Algebra 2, 10 points! factoring

Question

explain factoring 
1) z^6-17z^4+16z^2 

and 

2) (t+3)^2+8(t+3)+15

 

 

 

 

 

Answer 1

(1)

Given polynomial : z⁶ - 17z⁴ + 16z²

To find factors equate polynomial to zero.

z⁶ - 17z⁴ + 16z²

Take term z² as common

z² ( z⁴ - 17z² + 16 )

Solve above two terms separately.

z² =0  and  ( z⁴ - 17z² + 16 ) = 0

Step 1 :

Solve z² = 0

By using zero product property

z = 0 , 0

Step 2 :

Solve  z⁴ - 17z² + 16  = 0    ------------------- (1)

Consider z² = k

Then equation (1) become

k² - 17k + 16  = 0

k² - 16k - k + 16  = 0

k ( k - 16 ) - 1 ( k - 16 )  = 0

( k - 1 ) ( k - 16 ) = 0

Substitute k = z²

( z² - 1 ) ( z² - 16 ) = 0

Substitute z² - 1 = ( z - 1 ) ( z +1 )    and   z² - 16 = ( z - 4 ) ( z + 4 )

( z - 1 ) ( z +1 ) ( z - 4 ) ( z + 4 )= 0

By using zero product property

z = 1 , - 1 , 4 , - 4

Combined solution is 0 , 0 , 1 , - 1 , 4 , - 4

Solution :

Factors for polynomial  z⁶ - 17z⁴ + 16z² is 0 , 0 , 1 , - 1 , 4 , - 4.

Answer 2

(2)

(t+3)²+8(t+3)+15

Apply formula : (a+b)² = a² + b² + 2ab

t² + 3² + 2(t)(3) + 8(t+3) + 15

t² + 9 + 6t + 8t + 24 + 15

t² + 6t + 8t + 48

t ( t + 6 ) + 8 ( t + 6 )

( t + 6 ) ( t + 8 )

To find factors equate polynomial to zero.

( t + 6 ) ( t + 8 ) = 0

By using zero product property.

( t + 6 ) = 0   and  ( t + 8 ) = 0

t = - 6 , - 8