If f(x)=2x+1 and g(f(x))=4x^2+4x+3 find g(x) given that g(x)=ax^2+bx+c how do I do that?

I know that somebody has solved this before but I would like a better version of that one since I did not understand what was going on in the other one. Many thanks!

Aug 27, 2017

$g \left(x\right) = {x}^{2} + 2$

Explanation:

We know that $f \left(x\right) = 2 x + 1$ and $g \left(x\right) = a {x}^{2} + b x + c$.

$R i g h t a r r o w g \left(f \left(x\right)\right) = a {\left(2 x + 1\right)}^{2} + b \left(2 x + 1\right) + c$

Expanding the parentheses:

$R i g h t a r r o w g \left(f \left(x\right)\right) = a \left(4 {x}^{2} + 4 x + 1\right) + 2 b x + b + c$

$R i g h t a r r o w g \left(f \left(x\right)\right) = 4 a {x}^{2} + 4 a x + a + 2 b x + b + c$

Rearranging the variables:

$R i g h t a r r o w g \left(f \left(x\right)\right) = 4 a {x}^{2} + \left(4 a + 2 b\right) x + a + b + c$

Now, we are also given the fact that $g \left(f \left(x\right)\right) = 4 {x}^{2} + 4 x + 3$.

Comparing coefficients:

$R i g h t a r r o w 4 a = 4 \therefore a = 1$

$\mathmr{and}$

$R i g h t a r r o w 4 a + 2 b = 4 R i g h t a r r o w 4 \times \left(1\right) + 2 b = 4 R i g h t a r r o w 4 + 2 b = 4 R i g h t a r r o w 2 b = 0 \therefore b = 0$

$\mathmr{and}$

$R i g h t a r r o w a + b + c = 3 R i g h t a r r o w \left(1\right) + \left(0\right) + c = 3 R i g h t a r r o w 1 + c = 3 \therefore c = 2$

Therefore, $g \left(x\right) = \left(1\right) {x}^{2} + \left(0\right) x + \left(2\right) = {x}^{2} + 2$.