Question

If ( + 2)( + 7) = 15 + + 14 ax bx x cx 2 for all values of x, and a b + =8, what are the two possible values for c

Answer 1

If (ax+2)(bx+7) = 15x^2 + cx + 14 for all values of x, and a+b=8, what are the two possible values for c? 


I know that there must be a shorter way of doing this question..(I'm not a mathematician lol)... 
But here's one way (the long way) of doing it... 

First let's expand the left hand side by using the FOILING method..(FOIL => First, Outer, Inner, Last, (a + b)(c + d) = ac + ad + bc + bd)...so... 


(ax + 2)(bx + 7) 
= (ax)(bx) + (ax)(7) + (2)(bx) + (2)(7) 
= abx^2 + 7ax + 2bx + 14 <-- combine like terms.. 
= abx^2 + (7a + 2b)x + 14 

Now compare this to 15x^2 + cx + 14.. 

Equate the coefficients... like for example, compare the coefficients (number in front of the variable) of x^2...the coefficient on the left hand side is 'ab' and the coefficient on the right hand side is 15...so this means ab = 15... and same goes for the coefficients of 'x'... 

ab = 15 
7a + 2b = c 

We are given that a + b = 8... 
Ignore the 7a + 2b = c for now and let's just consider the two equations with just a and b...setting up a system of equations to solve for 'a' and 'b'... 

a + b = 8 
ab = 15 

Using the first equation 7a + 2b = 8, let's isolate "b"... subtracting a to both sides... 

b = 8 - a 
ab = 15 

Now substitute 8 - a for b in the second equation for b and solve for 'a'.. 

a (8 - a) = 15 <-- multiply.. 
8a - a^2 = 15 <-- subtract 8a and add a^2 to both sides of the equation.. 
8a - a^2 - 8a + a^2 = a^2 - 8a + 15 
0 = a^2 - 8a + 15 

Solving by factoring..let's look for two factors that multiply to the last term (15) and adds to the middle term (-8)..those two factors are -3 and -5...so... 

0 = (a - 3)(a - 5) 

Now set each factor equal to 0 and solve for 'a'... 

a - 3 = 0 
a - 3 + 3 = 0 + 3 
a = 3 

a - 5 = 0 
a - 5 + 5 = 0 + 5 
a = 5 


So the values of a are 3 and 5 (not surprisingly, there should be two solutions according to the question)...substitute both these values for 'a' in one of the two original equations and solve for 'b'.. 

ab = 15 
(3)b = 15 <-- divide both sides by 3... 
3b/3 = 15/3 
b = 5 

(5)b = 15 
5b/5 = 15/5 
b = 3 


So when a = 3, b = 5.. 
When a = 5, b = 3 


Recall that 7a + 2b = c from when we equated the coefficients of the variables right? Substitute these values for a and b and find the two values of 'c'... 


When a = 3 and b = 5... 
7(3) + 2(5) = c 
21 + 10 = c 
31 = c 

When a = 5 and b = 3... 
7(5) + 2(3) = c 
35 + 6 = c 
41 = c 


So the two values of 'c' are 31 and 41...