Algebra Know-It-ALL

686 Worked-Out Solutions to Exercises: Chapters 21 to 29

Chapter 25

1. Here’s the binomial-cubed equation that we’ve been told to multiply out:

(a+b)^3 = 0

Let’s start by rewriting the equation as a product of three identical binomials:

(ax+b)(ax+b)(ax+b)= 0

We can multiply the second two factors and then simplify to get

(ax+b)(a^2 x^2 + 2 abx+b^2 )= 0

When we multiply these two factors and then simplify, we obtain

a^3 x^3 + 3 a^2 bx^2 + 3 ab^2 x+b^3 = 0

2. Here’s the binomial-cubed equation that we’ve been told to multiply out:

(31/2x− 12 1/2)^3 = 0

We can rewrite this as

(31/2x− 12 1/2)(31/2x− 12 1/2)(31/2x− 12 1/2)= 0

Let’s multiply out the second two factors. We must pay attention to the signs! The expres-

sion evolves as follows:

(31/2x− 12 1/2)(31/2x− 12 1/2)

(31/2)^2 x^2 − (31/2× 12 1/2)x− (121/2× 3 1/2)x+ (121/2)^2

3 x^2 − (3 × 12)1/2x− (12 × 3)1/2x+ 12

3 x^2 − 36 1/2x− 36 1/2x+ 12

3 x^2 − 6 x− 6 x+ 12

3 x^2 − 12 x+ 12

Now, for the cubic, we have

(31/2x− 12 1/2)(3x^2 − 12 x+ 12) = 0

When we multiply these factors, we get

(31/2× 3)x^3 − (31/2× 12)x^2 + (31/2× 12)x

− (121/2× 3)x^2 + (121/2× 12)x− 12 1/2× 12