Factoring Polynomials Help

List the possible rational zeros.

• f(x) = 4 x ³ + 6 x ² − 2 x + 9

  The numerators in our list will be the divisors of 9: 1, 3, and 9 as well as their negatives, − 1, −3, and −9. The denominators will be the divisors of 4: 1, 2, and 4. The list of possible rational zeros is—

  

  This list could be written with a little less effort as , .

  We only need to list the numerators with negative numbers and not the denominators. The reason is that no new numbers are added to the list, only duplicates of numbers already there. For example, and are the same number.

• g(x) = 6 x ⁴ − 5 x ³ + 2 x − 8

  The possible numerators are the divisors of 8: ±1, ±2, ±4, and ±8. The possible denominators are the divisors of 6: 1, 2, 3, and 6. The list of possible rational zeros is—

  

  There are several duplicates on this list. There will be duplicates when the constant term and leading coefficient have common factors. The duplicates don’t really hurt anything, but they could waste time when checking the list for zeros.

Completely factor each polynomial.

• f(x) = 3 x ⁴ − 2 x ³ − 7 x ² − 2 x

  First we will factor x from each term: f(x) = x (3 x ³ − 2 x ² − 7x − 2). The possible rational zeros for 3 x ³ − 2 x ² − 7 x − 2 are ±1, ±2, .

  3(1) ³ − 2(1) ² − 7(1) − 2 ≠ 0

  3(−1) ³ − 2(−1) ² − 7(−1) −2 = 0

  We will use synthetic division to find the quotient for (3 x ³ − 2 x ² − 7 x − 2) ÷ ( x + 1).

  

  The quotient is 3 x ² − 5 x − 2 which factors into (3 x + 1)( x − 2).

  f(x) = 3 x ⁴ − 2 x ³ − 7 x ² − 2 x

  = x (3 x ³ − 2 x ² − 7 x − 2)

  = x ( x + 1)(3 x ² −5 x − 2)

  = x ( x + 1)(3 x + 1)( x − 2)

• h(x) = 3 x ³ + 4 x ² − 18 x + 5

  The possible rational zeros are ±1, ±5, , and .

  h (1) = 3(1 ³ ) + 4(1 ² ) − 18(1) + 5 ≠ 0

  h (−1) = 3(−1) ³ + 4(−1) ² − 18(−1) + 5 ≠ 0

  h (5) = 3(5 ³) + 4(5 ²) − 18(5) + 5 ≠ 0

  Continuing in this way, we see that h (−5) ≠ 0, and .

  

   

  

  We will find the zeros of x ² + 3 x − 1 using the quadratic formula.