Help with Polynomials.

(t-5)^2 = 2(5-t)

I didnt get very far just that

t^2 - 25 = 10 - 2t
-10

t^2 - 35 = -2t

Im not sure where to go from there, or what exactly im supposed to solve for, the problem just said to solve.
But, I think it wants two solutions, the lesson is factoring polynomials.

NO: (t - 5)^2 = t^2 - 10t + 25

Are you learning on your own, or in school?

colormekaylee said:

(t-5)^2 = 2(5-t)


I didnt get very far just that

t^2 - 25 = 10 - 2t
-10

t^2 - 35 = -2t

Im not sure where to go from there, or what exactly im supposed to solve for, the problem just said to solve.
But, I think it wants two solutions, the lesson is factoring polynomials.

Click to expand...

Solve for 't' - means you need to find the value of 't' for which the above equation is true.

There are many ways to skin this cat - since the lesson is in polynomials - I think they want you to follow following procedure.

(t - 5)[sup:3l3g0hlv]2[/sup:3l3g0hlv] = 2(5 - t)

(t - 5)[sup:3l3g0hlv]2[/sup:3l3g0hlv] - 2(5 - t) = 0

(t - 5)[sup:3l3g0hlv]2[/sup:3l3g0hlv] + 2(t - 5) = 0

(t-5)(t-5+2) = 0

(t-5)(t-3) = 0

If product of two "things" equals to zero - then one or both of those two things are individually equal to 0.

then

t - 5 = 0

and/or

t - 3 = 0

Now solve for "t".

Thank you for the help, to the first post's question, I am in school, but I am learning on my own. I don't have lessons or anything, I'm doing my school online, so its different than traditional school.

colormekaylee said:

Thank you for the help, to the first post's question, I am in school, but I am learning on my own. I don't have lessons or anything, I'm doing my school online, so its different than traditional school.

Click to expand...

Thanks. Makes it easier for us if we know this...