162 algebra De mystif ieD
More on Factoring Quadratic Polynomials
When the first term is not x^2 , we look to see if we can factor the coefficient of
x^2 from each term. If we can, then we are left with a quadratic whose first term
is x^2. For example each term in 2x^2 + 16x – 18 is divisible by 2: 2 x^2 + 16x – 18 =
2(x^2 + 8x – 9) = 2(x + 9)(x – 1).PRACTICE
Factor the quadratic polynomial by first dividing each term by the coeffi-
cient of x^2.- 4x^2 + 28x + 48 =
- 3x^2 − 9x − 54 =
- 9x^2 − 9x − 18 =
- 15x^2 − 60 =
- 6x^2 + 24x + 24 =
✔SOLUTIONS
- 4x^2 + 28x + 48 = 4(x^2 + 7x + 12) = 4(x + 4)(x + 3)
- 3x^2 − 9x − 54 = 3(x^2 − 3x − 18) = 3(x − 6)(x + 3)
- 9x^2 − 9x − 18 = 9(x^2 − x − 2) = 9(x − 2)(x + 1)
- 15x^2 − 60 = 15(x^2 − 4) = 15(x − 2)(x + 2)
- 6x^2 + 24x + 24 = 6(x^2 + 4x + 4) = 6(x + 2)(x + 2) = 6(x + 2)^2
The coefficient of the x^2 term will not always factor away. In order to factor
quadratic polynomials such as 4x^2 + 8x + 3, we need to try all combinations of
factors of 4 and of 3: (4x + __ )(x + __ ) and (2x + __ )(2x + __ ). The blanks will
be filled in with the factors of 3. These are all the possibilities for which the F
and L in FOIL will work. We then need to see which one (if any) make O + I
in FOIL work.
Of all the possibilities (4x + 1)(x + 3), (4x + 3)(x + 1), and (2x + 1)(2x + 3),
the last one is the one that works.PRACTICE
Factor the quadratic polynomial by first dividing each term by the coeffi-