One of the factors of 2x^3+42x^2+208x is x+b, where b is a positive constant. What is the smallest possible value of b ?

To find the smallest possible value of \( b \) such that \( x + b \) is a factor of \( 2x^3 + 42x^2 + 208x \), we can use the Factor Theorem. The Factor Theorem states that if \( x + b \) is a factor of a polynomial \( P(x) \), then \( P(-b) = 0 \).<br /><br />Let \( P(x) = 2x^3 + 42x^2 + 208x \). We need to find the smallest positive value of \( b \) such that \( P(-b) = 0 \).<br /><br />First, substitute \( -b \) into \( P(x) \):<br /><br />\[ P(-b) = 2(-b)^3 + 42(-b)^2 + 208(-b) \]<br /><br />Simplify the expression:<br /><br />\[ P(-b) = 2(-b^3) + 42(b^2) - 208b \]<br />\[ P(-b) = -2b^3 + 42b^2 - 208b \]<br /><br />Set this equal to zero and solve for \( b \):<br /><br />\[ -2b^3 + 42b^2 - 208b = 0 \]<br /><br />Factor out \( -2b \):<br /><br />\[ -2b(b^2 - 21b + 104) = 0 \]<br /><br />This gives us two factors:<br /><br />\[ -2b = 0 \quad \text{or} \quad b^2 - 21b + 104 = 0 \]<br /><br />The first factor \( -2b = 0 \) gives \( b = 0 \), which is not positive. We need to solve the quadratic equation \( b^2 - 21b + 104 = 0 \).<br /><br />Use the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), where \( A = 1 \), \( B = -21 \), and \( C = 104 \):<br /><br />\[ b = \frac{21 \pm \sqrt{(-21)^2 - 4 \cdot 1 \cdot 104}}{2 \cdot 1} \]<br />\[ b = \frac{21 \pm \sqrt{441 - 416}}{2} \]<br />\[ b = \frac{21 \pm \sqrt{25}}{2} \]<br />\[ b = \frac{21 \pm 5}{2} \]<br /><br />This gives us two solutions:<br /><br />\[ b = \frac{21 + 5}{2} = 13 \]<br />\[ b = \frac{21 - 5}{2} = 8 \]<br /><br />Since we are looking for the smallest positive value of \( b \), the smallest possible value is:<br /><br />\[ b = 8 \]<br /><br />Therefore, the smallest possible value of \( b \) is \( \boxed{8} \).