Question

can any one solve this

Answer 1

So you have to solve for x?...

Answer 2

-2 is one solution

Answer 3

only we have to find x values

Answer 4

so do you know how to do long division with polynomials?

Answer 5

Ohhh okay.

Answer 6

i know i will solve it ...

Answer 7

(2x^3-5x^2-14x+8)/(x+2)

Answer 8

How did you get the x+2..

Answer 9

-2 is a solution

Answer 10

so in factored form we know (x+2) is a factor

Answer 11

Would you explain HOW you got -2 as a solution? :P

Answer 12

divide the whole thing by that to find other factors

Answer 13

I guessed

Answer 14

lol -_-

Answer 15

one is 4 and other one is 1/2 is it right friends

Answer 16

there is an algorithm to solve these, google it.

Answer 17

correct @msingh

Answer 18

@zzr0ck3r
thank u and also thank u @Jhannybean

Answer 19

np

Answer 20

You can make some intelligent guesses about the roots of a polynomial if you know its coefficients. There's still some trial and error involved, but not so bad as just picking numbers out of the air. First, remember that any polynomial with roots \((r_1, r_2, ... r_n\)) can be written \[P(x) = (x-r_1)(x-r_2)...(x-r_n)\]Now think about what happens when you multiply two binomials:

\[(ax+b)(cx+d) = acx^2 + adx + bcx + bd = acx^2 + (ad+bc)x + bd\]Notice that the final term depends only on the roots of the polynomial! Doesn't matter how many of those root binomials you multiply together, the final term always is just a product of the roots. So, if you factor the last term, you've got a good starting point for trial roots.

Answer 21

Actually,... this is a difference in cubes i believe. \[2x ^{3}-5x^2-14x+8= \color{#ff6600}{(2x^3-5x^2)}-\color{#ff0000}{(14x+8)}=\color{#664488}{(x-4) (x+2) (2 x-1)=0}\]And from this you can solve for your x-values.
It's better than guess and check.