , x2+4x+4 x^2 + 4x + 4 and x2+3x+2 x^2 + 3x + 2

Answer

2x3+16=2(x3+8)=2(x+2)(x2−2x+4) 2x^3 + 16 = 2(x^3 + 8) = 2(x + 2)(x^2 - 2x + 4) ; x2+4x+4=(x+2)2 x^2 + 4x + 4 = (x + 2)^2 ; x2+3x+2=(x+1)(x+2) x^2 + 3x + 2 = (x + 1)(x + 2) .
(x + 2) at most twice: LCM =2(x+2)2(x+1)(x2−2x+4) = 2(x + 2)^2(x + 1)(x^2 - 2x + 4) .

e.

x6−1 x^6 - 1 , x3−1 x^3 - 1 and x4+x2+1 x^4 + x^2 + 1

Answer

x6−1=(x3−1)(x3+1)=(x−1)(x2+x+1)×(x+1)(x2−x+1) \begin{aligned} x^6 - 1 &= (x^3 - 1)(x^3 + 1) \\ &= (x - 1)(x^2 + x + 1) \\ &\quad \times (x + 1)(x^2 - x + 1) \end{aligned}
x3−1=(x−1)(x2+x+1) x^3 - 1 = (x - 1)(x^2 + x + 1) and x4+x2+1=(x2+x+1)(x2−x+1) x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1) : both are already inside the first.
LCM =(x2−1)(x4+x2+1) = (x^2 - 1)(x^4 + x^2 + 1) , which is x6−1 x^6 - 1 itself.

f.

x6−16x4 x^6 - 16x^4 , x5+6x4+8x3 x^5 + 6x^4 + 8x^3 and x4+8x3+16x2 x^4 + 8x^3 + 16x^2

Answer

x6−16x4=x4(x+4)(x−4) x^6 - 16x^4 = x^4(x + 4)(x - 4) ; x5+6x4+8x3=x3(x+2)(x+4) x^5 + 6x^4 + 8x^3 = x^3(x + 2)(x + 4) ; x4+8x3+16x2=x2(x+4)2 x^4 + 8x^3 + 16x^2 = x^2(x + 4)^2 .
The highest power of x is x4 x^4 , and (x + 4) at most twice:
LCM =x4(x+4)2(x−4)(x+2) = x^4(x + 4)^2(x - 4)(x + 2) , which can also be written x4(x2−16)(x2+6x+8) x^4(x^2 - 16)(x^2 + 6x + 8) .

g.

x4+8x2+144 x^4 + 8x^2 + 144 , x3+x(x+12)+3x2 x^3 + x(x + 12) + 3x^2 and x3+12x+4x2 x^3 + 12x + 4x^2

Answer

x4+8x2+144=(x2+12)2−16x2=(x2+4x+12)×(x2−4x+12) \begin{aligned} x^4 + 8x^2 + 144 &= (x^2 + 12)^2 - 16x^2 \\ &= (x^2 + 4x + 12) \\ &\quad \times (x^2 - 4x + 12) \end{aligned}
x3+x(x+12)+3x2=x3+4x2+12x=x(x2+4x+12) x^3 + x(x + 12) + 3x^2 = x^3 + 4x^2 + 12x = x(x^2 + 4x + 12) , and the third is the same.
LCM =x(x2+4x+12)(x2−4x+12) = x(x^2 + 4x + 12)(x^2 - 4x + 12) .

h.

x4−8x2+196 x^4 - 8x^2 + 196 , x3+x(x+14)+5x2 x^3 + x(x + 14) + 5x^2 and 2x2+12x+28 2x^2 + 12x + 28

Answer

x4−8x2+196=(x2+14)2−36x2=(x2+6x+14)×(x2−6x+14) \begin{aligned} x^4 - 8x^2 + 196 &= (x^2 + 14)^2 - 36x^2 \\ &= (x^2 + 6x + 14) \\ &\quad \times (x^2 - 6x + 14) \end{aligned}
x3+x(x+14)+5x2=x3+6x2+14x=x(x2+6x+14) x^3 + x(x + 14) + 5x^2 = x^3 + 6x^2 + 14x = x(x^2 + 6x + 14) ; 2x2+12x+28=2(x2+6x+14) 2x^2 + 12x + 28 = 2(x^2 + 6x + 14) .
LCM =2x(x2+6x+14)(x2−6x+14) = 2x(x^2 + 6x + 14)(x^2 - 6x + 14) .

i.

x4+10x2+169 x^4 + 10x^2 + 169 , x3+4x2+13x x^3 + 4x^2 + 13x and x3+x(x+13)+3x2 x^3 + x(x + 13) + 3x^2

Answer

x4+10x2+169=(x2+13)2−16x2=(x2+4x+13)×(x2−4x+13) \begin{aligned} x^4 + 10x^2 + 169 &= (x^2 + 13)^2 - 16x^2 \\ &= (x^2 + 4x + 13) \\ &\quad \times (x^2 - 4x + 13) \end{aligned}
x3+4x2+13x=x(x2+4x+13) x^3 + 4x^2 + 13x = x(x^2 + 4x + 13) , and the third is the same expression.
LCM =x(x2+4x+13)(x2−4x+13) = x(x^2 + 4x + 13)(x^2 - 4x + 13) .

j.

