Factor Theorem...2

[nycmathguy]

Section 2.3
Question 56

Please, do 56 showing steps.

 

[MathLover1]

56.

x^3 -52x-96=0 , x=-6

if zero is at x=-6, one factor of the given polynomial is (x-(-6))=(x+6)

and you can divide x^3 -52x-96 by (x+6)

use long division
here x^3 -52x-96 we missing x^2 and we can write it as 0*x^2

..........(x^2-6x-16
(x+6)| x^3 +0*x^2-52x-96
..........x^3+6x^2.........................subtract
.................-6x^2.........bring down next term
.................-6x^2-52x
.................-6x^2-36x.......................subtract
...........................-16x.......bring down next term
...........................-16x-96
...........................-16x-96
...................................0->reminder

result is x^2-6x-16 and we can factor it too
x^2-6x-16..........write -6x as 2x-8x
x^2+2x-8x-16
(x^2+2x)-(8x+16)
x(x+2)-8(x+2)
(x + 2) (x - 8)

so, x^2 + 6 x - 88 = (x + 2) (x - 8) (x + 6)

 

[nycmathguy]

56.

x^3 -52x-96=0 , x=-6

if zero is at x=-6, one factor of the given polynomial is (x-(-6))=(x+6)

and you can divide x^3 -52x-96 by (x+6)

use long division
here x^3 -52x-96 we missing x^2 and we can write it as 0*x^2

..........(x^2-6x-16
(x+6)| x^3 +0*x^2-52x-96
..........x^3+6x^2.........................subtract
.................-6x^2.........bring down next term
.................-6x^2-52x
.................-6x^2-36x.......................subtract
...........................-16x.......bring down next term
...........................-16x-96
...........................-16x-96
...................................0->reminder

result is x^2-6x-16 and we can factor it too
x^2-6x-16..........write -6x as 2x-8x
x^2+2x-8x-16
(x^2+2x)-(8x+16)
x(x+2)-8(x+2)
(x + 2) (x - 8)

so, x^2 + 6 x - 88 = (x + 2) (x - 8) (x + 6)

Again, why use long division if it is not required? By the way, the root tip extraction was not done yesterday morning. The crooked doctor requested $1350 for simply pulling a tooth tip. All I need is a simple root tip to be pulled out. Currently searching for a new oral surgeon. I lost a day of work for this nonsense.

 

[MathLover1]

sorry, automatically used same method as in previous problem

here is synthetic division:

Write the problem in a division-like format.
To do this:
Take the constant term of the divisor with the opposite sign and write it to the left.
Write the coefficients of the dividend to the right (missed terms are written with zero coefficients).

...|x^3 x^2 x^1 x^0
-6|1 0 -52 -96

Write down the first coefficient without changes:

-6|1 0 -52 -96
|
|________________
| 1

Multiply the entry in the left part of the table by the last entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend, and write down the sum.

-6|1 0 -52 -96
| -6*1=-6
|________________
| 1 0+-6=-6

Multiply the entry in the left part of the table by the last entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend, and write down the sum.

-6|1 0 -52 -96
| -6 -6*-6=36
|________________
| 1 -6 (-52+36)=-16

Multiply the entry in the left part of the table by the last entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend, and write down the sum.

-6|1 0 -5 -96
| -6 36 -16(-6)=96
|________________
| 1 -6 -16 -96+96=0



-6|1 0 -5 -96
| -6 36 96
|________________
| 1 -6 -16 0

We have completed the table and have obtained the following resulting coefficients: 1,-6,-16,0.
All the coefficients except the last one are the coefficients of the quotient, the last coefficient is the remainder.
Thus, the quotient is x^2-6x-16, and the remainder is 0.

 

[nycmathguy]

sorry, automatically used same method as in previous problem

here is synthetic division:

Write the problem in a division-like format.
To do this:
Take the constant term of the divisor with the opposite sign and write it to the left.
Write the coefficients of the dividend to the right (missed terms are written with zero coefficients).

...|x^3 x^2 x^1 x^0
-6|1 0 -52 -96

Write down the first coefficient without changes:

-6|1 0 -52 -96
|
|________________
| 1

Multiply the entry in the left part of the table by the last entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend, and write down the sum.

-6|1 0 -52 -96
| -6*1=-6
|________________
| 1 0+-6=-6

Multiply the entry in the left part of the table by the last entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend, and write down the sum.

-6|1 0 -52 -96
| -6 -6*-6=36
|________________
| 1 -6 (-52+36)=-16

Multiply the entry in the left part of the table by the last entry in the result row (under the horizontal line).
Add the obtained result to the next coefficient of the dividend, and write down the sum.

-6|1 0 -5 -96
| -6 36 -16(-6)=96
|________________
| 1 -6 -16 -96+96=0



-6|1 0 -5 -96
| -6 36 96
|________________
| 1 -6 -16 0

We have completed the table and have obtained the following resulting coefficients: 1,-6,-16,0.
All the coefficients except the last one are the coefficients of the quotient, the last coefficient is the remainder.
Thus, the quotient is x^2-6x-16, and the remainder is 0.

Wonderful notes! Impressive for sure. Putting math on hold for a few days.

 

[nycmathguy]

56.

x^3 -52x-96=0 , x=-6

if zero is at x=-6, one factor of the given polynomial is (x-(-6))=(x+6)

and you can divide x^3 -52x-96 by (x+6)

use long division
here x^3 -52x-96 we missing x^2 and we can write it as 0*x^2

..........(x^2-6x-16
(x+6)| x^3 +0*x^2-52x-96
..........x^3+6x^2.........................subtract
.................-6x^2.........bring down next term
.................-6x^2-52x
.................-6x^2-36x.......................subtract
...........................-16x.......bring down next term
...........................-16x-96
...........................-16x-96
...................................0->reminder

result is x^2-6x-16 and we can factor it too
x^2-6x-16..........write -6x as 2x-8x
x^2+2x-8x-16
(x^2+2x)-(8x+16)
x(x+2)-8(x+2)
(x + 2) (x - 8)

so, x^2 + 6 x - 88 = (x + 2) (x - 8) (x + 6)

You forgot to list all real solutions of the equation.

 

[MathLover1]

real solutions:
x =-2
x =8
domain:
{x element R : x≠-6}

 

[nycmathguy]

real solutions:
x =-2
x =8
domain:
{x element R : x≠-6}

Let me know if the precalculus journey is taking its toll on you. I don't want to stress you out.

 

[MathLover1]

it's relaxing

 

[nycmathguy]

it's relaxing

If you say so. I just don't want to trouble you. Also, keep in mind that answering questions on my lunch break on the overnight hours is never easy. Just recently moved from Queens to Brooklyn. Stressful time for me. However, I don't allow life to dictate what I should or shouldn't do.