HOW TO FIND THE PRODUCT OF THREE BINOMIALS

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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