Cubic Polynomial Roots 4
Let $ f(x) = x^3 + 3x^2 + 1 $. There exist real numbers $ a \neq 0 $ and $ b, $ such that
\[f(x) - f(a) = (x - a)^2 (x - b).\]Enter the ordered pair $ (a,b) $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
By the remainder theorem, $ f(x) - f(a) $ is divisible by $ x - a, $ so we can take out a factor of $ x - a $ accordingly:
\begin{align*}
f(x) - f(a) &= (x^3 + 3x^2 + 1) - (a^3 + 3a^2 + 1) \\
&= (x^3 - a^3) + 3(x^2 - a^2) \\
&= (x - a)(x^2 + ax + a^2) + 3(x - a)(x + a) \\
&= (x - a)(x^2 + ax + a^2 + 3x + 3a) \\
&= (x - a)(x^2 + (a + 3) x + a^2 + 3a).\end{align*}Thus, we want
\[x^2 + (a + 3) x + a^2 + 3a = (x - a)(x - b) = x^2 - (a + b) x + ab.\]Matching coefficients, we get
\begin{align*}
a + 3 &= -a - b, \\
a^2 + 3a &= ab.\end{align*}Since $ a \neq 0, $ we can divide both sides of the second equation by $ a, $ to get $ a + 3 = b $. Then $ -a - b = b, $ so $ a = -2b $. Then
\[-2b + 3 = 2b - b,\]which gives us $ b = 1 $. Then $ a = -2, $ so $ (a,b) = \boxed{(-2,1)} $.
Freeflows