polynomial and factor theorem

tommoturbo · #1 $Old$ November 2nd, 2009, 05:57 AM

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.

TBH i dont know where to start with this one

any guidance appreciated

mathgirlie · #2 $Old$ November 2nd, 2009, 06:49 AM

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)

tommoturbo · #3 $Old$ November 2nd, 2009, 07:10 AM

Right i see it now thanks for the help,

was my first part correct or was i way off?

regards

I4talent · #4 $Old$ November 2nd, 2009, 07:19 AM

Quote:

Originally Posted by tommoturbo $View Post$

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:

Quote:

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:

tommoturbo · #5 $Old$ November 2nd, 2009, 01:01 PM

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!