Knowing that p(n)= 2n^4+4n^3 -6n^2 what r the n roots

by the distributive property of all reals, we have that

p(n) = 2n^2(n^2 + 2n - 3)

by virtue of factoring, it is true that p(n) = 2n^2(n+3)(n-1).

should we want p(n) = 0, we know that at least one of 2, n^2, n+3, or n-1 needs to be zero by the zero factor property. 0 times anything is zero.

so the answer is n = 0, n = -3, and n = 1 as the roots because those are the values of n for which p(n) is 0.

it is true

step-by-step explanation:

all the combinations add up i used   calculator

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.