Math questions part2?

Question

x^4 + 2x^3 - 8x -16 = 0

36t^4 + 29t^2 - 7 =0

Answer 1

1. x^4 + 2x^3 - 8x - 16 = 0

Solve by grouping the first two terms and the second two terms

x^3(x + 2) - 8(x + 2) = 0

Since the x + 2 is in both terms, we can rearrange as

(x^3 - 8)(x + 2) = 0

Then you can factor the cubic formula, x^3 - 8, but I'm sorry I don't remember the formula.  You can look up in your Algebra book or google it.

2. 36t^4 + 29t^2 - 7 = 0

Factor

(36t^2 - 7)(t^2 + 1) = 0

36t^2 - 7 = 0 or t^2 = -1

t^2 = 7/36 or t = +-i

t = +-sqrt(7/36) or t = +-i

t = +-sqrt(7) / 6 or +-i

Hope this helps you!  Please email anytime for further questions.

Answer 2

x^4 + 2x^3 - 8x -16 = 0

x^3(x+2) -8(x+2)=0

(x+2)(x^3 -8)=0

by the zero property

x+2=0or x^3 -8 =0

x=-2 or x^3 = 8

so x=-2 or x=+2

36t^4 + 29t^2 - 7 =0

(36t^2 -7)(t^2+1)=0

by the zero property

either36t^2 -7 =0 or t^2+1=0

t^2 = 7/36  or t^2 = -1

t = +-(√7)/6 ot t = √ -1 which is +-i

4 solutions