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Mathematics 10 Online
OpenStudy (worldwarkatyy):

What is the product in simplest form? State any restrictions on the variable. 2x-8/x^2-16 • x^2+5x+4/x^2+8x+16

OpenStudy (cruffo):

Wow... this is gonna be a mess... how are your factoring skills ??

OpenStudy (cruffo):

Just to check ... is the problem the same as \[ \frac{2x-8}{x^2-16} \cdot \frac{x^2+5x+4}{x^2+8x+16}\]

OpenStudy (worldwarkatyy):

@cruffo yea that's the problem

OpenStudy (cruffo):

ok. I would suggest factoring all numerators and denominators first, in case we can simplify, and to check for those "restrictions" the question asked about.

OpenStudy (worldwarkatyy):

Can u explain or show me how to factor the numerators and denominators? Idk how to do that part.

OpenStudy (aveline):

\[\frac{ 2(x-4) }{ (x-4)(x+4) } \times \frac{ (x+4)(x+1) }{ (x+4)(x+4) }\]

jhonyy9 (jhonyy9):

use formula difference of two squares

OpenStudy (aveline):

Can you simplify it?

OpenStudy (cruffo):

What Aveline's answer is showing is the factored form. She had to use a number of techniques to do that. Do any of those factorization **not ** make sense?

OpenStudy (worldwarkatyy):

That makes sense, what do I do after its in factored form?

jhonyy9 (jhonyy9):

what mean restrictions in this case ? how you think ?

OpenStudy (worldwarkatyy):

Would the answer be 2=x+1?

OpenStudy (cruffo):

The "restrictions" that they are talking about are numbers that make the denominator in either fraction 0. Remember : #/0 is undefined.

OpenStudy (cruffo):

No, the goal is not to "solve", but to multiply.

jhonyy9 (jhonyy9):

yes cruffo you are right

OpenStudy (worldwarkatyy):

Ah okay sorry I'm so confused

jhonyy9 (jhonyy9):

the denominator zero make the fractions undefined

OpenStudy (cruffo):

For the point of Aveline's answer, what numbers make the denominator 0? Hint, there are two such numbers...

OpenStudy (aveline):

On the left side, cross off the two (x-4)'s on the top and bottom. They equal 1 anyway so they cancel out. On the right side, do the same thing with the two (x+4)'s: \[\frac{ 2 }{ (x+4) } \times \frac{ (x+1) }{ (x+4) }\]

OpenStudy (cruffo):

This is too busy...

OpenStudy (worldwarkatyy):

Okay so... 2 and 1 would be the restrictions?

OpenStudy (worldwarkatyy):

@cruffo am I right?

OpenStudy (aveline):

Well the equation simplifies to \[\frac{ (2)(x+1) }{ (x+4)(x+4) }\] If we plug in -4 for x, the bottom of the equation is equal to 0. Therefore, the restriction is x=-4 Take a look for yourself. X can not equal -4 https://www.desmos.com/calculator/ofx8rjslpa

OpenStudy (worldwarkatyy):

Okay I got it, thank you guys so much !! Aveline could you please help me with one more problem?

OpenStudy (worldwarkatyy):

@aveline please help !! Solve the following equation and check the solution. 4/x-2 = x-1/x-2

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