HELP! Divide, using synthetic division?

Question

already divided it but I don't understand Part II? (16x^3 + 80x^2 + x + 5) /( x+5)

Part II: Use this checklist to proceed through this problem: (8 points)

Answer 1

For x + 5 to be a factor, you must have x = -5  as a zero. Using this information, do the synthetic division with  x = -5 as the test zero on the left: Start out with the synthetic division algorithm.

Start as usual by bringing down the 16:

-5  |     16       80           1        5

     |  ___________________                                              

           16    

Multiply the 16 by -5 and put it diagonally above the 16 under the 80. 
Add 80 and -80, getting 0 and write this on the bottom of the line.

 

-5  |    16       80           1        5

     |                -80                             

          16        0    

 

Multiply the 0 by -5 and put it diagonally above the -80 under the 1.
Add 1 and 0 , getting 1 and write this on the bottom of the line.

 

-5 |     16       80           1        5

     |               -80           0  

           16        0          1

 

Multiply the 1 by -5 and put it diagonally above the 1 under the 5. 
Add 5 and -5, getting 0 as ramainder and write this on the bottom of the line.

 

-5 |    16       80           1        5

     |              -80           0       -5

          16         0           1        0

Since the remainder is zero, then  x = -5 is indeed a zero of 16x³ + 80x² + x + 5.

Thus, the quotient is  16x² + 1.

 

Comments

The related equation is 16x² + 1 = 0

Subtract each side by 1.

16x² + 1 - 1 = 0 - 1

Apply Addidve Inverse Property: a - a = 0.

16x² + 0 = - 1

Apply Additive Identity Proprty: a + 0 = a.

16x² = - 1

Diviide each side by 16.

16x² / 16 = - 1/16

Cancel common terms.

x² = - 1/16

x = ±√(-1/16)

x = ± [√(-1) / √(16)]

x = ± [ i / √(16) ]                   [Since i² = -1 ]

Write 16 as 4².

x = ± [ i / √(4²) ]

x = ± i / 4.

Check:

Substitute x = ± i / 4 in 16x² + 1 = 0.

16(± i / 4)² + 1 = 0

Take + i/4 for x.

16(+ i / 4)² + 1 ≟ 0

16(i² / 16) + 1 ≟ 0

i² + 1 ≟ 0

-1 + 1 ≟ 0

0 = 0 True

Take - i/4 for x.

16(- i / 4)² + 1 ≟ 0

16(i² / 16) + 1 ≟ 0

i² + 1 ≟ 0

-1 + 1 ≟ 0

0 = 0 True

Therefore the solution x = ± i / 4 is checked.