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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 $a + b + c$?

 Jun 23, 2021
 #1
avatar+524 
+1

\(x^2+13x+30 = x^2+3x+10x+30\)

                           \(=x(x+3)+10(x+3)\)

                           \(=(x+3)(x+10)\)

⇒ 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

 

\(a+b+c = 3+10+5 = 18\)

 Jun 23, 2021
 #3
avatar+795 
+1

@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 $b = 10$

\(\begin{align*} x^2+5x-50 &= x^2+10x-5x-50 \\ &= x(x+10) - 5(x+10) \\ &= (x+10)(x-5) \end{align*}\)

$\Rightarrow c = 5$

$a + b + c = 3 + 10 + 5 = \boxed{18}$

MathProblemSolver101  Jun 23, 2021
 #2
avatar+795 
+1

$x^2 + 13x + 30 = (x+10)(x+3)$

$x^2 + 5x - 50 = (x+10)(x-5)$

$10 + 3 + 5 = \boxed{18}$

 Jun 23, 2021

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