To find the product of three binomials, we can use the formula given below.
(x+a) (x+b) (x+c) = x3+(a+b+c)x2 + (ab+bc+ca)x + abc
Find the expansion of
Problem 1 :
(x + 1)(x + 4)(x + 7)
Solution :
(x + a)(x + b)(x + c) = x3+(a + b + c)x2 + (ab + bc + ca)x + abc
a = 1, b = 4 and c = 7
= x3+ (1 + 4 + 7)x2 + [1(4) + 4(7) + 7(1)]x + 1(4)(7)
= x3 + 12x2 + (4 + 28 + 7)x + 28
= x3 + 12x2 + 39x + 28
Problem 2 :
(p + 2)(p - 4)(p + 6)
Solution :
x = p, a = 2, b = -4 and c = 6
p3 + (2 - 4 + 6)p2 + [2(-4) + (-4)6 + 6(2)]p + (2)(-4)(6)
= p3 + 4p2 + (-8 - 24 + 12)p - 48
= p3 + 4p2 - 20p - 48
Problem 3 :
(x + 5)(x - 3)(x - 1)
Solution :
Here x = x, a = 5, b = -3 and c = -1
= x3 + (5 - 3 - 1)x2 + [5(-3) + (-3)(-1)(-1)5]x + 5(-3)(-1)
= x3 + x2 + (-15 + 3 - 5)x + 15
= x3 + x2 - 17x + 15
Problem 4 :
(x - a)(x - 2a)(x - 4a)
Solution :
Here x = 3x, a = -a, b = -2a and c = -4a
= x3 + (-a - 2a - 3a)x2 +
[(-a) (-2a) + (-2a) (-3a) + (-3a) (-a)]x + (-a) (-2a) (-3a)
= x3 - 6ax2 + (2a2 + 6a2 + 3a2)x - 6a3
= x3 - 6ax2 + 11a2x - 6a3
Problem 5 :
(3x + 1)(3x + 2)((3x + 5)
Solution :
Here x = 3x, a = 1, b = 2 and c = 5
= (3x)3 + (1 + 2 + 5) (3x)2 + (1.2 + 2.5 + 5.1) (3x) + 1.2.5
= 27 x3 + (8)9x2 + (2 + 10 + 5) (3x) + 1.2.5
= 27 x3+ 72x2 + 51x+ 10
Problem 6 :
(a + 1)(a - 1)(a2 + 1)
Solution :
= (a + 1)(a - 1)(a2 + 1)
Multiplying the first two binomials, (a + 1)(a - 1)
= a2 - 12
= a2 - 1
Multiplying all, we get
(a + 1)(a - 1)(a2 + 1) = (a2 - 1) (a2 + 1)
Using algebraic identity, we get
= (a2)2 - 12
= a4 - 1
Problem 7 :
(a + b)(a - b)(a2 + b2)
Solution :
= (a + b)(a - b)(a2 + b2)
Multiplying the first two binomials, (a + b)(a - b)
= a2 - b2
Multiplying all, we get
(a + b)(a - b)(a2 + b2) = (a2 - b2) (a2 + b2)
Using algebraic identity, we get
= (a2)2 - (b2)2
= a4 - b4
Problem 8 :
(2a + b)(2a - b)(4a2 + b2)
Solution :
= (2a + b)(2a - b)(4a2 + b2)
Multiplying the first two binomials, (2a + b)(2a - b)
= (2a)2 - b2
= 22a2 - b2
= 4a2 - b2
Multiplying all, we get
(2a + b)(2a - b)(4a2 + b2) = (4a2 - b2) (4a2 + b2)
Using algebraic identity, we get
= (4a2)2 - (b2)2
= 4a4 - b4
Problem 9 :
(3 - 2x)(3 + 2x)(9 + 4x2)
Solution :
= (3 - 2x)(3 + 2x)(9 + 4x2)
Multiplying the first two binomials, (3 - 2x)(3 + 2x)
= 32 - (2x)2
= 9 - 22x2
= 9 - 4x2
Multiplying all, we get
(3 - 2x)(3 + 2x)(9 + 4x2) = (9 - 4x2) (9 + 4x2)
Using algebraic identity, we get
= 92 - (4x2)2
= 81 - 16x4
Problem 10 :
(3x - 4y)(3x + 4y)(9x2 + 16y2)
Solution :
= (3x - 4y)(3x + 4y)(9x2 + 16y2)
Multiplying the first two binomials, (3x - 4y)(3x + 4y)
= (3x)2 - (4y)2
= 32x2 - 42y2
= 9x2 - 16y2
Multiplying all, we get
(3x - 4y)(3x + 4y)(9x2 + 16y2) = (9x2 - 16y2) (9x2 + 16y2)
Using algebraic identity, we get
= (9x)2 - (16y2)2
= 81x2 - 256y4
Problem 11 :
Given (x + 3)(x + a)(x + 7) = x³ + 15x² + 71x + 105, find a.
Solution :
(x + 3)(x + a)(x + 7) = x³ + 15x² + 71x + 105
(x + 3)(x + a) = x2 + ax + 3x + 3a
= x2 + x(a + 3) + 3a
(x2 + x(a + 3) + 3a) (x + 7)
= x(x2 + x(a + 3) + 3a) + 7(x2 + x(a + 3) + 3a)
= x3 + x2(a + 3) + 3ax + 7x2 + 7x(a + 3) + 21a
= x3 + x2(a + 3) + 7x2 + 3ax + 7x(a + 3) + 21a
= x3 + x2(a + 3 + 7) + [3a + 7(a + 3)] x + 21a
= x3 + x2(a + 10) + [3a + 7a + 21] x + 21a
= x3 + x2(a + 10) + [10a + 21] x + 21a
Equating the expansion to the given expression, we get
x3 + x2(a + 10) + [10a + 21] x + 21a = x³ + 15x² + 71x + 105
(a + 10) = 15, 10a + 21 = 71 and 21a = 105
a = 15 - 10
a = 5
So, the value of a is 5.
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