Quadratic Equation Problem: (3x-2)(x+7)

Question

Let A, B, and C be numbers such that

(3x-2)(x+7) = Ax²+Bx+C

for all values of x.

a) Find A, B, and C

b) How is A related to the constants and coefficients of x in 3x-2 and x+7?

c) Find all solutions to the equation 3x²+19x-14=0

Answer 1

a) To find A, B, and C, we use the FOIL method to express the polynomial in standard form.

F) First: 3x *x = 3x²

O) Outer: 3x*7= 21x

I) Inner: -2*x = -2x

L) Last -2*7 = -14

We now combine like terms

3x²+21x-2x-14= 3x²+19x-14

A=3 , B=19, C=-14

b) The two x terms in the binomials combine to give the squared term in the quadratic (3x)*(x) = Ax²

so A is the product of the coefficients of the x terms in the binomial.

c) To find the solutions to the given equation, we already know the polynomial equals (3x-2)(x+7)= 0

Therefore, we must have that 3x-2=0 or x+7=0

The first gives us x =2/3 and the second gives us x=-7 which are our two solutions.