cubed polynomial equation: 16a = a[cubed]

[Thomas]

My math lesson in the chapter 'Factoring' had this problem that I can't find any help with in the lesson or extra problems, including the notes.

16a = a[cubed] I didn't know how else to type that)

Any help in knowing how to do problems like this would be so appreciated.

Thanks

Thomas

 

[stapel]

My math lesson in the chapter 'Factoring' had this problem...

16a = a[cubed] I didn't know how else to type that)

To learn how to format, try following either of the formatting-article links provided in the "Read Before Posting" thread that you read before posting. :wink:

I will guess that the instructions were something along the lines of "solve the equation", or maybe "solve by factoring".

The customary first step will be to gather the terms together on one side, so subtract the 16a over to the left-hand side.

Then take the common factor out front. This will leave you with a difference of squares, which you can factor.

Then set all three factors equal to zero, and solve the three linear equations.

If you get stuck, please reply showing how far you have gotten in the above steps. Thank you!

 

[TomF]

16a = a^3

Since there is an "a" in all terms, my first step would be to divide through by "a", so that the equation becomes

16 = a^2

Then it is simply a matter of taking the square root of both sides.

i.e.
a = 4 or -4

 

[Mrspi]

Tom, you don't know what the value of "a" is....


SO, dividing both sides of the equation

16a[sup:14hawj76]3[/sup:14hawj76] = 4a

by "a" is a VERY RISKY proposition. What if a = 0?? Is it OK to divide by 0? I think not.

Here's a much better approach.

You've got

16a[sup:14hawj76]3[/sup:14hawj76] = 4a

Subtract 4a from both sides of the equation:

16a[sup:14hawj76]3[/sup:14hawj76] - 4a = 4a - 4a

16a[sup:14hawj76]3[/sup:14hawj76] - 4a = 0

Now, factor the expression on the left side. Start by removing a common factor of 4a:

4a (4a[sup:14hawj76]2[/sup:14hawj76] - 1) = 0

Now look at the expression inside the parentheses. Do you see that 4a[sup:14hawj76]2[/sup:14hawj76] - 1 is a difference of two squares? 4a[sup:14hawj76]2[/sup:14hawj76] is (2a)[sup:14hawj76]2[/sup:14hawj76] - 1[sup:14hawj76]2[/sup:14hawj76]

So, (2a)[sup:14hawj76]2[/sup:14hawj76] - 1[sup:14hawj76]2[/sup:14hawj76] factors as (2a + 1)(2a - 1)

And your factored expression is

4a (2a + 1)(2a - 1) = 0

If the product of two or more factors is 0, then at least ONE of those factors must have a value of 0. So, if
4a (2a + 1) (2a - 1) = 0
then
4a = 0
OR
2a + 1 = 0
OR
2a - 1 = 0

Solve each of those simple equations. You'll end up with THREE solutions.

 

[Denis]

Here's a much better approach.
You've got
16a[sup:wyuszqke]3[/sup:wyuszqke] = 4a

That should be 16a = a^3

 

[Mrspi]

Here's a much better approach.
You've got
16a[sup:ky4l8otp]3[/sup:ky4l8otp] = 4a

That should be 16a = a^3

Aw gee....I can't believe I did that! Oh well, the method is the same, at any rate. And I guess my point (that dividing through by a causes a solution to be "lost") is still valid.

 

[Denis]

16a = a^3
Dunno, Mrs pi.
I look at these this way:
0 is obviously one solution.
For solutions where a<>0:
divide by a: 16 = a^2 ; as per Fern

 

[Deleted member 4993]

But the problem was not "solve" - it was "factor"

My math lesson in the chapter 'Factoring' had this problem that I can't

 

[Denis]

Probably SOLVE by factoring Subhotosh; if factoring only, we'd have no equation:
a^3 - 16a

No?

 

[TomF]

What if a = 0??

Oops. I forgot to explicitly state that: 0 is the third solution.

So . . .
The three solutions are -4, 0, and 4.

Before we solve the equation, we may not know if "a" is 0 or not, but because we are doing the division algebraically, it is okay. If we were doing a divide by 0 in a computer program, we would cause an overflow error, and our program might crash.