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 :
(2x+3)(2x-5)(2x-7)
Solution :
Here x = 2x, a = 3, b = -5 and c = -7
= (2x)3 + (3-5-7)(2x)2 + [(3)(-5)+(-5)(-7)+(-7)(3)] (2x) + (3)(-5)(-7)
= 8x3 + (-9)(4x2) + (-15+35-21)(2x) + 105
= 8x3 - 36 x2 - 2x + 105
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