Answer
False
Work Step by Step
The trinomial $4x^2+8x+3$ has the discriminant $8^2-4(4)(3)=16$, so its roots are:
$$\begin{align*}
x&=\dfrac{-8\pm\sqrt{16}}{2(4)}=\dfrac{-8\pm 4}{8}\\
x_1&=-\dfrac{12}{8}=-\dfrac{3}{2}\\
x_2&=-\dfrac{4}{8}=-\dfrac{1}{2}.
\end{align*}$$
We factor the trinomial:
$$4x^2+8x+3=4\left(x+\dfrac{3}{2}\right)\left(x+\dfrac{1}{2}\right)=(2x+3)(2x+1).$$
The trinomial $4x^2+8x+1$ has the discriminant $8^2-4(4)(1)=48$, so its roots are:
$$\begin{align*}
x&=\dfrac{-8\pm\sqrt{48}}{2(4)}=\dfrac{-8\pm 4\sqrt 3}{8}=\dfrac{-2\pm \sqrt 3}{2}\\
x_1&=\dfrac{-2-\sqrt 3}{2}\\
x_2&=\dfrac{-2+\sqrt 3}{2}.
\end{align*}$$
We factor the trinomial:
$$4x^2+8x+1=4\left(x+\dfrac{2+\sqrt 3}{2}\right)\left(x+\dfrac{2-\sqrt 3}{2}\right)=(2x+2+\sqrt 3)(2x+2-\sqrt 3).$$
Therefore the given statement is FALSE.
