how do u solve (2a-4b+c)(a3-b2+5c4)?

Answer 1

You need the same number of equations as you have variables to solve any set of linear algebraic equations. In your case you have three variables, but only one relationship...which is not even an equation. You need three more equations, such as initial conditions, to completely solve for a, b, and c.

Or...was your question, 'How do I expand the relation...'?

Answer 2

FOIL method. First outside inside last. Basically start by multiplying the 2a by the a3 then b2 then 5c4, then do the same with the 4b... when finished add up all similar terms. Don't forget a negative times anegative is a positive.

Answer 3

You can't "solve" that expression, it is not an equation (no equal sign), but you can multiply it out.

(2a-4b+c)(a^3-b^2+5c^4) = 2a*a^3 - 2a*b^2 + 2a*5c^4 -4b*a^3 +4b*b^2 -4b*5c^4 + c*a^3 -c*b^2 +c*5c^4
=2a^4-2a(b^2)+10a(c^4)
-4(a^3)b+4b^3-20b(c^4)
+(a^3)c-(b^2)c+5c^5

Answer 4

use the distributive method. Multiply 2a by everything in the second () and then mulitply -4b by everything in the (), and so on

Answer 5

No. I purely ask your self, nonetheless, if, like kangaroos and Pandas they are able to hold an embryo for a protracted time till now they permit it develop. it would be exciting to be certain how long they have had that distinctive anteater in that zoo. purely a thought.

Answer 6

what are you looking for?
(2a-4b+c)(a3-b2+5c4) = (2a-4b+c)(a3-b2+20c)
= 2a(a3) + 2a(-2b) + 2a(20c) - 4b(a3) - 4b(-2b) - 4b(20c) + c(a3) + c(-2b) + c(20c) = 6a(squared) + 8b(squared) + 20c(squared) - 16ab + 23ac - 82bc.

Answer 7

combine like terms i think. im not sure. sorry =(