(y+3)2−9y−27 (y + 3)^2 - 9y - 27 , y3−2y2−15y y^3 - 2y^2 - 15y and y5−13y3+36y y^5 - 13y^3 + 36y

Answer

(y+3)2−9y−27=(y+3)2−9(y+3)=(y+3)(y−6) (y + 3)^2 - 9y - 27 = (y + 3)^2 - 9(y + 3) = (y + 3)(y - 6) .
y3−2y2−15y=y(y2−2y−15)=y(y−5)(y+3) y^3 - 2y^2 - 15y = y(y^2 - 2y - 15) = y(y - 5)(y + 3) .
y5−13y3+36y=y(y4−13y2+36)=y(y2−4)(y2−9) \begin{aligned} y^5 - 13y^3 + 36y &= y(y^4 - 13y^2 + 36) \\ &= y(y^2 - 4)(y^2 - 9) \end{aligned}
LCM =y(y2−9)(y2−4)(y−5)(y−6) = y(y^2 - 9)(y^2 - 4)(y - 5)(y - 6) .

Question 7

a.

m2−10m+24+6n−9n2 m^2 - 10m + 24 + 6n - 9n^2 , m2+6mn+9n2−36 m^2 + 6mn + 9n^2 - 36 and m2+3mn−6m m^2 + 3mn - 6m

Answer

m2−10m+24+6n−9n2=(m−5)2−(3n−1)2=(m+3n−6)(m−3n−4) \begin{aligned} &m^2 - 10m + 24 + 6n - 9n^2 \\ &= (m - 5)^2 - (3n - 1)^2 \\ &= (m + 3n - 6)(m - 3n - 4) \end{aligned}
m2+6mn+9n2−36=(m+3n)2−62=(m+3n+6)(m+3n−6) \begin{aligned} &m^2 + 6mn + 9n^2 - 36 \\ &= (m + 3n)^2 - 6^2 \\ &= (m + 3n + 6)(m + 3n - 6) \end{aligned}
m2+3mn−6m=m(m+3n−6) m^2 + 3mn - 6m = m(m + 3n - 6) .
LCM = m(m + 3n − 6)(m − 3n − 4)(m + 3n + 6).

b.

x4−8x2−33−14y−y2 x^4 - 8x^2 - 33 - 14y - y^2 , x4+2x2y−9+y2 x^4 + 2x^2y - 9 + y^2 and x3+xy+3x x^3 + xy + 3x

Answer

x4−8x2−33−14y−y2=(x2−4)2−(y2+14y+49)=(x2−4)2−(y+7)2=(x2+y+3)(x2−y−11) \begin{aligned} &x^4 - 8x^2 - 33 - 14y - y^2 \\ &= (x^2 - 4)^2 - (y^2 + 14y + 49) \\ &= (x^2 - 4)^2 - (y + 7)^2 \\ &= (x^2 + y + 3)(x^2 - y - 11) \end{aligned}
x4+2x2y−9+y2=(x2+y)2−32=(x2+y+3)(x2+y−3) \begin{aligned} x^4 + 2x^2y - 9 + y^2 &= (x^2 + y)^2 - 3^2 \\ &= (x^2 + y + 3)(x^2 + y - 3) \end{aligned}
x3+xy+3x=x(x2+y+3) x^3 + xy + 3x = x(x^2 + y + 3) .
LCM =x(x2+y+3) = x(x^2 + y + 3) (x2−y−11)(x2+y−3) (x^2 - y - 11)(x^2 + y - 3) .

c.

a4+b2(2a2−1)+b4 a^4 + b^2(2a^2 - 1) + b^4 , a3−b(a+1)(a−b)−b3 a^3 - b(a + 1)(a - b) - b^3 and a3−b(a−1)(a−b)−b3 a^3 - b(a - 1)(a - b) - b^3

Hint

Put a3 a^3 and −b3 -b^3 together: both the second and third have a3−b3 a^3 - b^3 and a multiple of (a − b).

Answer

a4+b2(2a2−1)+b4=a4+2a2b2+b4−b2=(a2+b2)2−b2=(a2+b2+b)(a2+b2−b) \begin{aligned} &a^4 + b^2(2a^2 - 1) + b^4 \\ &= a^4 + 2a^2b^2 + b^4 - b^2 \\ &= (a^2 + b^2)^2 - b^2 \\ &= (a^2 + b^2 + b)(a^2 + b^2 - b) \end{aligned}
a3−b3−b(a+1)(a−b)=(a−b)(a2+ab+b2−ab−b)=(a−b)(a2+b2−b) \begin{aligned} &a^3 - b^3 - b(a + 1)(a - b) \\ &= (a - b)(a^2 + ab + b^2 - ab - b) \\ &= (a - b)(a^2 + b^2 - b) \end{aligned}
a3−b3−b(a−1)(a−b)=(a−b)(a2+ab+b2−ab+b)=(a−b)(a2+b2+b) \begin{aligned} &a^3 - b^3 - b(a - 1)(a - b) \\ &= (a - b)(a^2 + ab + b^2 - ab + b) \\ &= (a - b)(a^2 + b^2 + b) \end{aligned}
LCM =(a−b)(a2+b2−b) = (a - b)(a^2 + b^2 - b) (a2+b2+b) (a^2 + b^2 + b) .