How would I solve these? Please give as much detail as possible?

Question

Solve for x:

x^2-16=0

x^3+14x^2+40x=0

x^4-5x^2+4=0

Thank You!

Answer 1

x^2-16=0
(x+4)(x-4)=0
x+4=0
x=-4

x-4=0
x=-4

x^3+14x^2+40x=0
x(x^2-14x+40)=0
x(x-10)(x-4)=0
x=0

x-10=0
x=10

x-4=0
x=4

x^4-5x^2+4=0
(x^2-4)(x^2-1)=0
(x-2)(x+2)(x-1)(x+1)=0
xi2=0
x=2
x+2=0
x=-2
x-1=0
x=1
x+1=0
x=-1

Answer 2

x^2-16=0
(x-4)^2=0
x=4 (twice)

x^3+14x^2+40x=0
x(x^2+14x+40)=0
x(x+10)(x+4)=0
x=0, 10 or 4


x^4-5x^2+4=0
put y=x^2
y^2-5y+4=0
(y-1)(y-4)=0
y=1, or y=4
x^2=1 or x^2=4
x=+/- 1 or x=+/-2

Answer 3

Easy.
1. They are both perfect squares so it would be (x-4)(x+4)=0

2. This time you have a GCF of x. After you do that your equation is x^2+14x+40=0. Then your answer is (x+4)(x+10)=0 because 4+10 equals 14 and 4*10=40.

3. This one I don't know how to do.

Pleeease give me the ten points if this helped. Thank you.

Answer 4

x^2-16=0 factor

(x-4)(x+4)=0 set each equal to zero

x-4=0
x=4

x+4=0
x=-4

x^3+14x^2+40x=0 factor

x(x+4)(x+10)=0 set each equal to zero

x=0

x+4=0
x=-4

x+10=0
x=-10


x^4-5x^2+4=0 factor

(x-2)(x-1)(x+1)(x+2) set each equal to zero

x-2=0
x=2

x-1=0
x=1

x+1=0
x=-1

x+2=0
x=-2