Factor completely c^2 + 6c + 9

In it's current state, this trinomial is not factorable using the GCF.

Knowing the squares of binomials rule, we could apply this here, but I'd rather try splitting the middle term.

Squares of binomials rule:

(a + b)^2 = (a^2 + ab + c^2)

Knowing this rule, we can automatically surmise that our factored form is

(c + 3)^2, but let's also factor by splitting the middle term.

Check factors of 9

3 * 3

These digits satisfy the criteria.

Split the middle term.

c^2 + 3c + 3c + 9

(Because the leading coefficient is 1, we know the factors are just going to be (c + 3)(c + 3) or (c + 3)^2)

Factor by grouping.

c(c + 3) + 3(c + 3)

(c + 3)(c + 3)

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!