Example

Determine whether each of the following is a perfect-square trinomial.a) x2 + 8x + 16 b) t2 9t 36 c) 25x2 + 4 20xSolutiona) x2 + 8x + 16 1. Two terms, x2 and 16, are squares. 2. Neither x2 or 16 is being subtracted. 3. The remaining term, 8x, is 2 x 4, where x and 4 are the square roots of x2 and 16.

b) t2 9t 36 1. Two terms, t2 and 36, are squares. But 2. Since 36 is being subtracted, t2 9t 36 is not a perfect-square trinomial.

c) 25x2 + 4 20x It helps to write it in descending order 25x2 20x + 4 1. Two terms, 25x2 and 4, are squares. 2. There is no minus sign before 25x2 or 4. 3. Twice the product of 5x and 2, is 20x, the opposite of the remaining term, 20x. Thus 25x2 20x + 4 is a perfect-square trinomial.

= (x + 4)2

= (5x – 2)2

Example Factor: 16a2 24ab + 9b2

Solution

16a2 24ab + 9b2 = (4a 3b)2

Example Factor: 12a3 108a2 + 243a

SolutionAlways look for the greatest common factor. This time there is one. We factor out 3a.

12a3 108a2 + 243a = 3a(4a2 36a + 81) = 3a(2a 9)2

Note that in order for a term to be a perfect square, its coefficient must be a perfect square and the power(s) of the variable(s) must be even.

Differences of Squares

An expression, like 25x2 36, that can be written in the form A2 B2 is called a difference of squares. Note that for a binomial to be a difference of squares, it must have the following.

Example

Factor: a) x2 9 b) y2 16w2

c) 25 36a12 d) 98x2 8x8

Solutiona) x2 9 = x2 32 = (x + 3) (x 3)

A2 B2 = (A + B)(A B)

b) y2 16w2 = y2 (4w)2 = (y + 4w) (y 4w)

A2 B2 = (A + B) (A B)

c) 25 36a12 = (5 + 6a6)(5 6a6)

d) 98x2 8x8 = 2x2(49 4x6) Greatest Common Factor = 2x2(7 + 2x3)(7 2x3)

A2 + B2 is Prime!!

More Factoring by Grouping

Sometimes when factoring a polynomial with four terms, we may be able to factor further.

Example Factor: x3 + 6x2 – 25x – 150.

Solution

Grouping x3 + 6x2 – 25x – 150 = (x3 + 6x2) – (25x + 150)

= x2(x + 6) – 25(x + 6) = (x + 6)(x2 – 25) = (x + 6)(x + 5)(x – 5)

Must change the sign of each term.

Example Factor: x2 + 8x + 16 – y2.

Solution x2 + 8x + 16 – y2 = (x2 + 8x + 16) – y2

= (x + 4)2 – y2

= (x + 4 + y)(x + 4 – y)

Example Solve: x3 + 6x2 = 25x + 150.

Solution—Algebraic x3 + 6x2 = 25x + 150

x3 + 6x2 – 25x – 150 = 0(x3 + 6x2) – (25x + 150) = 0 x2(x + 6) – 25(x + 6) = 0 (x + 6)(x2 – 25) = 0 (x + 6)(x + 5)(x – 5) = 0

x + 6 = 0 or x + 5 = 0 or x – 5 = 0 x = –6 or x = –5 or x = 5The solutions are x = –6, –5, or 5.

Solving Equations

We can now solve polynomial equations involving differences of squares and perfect-square trinomials.

Graphical Solution

Let 1 ^ 3 6 ^ 2 (25 150)y x x x

The solutions are x = –6, –5, or 5.