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 :

(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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