If (ax - 2)(bx - 5) = 16x^2 + cx +10 for all values of x and a + b = 10 , what are the two possible values of c ?

Question

Answer: -26 and - 44

Answer 1

FOILING the left side:

 

  ABx^2 - 5ax - 2bx + 10 = 16x^2 + cx + 10

 

  ABx^2 - (5a+2b)x + 10 = 16x^2 + cx + 10 <---- factors out -x

 

Equating the coefficients

  AB = 16 <---- equation I

 

 -(5a+2b )= c  <---- equation II

 

 a+b = 10 ---> b = 10 - a <--- equation III

 

Substitutes equation III into equation I

 

a(10 - a) = 16

 

10a - a^2 = 16 <--- distributive

 

0 = a^2 - 10a + 16  <---- everything to right side

 

0 = (  a    -    2)( a   -     8  ) <--- factors

 

a-2 = 0 ----> a = 2 ---> b = 10 - 2 = 8 ----> c =-( 5a + 2b) = - [5(2) + 2(8)] =-[ 10 + 16] = -26

 

So the quadratic is 16x^2 - 26x + 10 = 2 ( 8x^2 - 13x + 5) = 2 ( 8x     -    5)(x   -     1 ) = (8x-5)(2x-2)

 

 

 

a-8  = 0 ---> a = 8 ----> b = 10-8 = 2 ----> c =-[ 5a + 2b] = -[5(8) + 2(2)] = -[40 +4] = -44 

  The quadratic in this case is 16x^2 - 44x + 10 = 2 ( 8x^2 - 22x + 5) = 2 ( 4x  -     1)(2x  -     5 ) = (8x-2)(2x-5)

 

 

c = -26 or c=-44

     

Answer 2

Foil (ax-2)(bx-5)=abx²-5ax-2bx+10

combine like term , abx²+(-5a-2b)x+10=16x²+cx+10 for all value x, which means the coefficient of like terms on both side must equal 

ab=16 and a+b=10, solve the systems of equation, a=2,b=8. Or a=8,b=2

then solve for c

c=-5a-2b=-26,  or -44