factor completely, if polynomial is not factorable write prime A. b^2 -16b +16 B. 10c^2 -17c +3 C. p^2+2p-24 D. c^3 + 8
Hi. haven't you already posted the first of these four problems? I gave you feedback on A.
I know but my computer started acting up
could you help
b^2 -16b +16 has the coefficients a=1, b=-16 and c=16. Look up the quadratic formula. Substitute these values of a, b and c into that formula. This will give you the roots of b^2 -16b +16.
what quadratic formula
Cookie: the quadratic formula is really important in algebra. If you haven't already seen it, you will see it very soon. If y ou have seen it before, please review it.
i havent
b^2 -16b +16 can be factored, but the factors are not 'pretty.' I've used the quadratic formula to find the two 'roots' or 'solutions' of b^2 -16b +16 = 0.
Are you in an algebra class or are you studying math online?
class
I'd suggest you ask your teacher tomorrow whether or not you can use the quadratic formula to find the roots of b^2 -16b +16 = 0, and thus find the factors. b^2 -16b +16 = 0 is not easily factored without the q. formula.
I dont think we can
Could you help with the next one though
The quadratic formula looks like\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
okay
@mathmale
b^2 -16b +16 has the coefficients a=1, b=-16 and c=16.
Substitute these values into the quadratic formula: a=1, b=-16 and c=16.
Solve for x.
Example: Were you to obtain the two roots x and y, then the corresponding factors would be (b-x) and (b-y).
could we skip a for now
I'm sorry, Cookie, but I have to get off the computer. If you continue working on this problem, tag another moderator for help, or wait 'til I return later this evening.
This would be a good time for us to stop. My suggestion is that you post (b) as a new question. Someone will surely respond.
Forgive me, but I'm saying 'bye' for the present.
k
Join our real-time social learning platform and learn together with your friends!