Factor by any method. See Example 1, Example 2, Example 3, Example 4, Example 5, Example 6...

Factor by any method. See Example 1, Example 2, Example 3, Example 4, Example 5, Example 6 and Example 7.

EXAMPLE Factoring Out the greatest Common Factor

Factor out the greatest common factor from each polynomial.

(a) 9y5 + y2

(b) 6x2t + 8xt + 12t

(c) 14(m + 1)3 − 28(m + 1)2 – 7(m + 1)

SOLUTION

(a)

(b)

(c)

EXAMPLE Factoring by grouping

Factor each polynomial by grouping.

(a) mp2 + 7m + 3p2 + 21

(b) 2y2 + az − 2z − ay2

(c) 4x3 + 2x2 − 2x − 1

SOLUTION

(a)

(b)

(c)

EXAMPLE Factoring Trinomials

Factor each trinomial, if possible.

(a) 4y2 − 11y +6

(b) 6p2− 7p − 5

(c) 2x2 + 13x − 18

(d) 16y3 + 24y2 − 16y

SOLUTION

(a) To factor this polynomial, we must find integers a, b, c, and d such that

Using FOIL, we see that ac = 4 and bd = 6. The positive factors of 4 are 4 and 1 or 2 and 2. Since the middle term has a negative coefficient, we consider only negative factors of 6. The possibilities are −2 and −3 or −1 and −6.

Now we try various arrangements of these factors until we find one that gives the correct coefficient of y.

Therefore, 4y2 − 11y + 6 factors as (y − 2)(4y – 3)

(b) Again, we try various possibilities to factor 6p2 − 7p − 5. The positive factors of 6 could be 2 and 3 or 1 and 6. As factors of -5 we have only −1 and 5 or −5 and 1.

Thus, 6p2 − 7p − 5 factors as (3p – 5)(2p + 1).

(c) If we try to factor 2x2 + 13x - 18 as above, we find that none of the pairs of factors gives the correct coefficient of x.

Additional trials are also unsuccessful. Thus, this trinomial cannot be factored with integer coefficients and is prime.

(d)

EXAMPLE Factoring perfect Square Trinomials

Factor each trinomial.

(a) 16p2 − 40pq + 25q2

(b) 36x2y2 + 84xy + 49

SOLUTION

(a) Since 16p2 = (4p)2and 25q2 = (5q)2, we use the second pattern shown in the box with 4p replacing x and 5q replacing y.

Make sure that the middle term of the trinomial being factored, −40pq here, is twice the product of the two terms in the binomial 4p − 5q.

(b)

EXAMPLE Factoring Differences of Squares

Factor each polynomial.

(a) 4m2 − 9

(b) 256k4− 625m4

(c) (a + 2b)2 − 4c2

(d) x2 − 6x + 9 − y4

(e) y2 − x2 + 6x − 9

SOLUTION

(a)

(b)

(c)

(d)

(e)

EXAMPLE Factoring Sums or Differences of Cubes

Factor each polynomial.

(a) x3 + 27

(b) m3 − 64n3

(c) 8q6+125p9

Solution

(a)

(b)

(c)

EXAMPLE Factoring by Substitution

Factor each polynomial.

(a) 10(2a − 1)2 − 19(2a − 1) − 15

(b) 12a − 123 + 8

(c) 6z4 − 13z2 − 5

SOLUTION

(a)

(b)

(c)

(x + y)3 − (x − y)3