One of the factors of 3x^3+36x^2+105x is x+b where b is a positive constant. What is the smallest possible value of b? square

To find the smallest possible value of \( b \) for which \( x + b \) is a factor of \( 3x^3 + 36x^2 + 105x \), we need to use the Factor Theorem. According to the Factor Theorem, if \( x + b \) is a factor of the polynomial \( P(x) \), then \( P(-b) = 0 \).<br /><br />Given \( P(x) = 3x^3 + 36x^2 + 105x \), we need to find \( b \) such that \( P(-b) = 0 \).<br /><br />First, let's factor out the common term \( x \) from the polynomial:<br />\[ P(x) = 3x^3 + 36x^2 + 105x = 3x(x^2 + 12x + 35) \]<br /><br />Now, we need to find the roots of the quadratic equation \( x^2 + 12x + 35 = 0 \). We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 12 \), and \( c = 35 \).<br /><br />Calculate the discriminant:<br />\[ \Delta = b^2 - 4ac = 12^2 - 4 \cdot 1 \cdot 35 = 144 - 140 = 4 \]<br /><br />Now, find the roots:<br />\[ x = \frac{-12 \pm \sqrt{4}}{2 \cdot 1} = \frac{-12 \pm 2}{2} \]<br /><br />This gives us two solutions:<br />\[ x = \frac{-12 + 2}{2} = \frac{-10}{2} = -5 \]<br />\[ x = \frac{-12 - 2}{2} = \frac{-14}{2} = -7 \]<br /><br />Since \( b \) must be positive, we take the negative of these roots:<br />\[ b = 5 \quad \text{or} \quad b = 7 \]<br /><br />The smallest possible value of \( b \) is:<br />\[ \boxed{5} \]