help me simplify this expression?

Question

I just can't get it.
The expression is (y-1)(y^+2y+1)
Thanks, and please eplain in detail.

and slove this one?
5[1-2(x+2)]=4x

I already missed both of them, I just can't figure out how to correct them.

thanks again.

Thank you all!
I understand both of them now.
Thanks again. =)

Answer 1

distribute each member of the binomial through then collect like terms
y^3+2y²+y-y²-2y-1 = y^3+y²-y-1

5[1-(2x+2)]=4
5(1-2x-2)=
5(-2x-1)
-10x-5=4
-10x=9
x=-9/10

Answer 2

Factor:
= (y - 1) (y^2 + 2y + 1)
= (y - 1) (y + 1)^2

Simplify:
5([1 - 2] [x + 2]) = 4x
5(- 1[x + 2]) = 4x
5(- x - 2) = 4x
- 5x - 10 = 4x
- 9x = 10
x = - 10/9

Answer: - 10/9 or - 1 1/9

Proof:
5([1 - 2] [- 10/9 + 2]) = 4(- 10/9)
5(- 1[8/9]) = - 40/9
5 * - 8/9 = - 40/9
- 40/9 = - 40/9

Answer 3

The expression is (y-1)(y^2+2y+1)<-- I think you meant this.
= (y-1)/(y+1)^2

5(1-2(x+2)=4x
5(1-2x-2) = 4x
5 -10x -10 = 4x
-5 = 14x
x = -14/5

Answer 4

ok, to solve this type of equation (polinomial and of order two), you have the following formula:
x=(-b±sqrt((-b^2) -4*a*c))/2*a
identifying: a*x^2+b*x+c=0
this serves you to factorize the equation because you're going to obtain two solutions x1 and x2, and you can rewrite the euqation this way:
a*x^2+b*x+c = (x-x1)*(x-x2)
if any solution is negative, you have to change the sign in the factorization of the solution in case.
for your expression and employing the formula:
y^2+2y+1=0
y= -2±sqrt(2^2-4*1*1) / 2*1 = -2±0 / 2 = -1
in this case, when the result of what you have inside the square root is zero, you have a double square and that is the solution twice, or the solution powered to two:
(x+1)*(x+1)=(x+1)^2
if you obtain a negative number inside the square root, we will
be talking about complex numbers, but this is another issue.

Answer 5

Looks like it's factored from a difference of cubes... I would personally leave it how it is but if you want to factor that quadratic part..

(y-1)(y+1)^2


5[1-2(x+2)]=4x
5(1-2x-4)-4x=0
5-10x-20-4x=0
-14x=15
x=-15/14

Answer 6

y^2 +2y +1 = (y + 1)^2
so (y - 1)(y^2 +2y + 1) = (y - 1)(y + 1)(y + 1)