Algebra Know-It-ALL

Adding x^2 to each side and then transposing the left and right sides gives us a cubic equa-

tion in polynomial standard form:

2 x^3 +x^2 = 0

It’s tempting to divide this equation through by x^2. But if x= 0 happens to be a root,

dividing through by x^2 will blind us to the existence of that root (and might cause other

problems, too). We can use a two-step trick to avoid that trouble. First, let’s check to see

ifx= 0 is a root by plugging it in and doing the arithmetic. We get

2 × 03 + 02 = 0

0 + 0 = 0

0 = 0

This cubic does have the root x= 0! Now that we’re aware of this fact, the second step in

our trick is to see if the equation has any other roots. Let’s impose a temporary restric-

tion on x: It can have any value except 0. That makes it “legal” to divide through by x^2 ,

obtaining the equation

(2x^3 +x^2 ) / x^2 = 0/x^2

We can rewrite this as

2 x^3 /x^2 +x^2 /x^2 = 0

which simplifies to

2 x+ 1 = 0

Subtracting 1 from each side and then dividing through by 2 tells us that x=−1/2. We

can now remove the temporary restriction on the value of x, making sure we include the

root x= 0. The cubic equation we got by morphing and mixing therefore has two roots:

x= 0 or x=−1/2

We can plug these into either of the original functions to get the y-values. Let’s use the

second one. When x= 0, we have

y= 2 x^3

= 2 × 03

= 2 × 0

= 0

Chapter 27 705