Algebra Know-It-ALL

516 Review Questions and Answers

Answer 25-2

An equation in binomial-cubed form can always be written like this when x is the variable:

(ax+b)^3 = 0

where a is a nonzero real number, and b is a real number.

Question 25-3

What does the binomial-cubed equation from Answer 25-2 look like when multiplied out

into polynomial standard form?

Answer 25-3

Let’s begin by separating the left side of the equation into a product of three identical binomials:

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

When we multiply the second two binomials together and then simplify the result into a

trinomial, we get the equation

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

Multiplying the binomial by the trinomial and simplifying gives us

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

Question 25-4

How many real roots does the equation stated in Answer 25-2 have? What is that root, or what

are they? What is the real solution set X?

Answer 25-4

There is one real root with multiplicity 3. It is the solution to the equation we obtain when

we set the binomial equal to 0:

ax+b= 0

When we subtract b from both sides and then divide through by a (which is okay because we

know that a≠ 0), we get

x=−b/a

The real solution set is therefore

X= {−b/a}