# Determine f(x) if f(3x+2) = 27x^3 + 30x + 9?

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f(3x+2) = 27x^3 + 30 x + 9

Let y = 3x + 2

==> x = (y-2)/3

Now substitute:

f(y) = 27[(y-2)/3]^3 + 30 [(y-2)/3] + 9

f(y) = 27(y-2)^3 / 27 + 10(y-2) + 9

= (y-2)^3 + 10(y-2) + 9

Let us open brackets:

f(y) = y^3 -6y^2 + 12y -8 + 10y -20 + 9

= y^3 -6y^2 + 22y -19

==> **f(x) = x^3 - 6x^2 + 22x -19**

We'll make the substitution 3x+2 = t.

From this relation, we'll isolate x to the left side:

3x = t-2

We'll divide by 3 both sides:

x = (t-2)/3

So, f(x) = f((t-2)/3)

f((t-2)/3) = 27*[(t-2)/3]^2 + 30*(t-2)/3 + 9

f((t-2)/3) = 3*(t-2)^2 + 10*(t-2) + 9

We'll factorize:

f((t-2)/3) = (t-2)*[3(t-2) + 10] + 9

f((t-2)/3) = (t-2)*(3t+4) + 9

We'll remove the brackets:

f((t-2)/3) = 3t^2 - 2t - 8 + 9

f((t-2)/3) = 3t^2 - 2t + 1

Because the notation is arbitrary, we'll consider:

**f(x) = 3x^2 - 2x + 1**

f(3x+2) = 27x^3+30x+9

To determine f(x).

Solution:

3x+2|27x^3+30x+9(9x^2 - 6x +14

27x^2+18x^2

------------------------------

-18x^2+ 30x

-18x^2 -12x

--------------------------

42x +9

42x+28

---------------------------

**-19 .............................R1.**

Again devide 9x^2-14x+14 by 3x+2.

3x+2|9x^2-6x+14( 3x -4.

9x^2+6x

------------------------

-12x+14

-12x-8

----------------------------

** 22...........................R2**

Again divide the quotient by 3x-4 by 3x+2:

3x-2|3x-4( **1 = Rsay.**

3x+2

---------------------

-**6 --------------R3**

Therefore

f(3x+2) = Rt^3+R1*t^2+R2 * t+R1, where t =3x+2 and R1 ,R2,R3 are the successive remainders.

So f(3x+2) = (3x+2)^2 -6(3x+2)^2 + 22(3x+2) -19.