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January 31, 2015

January 31, 2015

Posted by **Yvonne** on Wednesday, April 24, 2013 at 11:49pm.

- Algebra -
**Reiny**, Thursday, April 25, 2013 at 8:23amfactor each part

here is the hardest part

x^3 + 2x^2 - 4x - 8 , using grouping

= x^2(x+2) - 4(x+2)

= (x^2 - 4)(x+2)

= (x+2)(x-2)(x+2) or (x-2)(x+2)^2

so...

(x^3+2x^2-4x-8)/(x^4-16)x(3x^2+8x+5)/3x^2+11x+10)

= (x-2)(x+2)^2 /((x-2)(x+2)(x^2 + 4)) * (x+1)(3x+5)/((3x+5)(x+2))

=**(x+1)/(x^2 + 4)**, x ≠ ±2, -5/3

check by picking any value of x other than the restricted values

e.g. let x = 1

in original:

= (-9)/(-15)*(16/24) = 2/5

value of (x+1)/(x^2+4) = 2/5

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