How do you find $g (- 4 n)$ $g(-4n)$ given $g (n) = n^{3} - 5 n^{2}$ $g(n)=n^3-5n^2$ ?

Question

How do you find $g (- 4 n)$ $g(-4n)$ given $g (n) = n^{3} - 5 n^{2}$ $g(n)=n^3-5n^2$ ?

1 Answer

Impact of this question

10550 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-24T11:15:32

$- (64 n^{3} + 80 n^{2})$ $-(64n^3+80n^2)$

Explanation:

$g (n) = n^{3} - 5 n^{2}; g (- 4 n) = {(- 4 n)}^{3} - 5 \cdot {(- 4 n)}^{2} = - 64 n^{3} - 5 \cdot 16 n^{2} = - 64 n^{3} - 80 n^{2} = - (64 n^{3} + 80 n^{2})$ $g(n) = n^3-5n^2 ; g(-4n) = (-4n)^3 -5*(-4n)^2 = -64n^3- 5*16n^2= -64n^3-80n^2 = -(64n^3+80n^2)$ [Ans]

Science

Math

Humanities

... and beyond

How do you find $g (- 4 n)$ $g(-4n)$ given $g (n) = n^{3} - 5 n^{2}$ $g(n)=n^3-5n^2$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find g(−4n)g(-4n) given g(n)=n3−5n2g(n)=n^3-5n^2?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find $g (- 4 n)$ $g(-4n)$ given $g (n) = n^{3} - 5 n^{2}$ $g(n)=n^3-5n^2$ ?