Algebra Demystified 2nd Ed

164 algebra De mystif ieD

You can see that when the constant term and the coefficient of x^2 have many

factors, this list of factorizations to check can grow rather long. Fortunately

there is a way around this problem as we shall see in chapter 10.

PRACTICE

Factor the quadratic polynomial.

1. 6x^2 + 25x − 9 =


2. 18x^2 + 21x + 5 =


3. 8x^2 − 35x + 12 =


4. 25x^2 + 25x − 14 =


5. 4x^2 − 9 =


6. 4x^2 + 20x + 25 =


7. 12x^2 + 32x − 35 =

✔SOLUTIONS

1. 6x^2 + 25x − 9 = (2x + 9)(3x − 1)


2. 18x^2 + 21x + 5 = (3x + 1)(6x + 5)


3. 8x^2 − 35x + 12 = (8x − 3)(x − 4)


4. 25x^2 + 25x − 14 = (5x − 2)(5x + 7)


5. 4x^2 − 9 = (2x − 3)(2x + 3)


6. 4x^2 + 20x + 25 = (2x + 5)(2x + 5) = (2x + 5)^2


7. 12x^2 + 32x − 35 = (6x − 5)(2x + 7)

Quadratic-Type Expressions

An expression with three terms where the power of the first term is twice that

of the second and the third term is a constant is called a quadratic-type expres-

sion. These expressions factor in the same way as quadratic polynomials. The

power on x in the factorization will be the power on x in the middle term. To

see the effect of changing the exponents, let us look at the factors of x^2 – 2x – 3 =

(x – 3)(x + 1). Each of the expressions below, factors the same way.

x^4 – 2x^2 – 3 = (x^2 – 3)(x^2 + 1)

x^6 – 2x^3 – 3 = (x^3 – 3)(x^3 + 1)

PRACTICE

Factor the quadratic polynomial.