Find all the real roots of the equation :

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- June 11th 2009, 12:44 AMthereddevilsfactor theorem
Find all the real roots of the equation :

- June 11th 2009, 02:05 AMAmer

the possible root is

but no one of the possible root is a root of the polynomial that you wrote

there is no real root can be found without calculator see what I have when I put them in the calculator I think there is wrong in your equation

Attachment 11857

Edited : - correct something yeongil mentioned - June 11th 2009, 02:17 AMyeongil
- June 11th 2009, 06:36 AMthereddevils
Btw the answer is

- June 11th 2009, 07:37 AMyeongil
No, they are

**not**. If you plug either or into the original equation:

you get 8 = 0, so neither solution works.

However, if your original equation was supposed to be

then yes, those two solutions work. In fact, it makes solving this equation much easier, as you can solve it by factoring by grouping:

Set each factor equal to zero:

(the two real solutions)

(two non-real solutions)

(two more non-real solutions)

So it looks like you copied the problem wrong? :confused:

01 - June 11th 2009, 07:50 AMthereddevils
oh thanks a lot .. nope , i copied it down correctly .. i think its the book's mistake

Btw i hv a question here .. For questions like this where the leading coefficient is not 1 , is it a must for me to reduce the coefficient to 1 before i find the factors : example in this case

, is this step necessary

Before i start substituting the factors of the constant term to find the factors of the equation .

THanks . - June 12th 2009, 04:44 AMgreat_math