# Thread: polynomial and factor theorem

1. ## polynomial and factor theorem

Hi guys could you check over what i have done and point me in the right direction with part ii

i had to use polinomial division to find the remainder 3x-1 / 9x^5-4x^4

i got the remainder to be 1x or (x) is this correct?
part ii

show using factor theorem that 2x-1 is a factor of 2x^4-x^3-6x^2+5x-1
and express 2x^4-x^3-6x^2+5x-1 as a linear and cubic factor.

any guidance appreciated

2. try using long division... divide the polynomial by (2x-1), and if there is 0 remainder, that means it is a factor. Also, the division should yield a cubic function, so expressing it as a product of a linear factor and a cubic factor should look like this:

(2x-1)(the cubic function you get from division)

3. Right i see it now thanks for the help,

was my first part correct or was i way off?

regards

4. Originally Posted by tommoturbo
Show using factor theorem that $2x-1$ is a factor of 2x^4-x^3-6x^2+5x-1
The factor theorem states that a polynomial $p(x$) has a factor $(ax - k)$ if and only if $p\left(\frac{k}{a}\right)$ = 0. So to show that $2x-1$ is a factor of $2x^4-x^3-6x^2+5x-1$, you plug in $\frac{1}{2}$ for x in $2x^4-x^3-6x^2+5x-1$ and see if it yields to zero. If it does (which it hopefully will), then by the factor theorem, $2x-1$ is a factor of $2x^4-x^3-6x^2+5x-1$; but if it doesn't, then $2x-1$ is a not a factor of $2x^4-x^3-6x^2+5x-1$. Was that clear? If not, see the spoiler:

Spoiler:
First, solve $2x-1$: $2x-1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$

Then substitute $\left(\frac{1}{2}\right)$ for $x$ in $2x^4-x^3-6x^2+5x-1$:

$2\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^3-6\left(\frac{1}{2}\right)^2+5\left(\frac{1}{2}\rig ht)-1$ = $\frac{1}{8}-\frac{1}{8}-\frac{3}{2}+\frac{5}{2} = 0$

Thus, by the factor theorem, $2x-1$ is a factor of $2x^4-x^3-6x^2+5x-1$.

Express $2x^4-x^3-6x^2+5x-1$ as a linear and cubic factor.
It's asking you to express it in the form $(ax^3+bx^2+cx+d)(ex+f)$. Like this:

$2x^4-x^3-6x^2+5x-1$ = $(ax^3+bx^2+cx+d)(ex+f)$

We know that $2x-1$ is a factor $2x^4-x^3-6x^2+5x-1$, so we have:

$2x^4-x^3-6x^2+5x-1 = (ax^3+bx^2+cx+d)(2x-1)$

Now, find $ax^3+bx^2+cx+d$ and you are done.

Hint:
Spoiler:
$\frac{2x^4-x^3-6x^2+5x-1}{2x-1} = ax^3+bx^2+cx+d$

5. Wow thanks very much that has explained it fully, the notes in my lesson where limited to two examples and i hadnt grasped it at all

thank you!