+47 The expression $x^2 + 13x + 30$ can be written as $(x + a)(x + b),$ and the expression $x^2 + 5x - 50$ written as $(x + b)(x - c)$, where $a$, $b$, and $c$ are integers. What is the value of $a + b + c$?
+526 \(x^2+13x+30 = x^2+3x+10x+30\)
\(=x(x+3)+10(x+3)\)
\(=(x+3)(x+10)\)
⇒ a = 3 and b = 10
\(x^2+5x-50 = x^2+10x-5x-50\)
\(=x(x+10)- 5(x+10)\)
\(=(x+10)(x-5)\)
⇒ c = 5
\(a+b+c = 3+10+5 = 18\)
+874 @amygdaleon305, nice solution! Next time, consider using the \begin{align*} environment.
Your solution in that environment(you can view the code by double-clicking the TeX and clicking view TeX commands):
\(\begin{align*} x^2+13x+30 &= x^2+3x+10x+30 \\ &= x(x+3) + 10(x+3) \\ &= (x+3)(x+10) \end{align*}\)
$\Rightarrow$ $a = 3$ and $b = 10$
\(\begin{align*} x^2+5x-50 &= x^2+10x-5x-50 \\ &= x(x+10) - 5(x+10) \\ &= (x+10)(x-5) \end{align*}\)
$\Rightarrow c = 5$
$a + b + c = 3 + 10 + 5 = \boxed{18}$
+874 $x^2 + 13x + 30 = (x+10)(x+3)$
$x^2 + 5x - 50 = (x+10)(x-5)$
$10 + 3 + 5 = \boxed{18}$
