Factor of Polynomial Function 2

[mathdad]

Use synthetic division to determine if x - c is a factor of the given polynomial function.

x^6 - 16x^4 + x^2 - 16; x + 4

Change 4 to - 4. Plug into function.

(-4)^6 - 16(-4)^4 + (-4)^2 - 16

4096 - 16(256) + 16 - 16

4096 - 4096 + 16 - 16 = 0

So, x + 4 is a factor of the polynomial function.

Yes?

 

[HallsofIvy]

Do you understand what "synthetic division" is?

How does synthetic division of polynomials work?

Synthetic division is a shorthand method for dividing a polynomial by a linear factor such as x + 3, and it's much simpler and faster.

www.purplemath.com

 

[Harry_the_cat]

What you've done is correct, but you have used the factor theorem to show that x+4 is a factor, ie (x-c) is a factor of f(x) iff f(c)=0.
The question specifies to use synthetic division.

 

[mathdad]

What you've done is correct, but you have used the factor theorem to show that x+4 is a factor, ie (x-c) is a factor of f(x) iff f(c)=0.
The question specifies to use synthetic division.

I did not use synthetic division because it is not easy to express using the keyboard.

 

[MarkFL]

It's pretty easy if you're willing to use \(\LaTeX\):

[MATH]\begin{array}{c|rr}& 1 & 0 & -16 & 0 & 1 & 0 & -16 \\ -4 & & -4 & 16 & 0 & 0 & -4 & 16 \\ \hline & 1 & -4 & 0 & 0 & 1 & -4 & 0 \end{array}[/MATH]
We see the remainder is zero, thus the given linear polynomial is indeed a factor of the given 6th degree polynomial:

[MATH]x^6-16x^4+x^2-16=(x+4)\left(x^5-4x^4+x-4\right)[/MATH]

 

[mathdad]

It's pretty easy if you're willing to use \(\LaTeX\):

[MATH]\begin{array}{c|rr}& 1 & 0 & -16 & 0 & 1 & 0 & -16 \\ -4 & & -4 & 16 & 0 & 0 & -4 & 16 \\ \hline & 1 & -4 & 0 & 0 & 1 & -4 & 0 \end{array}[/MATH]
We see the remainder is zero, thus the given linear polynomial is indeed a factor of the given 6th degree polynomial:

[MATH]x^6-16x^4+x^2-16=(x+4)\left(x^5-4x^4+x-4\right)[/MATH]

LaTex with a cell phone?

 

[MarkFL]

LaTex with a cell phone?

Yes, I know several who do it.

 

[mathdad]

Yes, I know several who do it.

What about MathMagic Lite?

 

[MarkFL]

What about MathMagic Lite?

That's certainly better than plain text, but not as good as \(\LaTeX\), because it cannot be quoted and edited. But, whatever works for you is fine.

 

[mathdad]

That's certainly better than plain text, but not as good as \(\LaTeX\), because it cannot be quoted and edited. But, whatever works for you is fine.

I definitely need to use LaTex or MathMagic Lite as equations get more complicated, especially later on using David Cohen's precalculus textbook.

 

[MarkFL]

I definitely need to use LaTex or MathMagic Lite as equations get more complicated, especially later on using David Cohen's precalculus textbook.

I can't speak for all math helpers, but I know I am much more likely to provide help if the posted question is relatively easy to read.

 

[mmm4444bot]

I did not use synthetic division because it is not easy to express using the keyboard.

Take a picture of your work on paper and attach it to your post.

?

 

[mathdad]

Take a picture of your work on paper and attach it to your post.

?

Trust me, this I will do.