Question 1209678: Find an ordered pair of constants (a,b) such that the polynomial f(x)=x^3+ax^2+(b+2)x+1 is divisible by x^2-3.
Enter your answer as an ordered pair in the format $(a,b)$, where $a$ and $b$ are replaced by appropriate numbers.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Since $f(x)$ is divisible by $x^2-3$, the roots of $x^2-3=0$ must also be roots of $f(x)$. The roots of $x^2-3=0$ are $x=\sqrt{3}$ and $x=-\sqrt{3}$.
Therefore, $f(\sqrt{3})=0$ and $f(-\sqrt{3})=0$.
$f(\sqrt{3}) = (\sqrt{3})^3 + a(\sqrt{3})^2 + (b+2)\sqrt{3} + 1 = 0$
$3\sqrt{3} + 3a + (b+2)\sqrt{3} + 1 = 0$
$(3+b+2)\sqrt{3} + 3a + 1 = 0$
$(b+5)\sqrt{3} + 3a + 1 = 0$
$f(-\sqrt{3}) = (-\sqrt{3})^3 + a(-\sqrt{3})^2 + (b+2)(-\sqrt{3}) + 1 = 0$
$-3\sqrt{3} + 3a - (b+2)\sqrt{3} + 1 = 0$
$-(b+5)\sqrt{3} + 3a + 1 = 0$
We have a system of two equations:
1) $(b+5)\sqrt{3} + 3a + 1 = 0$
2) $-(b+5)\sqrt{3} + 3a + 1 = 0$
Adding the two equations gives:
$2(3a+1) = 0$
$3a+1 = 0$
$3a = -1$
$a = -\frac{1}{3}$
Substituting $a=-\frac{1}{3}$ into the first equation:
$(b+5)\sqrt{3} + 3(-\frac{1}{3}) + 1 = 0$
$(b+5)\sqrt{3} - 1 + 1 = 0$
$(b+5)\sqrt{3} = 0$
$b+5 = 0$
$b = -5$
Thus, the ordered pair of constants is $(-\frac{1}{3}, -5)$.
Final Answer: The final answer is $\boxed{\left(-\frac{1}{3}, -5\right)}$
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