2018-I-12

$\begin{array}{rcl}
f(3) & = & 0 \\
4(3)[(3) + 1]^2 + a(3) + b & = & 0 \\
3a + b + 192 & = & 0 \ \ldots \unicode{x2460}
\end{array}$

Since the remainder is $2b + 165$ when $f(x)$ is divided by $x + 2$, by remainder theorem we have

$\begin{array}{rcl}
f(-2) & = & 2b + 165 \\
4(-2)[(-2) + 1]^2 + a(-2) + b & = & 2b + 165 \\
-2a + b – 8 & = & 2b + 165 \\
2a + b +173 & = & 0 \ \ldots \unicode{x2461}
\end{array}$

$\unicode{x2460} – \unicode{x2461}$, we have

$\begin{array}{rcl}
a +19 & = & 0 \\
a & = & -19 \ \ldots \unicode{x2462}
\end{array}$

Sub. $\unicode{x2462}$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
3(-19) + b + 192 & = & 0 \\
b & = & -135
\end{array}$

$\begin{array}{rcl}
f(x) & = & 0 \\
4x(x+1)^2 – 19x – 135 & = & 0 \\
4x^3 + 8x^2 + 4x – 19x – 135 & = & 0 \\
4x^3 + 8x^2 – 15x – 135 & = & 0 \\
(x – 3)(4x^2 + 20x + 45) & = & 0
\end{array}$

Therefore $x=3$ or $4x^2 + 20x + 45 = 0$.

Consider the discriminant of $4x^2 + 20x + 45 = 0$, we have

$\begin{array}{rcl}
\Delta & = & 20^2 – 4 \times 4 \times 45 \\
& = & -320 \\
& < & 0 \end{array}$

Therefore, equation $4x^2 + 20x + 45 = 0$ has not real roots.

Note that $3$ is not an irrational number. Hence, there is no irrational root for $f(x) = 0$.

I don’t agree with the claim.