# [SOLVED]Find F(x)

#### dwsmith

##### Well-known member
Find $F(x)$

$G(x) = 3 + x + x^2\qquad\qquad G[F(x)] = x^2 - 3x + 5$

I don't see what the composition should be.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
I don't see what the composition should be.
The composition is given.

Try to search for F(x) in the form of a polynomial.

#### soroban

##### Well-known member
Hello, dwsmith!

$G(x) = x^2 + x + 3\qquad G[F(x)] = x^2 - 3x + 5$

$\text{Find }F(x).$

Let $F(x) \:=\:ax+b.$

Then: .$G\left(F(x)\right) \;=\;G[ax+b]$

. . . . . . . . . . . . $$=\; (ax+b)^2 + (ax+b) + 3$$

. . . . . . . . . . . . $$=\;a^2x^2 + 2abx + b^2 + ax + b + 3$$

. . . . . . . . . . . . $$=\;a^2x^2 + a(2b+1)x + (b^2+b+3)$$

We are told that: $$G\left(F(x)\right) \:=\:x^2-3x+5$$

Hence: .$$a^2x^2 + a(2b+1)x + (b^2+b+3) \;=\;x^2 -3x+5$$

Equate coefficients: .$$\begin{Bmatrix}a^2 \:=\:1 \\ a(2b+1) \:=\:-3 \\ b^2-3b+3\:=\:5 \end{Bmatrix}$$

Solve the system: .$$\begin{Bmatrix}a &=& 1\\b &=& \text{-}2 \end{Bmatrix}\;\text{ or }\;\begin{Bmatrix}a&=&\text{-}1 \\ b&=& 1\end{Bmatrix}$$

Therefore: .$$F(x) \;=\;\begin{Bmatrix}x - 2 \\ \text{or} \\ -x + 1 \end{Bmatrix}$$