Section 5.5 Day 2 – Factoring Polynomials EQ: How do I factor polynomials in order to solve for the zeros?

Example 1 Factor each polynomial. If the polynomial cannot be factored, write prime. a) x4 – 6x2 – 27 Factor the polynomial as you normally would using the product sum method. However, just know that your first term will be 𝑥 2 . 𝑥 2 −9 𝑥 2 +3 (𝑥−3)(𝑥+3)( 𝑥 2 +3) Factor further

b) 4x4 – 5x2 – 6 Factor the polynomial as you normally would using one of the factoring methods. In this case we are going to use the slide and divide method. Again your first term will be 𝑥 2 . 𝑥 4 −5 𝑥 2 −24 𝑥 2 −8 𝑥 2 +3 ( 𝑥 2 − 8 4 )( 𝑥 2 + 3 4 ) 𝑥 2 −2 4𝑥 2 +3

c) 4x6 – 20x4 + 24x2 4 𝑥 2 ( 𝑥 4 −5 𝑥 2 +6) 4𝑥 2 ( 𝑥 2 −3)( 𝑥 2 −2)

Solve Polynomials by Factoring Put the polynomial in standard form Factor as far as you can – starting with the GCF Set all the factors with variables equal to zero Solve these new equations

Example 2 Solve a) x2 + 2x = 0 𝑥 𝑥+2 =0 𝑥=0, 𝑥+2=0 𝑥=0, −2

b) 54x3 – 2 = 0 2 27 𝑥 3 −1 =0 2( 3𝑥 3 − 1 3 )=0 2 3𝑥−1 9𝑥 2 +3𝑥+1 =0 2( 3𝑥 3 − 1 3 )=0 2 3𝑥−1 9𝑥 2 +3𝑥+1 =0 3𝑥−1=0 𝑥= 1 3 9𝑥 2 +3𝑥+1 𝑥= −3± 3 2 −4(9)(1) 2(9) 𝑥= −1±𝑖 3 6

c) 3x3 + 7x2 = 12x 3𝑥 3 +7 𝑥 2 −12𝑥=0 𝑥( 3𝑥 2 +7𝑥−12) 𝑥= −7± 7 2 −4(3)(−12) 2(3) 𝑥= −7± 193 6 𝑥=0