# Factor of the polynomial

• Jan 1st 2009, 02:34 PM
djmccabie
Factor of the polynomial
a) Show that 2x-1 is a factor of the polynomial 2x³ - 3x² - 11x + 6
Find the other factors of the polynomial

b) Find, correct to three decimal places, the values of t satisfying

2e^3t - 3e^2t - 11e^t + 6 = 0

Could someone hellp me start this question because I've no idea at all what to do. If someone could start me off i might be able to figure it out.

thanks
• Jan 1st 2009, 03:08 PM
skeeter
Quote:

Originally Posted by djmccabie
a) Show that 2x-1 is a factor of the polynomial 2x³ - 3x² - 11x + 6
Find the other factors of the polynomial

b) Find, correct to three decimal places, the values of t satisfying

2e^3t - 3e^2t - 11e^t + 6 = 0

a) if $\displaystyle 2x-1$ is a factor, then $\displaystyle f\left(\frac{1}{2}\right) = 0$ ... show this by synthetic division

Code:

1/2]      2    -3    -11    6               1    -1    -6 ------------------------------         2    -2    -12    0
the depressed polynomial is

$\displaystyle 2x^2 - 2x - 12 = 2(x^2 - x - 6) = 2(x - 3)(x + 2)$

the other two roots are $\displaystyle x = 3$ and $\displaystyle x = -2$

b) note the similarity between this equation and the one in part (a)

$\displaystyle e^t = \frac{1}{2}$

$\displaystyle e^t = 3$

$\displaystyle e^t = -2$

two of the three above equations will yield a real solution.
• Jan 1st 2009, 03:15 PM
djmccabie
Quote:

Originally Posted by skeeter
Code:

1/2]      2    -3    -11    6               1    -1    -6 ------------------------------         2    -2    -12    0
the depressed polynomial is

$\displaystyle 2x^2 - 2x - 12 = 2(x^2 - x - 6) = 2(x - 3)(x + 2)$

Hi thanks for your help. I don't really understand this part of your answer. I've never learned what a depressed polynomial is. how did you deduce the 1 -1 -6 part? thanks
• Jan 1st 2009, 03:21 PM
djmccabie
Hi i just found a tutorial for synthetic division online. I've never even heard of it. Thanks!
• Jan 1st 2009, 03:22 PM
11rdc11
Quote:

Originally Posted by djmccabie
Hi thanks for your help. I don't really understand this part of your answer. I've never learned what a depressed polynomial is. how did you deduce the 1 -1 -6 part? thanks

Try googling synthetic division and it should make sense. I would explain to you but just not sure how to.
• Jan 2nd 2009, 09:38 AM
Soroban
Hello, djmccabie!

Quote:

a) Show that $\displaystyle 2x-1$ is a factor of the polynomial $\displaystyle 2x^3 - 3x^2 - 11x + 6$

Find the other factors of the polynomial

What's stopping you from using long division?

. . $\displaystyle \begin{array}{cccccc} & & & x^2 & -x & -6 \\ & & -- & -- & -- & -- \\ 2x-1 & ) & 2x^3 & -3x^2 & -11x & +6 \\ & & 2x^3 & -x^2 \\ & & -- & -- \\ & & & -2x^2 & -11x \\ & & & -2x^2 & +x \\ & & & -- & -- \\ & & & & -12x & +6 \\ & & & & -12x & +6 \\ & & & & -- & --\end{array}$

Therefore: .$\displaystyle 2x^3-3x^2-11x+6 \:=\:(2x-1)(x^2-x-6) \:=\:\boxed{(2x-1)(x+2)(x-3)}$