The expression $x^2 + 13x + 30$ can be written as $(x + a)(x + b),$ and the expression $x^2 - 9$ written as $(x + b)(x - c)$, where $a$, $b$, and $c$ are integers. What is the value of $a + b + c$?
+28 You solve this by factoring the first polynomial x^2+13x+30.
When factoring that into (x+a)(x+b), you want ab to equal 30, and a+b to equal 13.
This is the case for all quadratics, ax^2+bx+c, into the factored form (x+m)(x+n). m+n=b, and mn=c.
To start factoring the first polynomial, find the factors of 30, and then add those factors together to see if they equal 13.
The factors cannot be negative in this case because both b and c are positive.
3*10=30, and 3+10=13, so 10 and 3 are a and b. (However, we don't know the order yet).
x^2-9, is a special case. Both terms are squares, and they are being subtracted.
m^2-n^2=(m+n)(m-n)
In this case, x is our 'm', and 3 is our 'n', therefore x^2-9=(x+3)(x-3).
(x+3) is the overlapping factor between the two polynomials, meaning that b is 3, a is 10, and c is -3.
a+b+c=10+3-3=10
