could someone help me with math?

Question

Factor the following.

 

16x^4-81

 

i got this wrong and i want to learn from my mistake. how to do this?

Answer 1

See --  https://www.wyzant.com/resources/answers/193173/could_someone_help_me_with_this_math_problem

Answer 2

 

16x⁴  - 81 = 2⁴ x⁴ - 3⁴ = (2x)⁴ - 3⁴ = [(2x)² - 3² ][(2x)² + 3² ] = (2x - 3)(2x + 3)(4x² + 9)

Answer 3

Hi Candice.

 

This one is pretty tricky! What you actually have is a perfect square binomial of the form:

 

a² - b²

 

which factors to:

 

(a + b)(a - b)

 

Normally you see things like

 

x² - 4

 

so that

 

a = x, b = 2

 

or

 

x² - 49

 

so that

 

a = x, b = 7

 

But in this one you have something a lot more complicated:

 

16x⁴ - 81

 

If a² = 16x⁴, that means that a = √(16x⁴) = 4x²

 

If b² = 81, that means that b = √(81) = 9

 

So what you have here is:

 

(4x²)² - 9²

 

You can factor this just like you factor any other perfect square binomial.

 

(4x² + 9)(4x² - 9)

 

Unfortunately, you're not quite done yet. Your second term, (4x² - 9), is also able to be factored, since it again takes the form of

 

a² - b²

 

Except this time, following the same procedure above:

 

a = 2x

b = 3

 

So your final answer will be:

 

(4x² + 9)(2x + 3)(2x - 3)

 

This is fully factored and can't be pulled apart any more.

 

Hope this helps!

Answer 4

Just think of it the same as a quadratic since the exponent is even. This would be a difference of squares equation and would factor into (4x^2-9)*(4x^2+9). This can be further factored since you have 2 more difference of squares equations. The first factors into (2x-3)(2x+3) and the second into (2x-3i)(2x+3i). Therefore the final factored equation is (2x-3)(2x+3)(2x-3i)(2x+3i).