SIGN UP

Maths question - solve (x-2)(x^2+3x+1)=0

Avatar Image
ATB_roo | 14:49 Sun 25th Nov 2007 | Science
4 Answers
How can I solve (x-2)(x^2+3x+1)=0

From the question:

a) Show that (x-2) is a factor of f(x)=x^3 + x^2 -5x -2.

b) Hence find the exact solutions of the equation f(x) = 0

I've managed to get to the last bit of part b, and I think I've done it correctly, but I can't see how to solve (x-2)(x^2+3x+1)=0

Thanks

Answers

1 to 4 of 4rss feed

Best Answer

No best answer has yet been selected by ATB_roo. Once a best answer has been selected, it will be shown here.

For more on marking an answer as the "Best Answer", please visit our FAQ.
(a) Put x=2 into expression x^3 = 8 x^2 = 4 5x=10
so 8+4-10-2=0 thereore x=2 is a factor

(b) x^2+3x+1 has to factorised using formula
x = (-1+/-(3^2-4.1.1)^0.5)/2.1 = -1+/-2^0.5
therefore solutions are x=2, x=-1+2^0.5, x=-1-2^0.5

Hope this helps
Question Author
hmm
isnt the formula x=(-b+/- (b^2-4ac)^0.5)/2a
so wouldnt it be x=(-3+/- (3^2-4.1.1)^0.5)/2.1
and I guess i just work that out on the calculator seeing as it asks for exact solutions, right"?

But, yeah, thanks for the help. I'm not sure why I didnt think to use the formula lol =]

Cheers
In the expression given, either the first bracket = 0 or the second does.
So the first solution for x is x - 2 = 0, which gives x = 2.


Moving to the second bracket, we use the formula

x = minus b plus-or-minus square root of b^2 minus 4ac, all over 2a

With a = 1, b =3, c = 1, this gives:

-3 plus or minus square root of (9 minus 4) all over 2

-3 plus or minus root 5, all over 2

-3 plus 2.236, divided by 2 = - 0.382 .... Or....

-3 minus 2.236, divided by 2 = - 2.668

So the answers are x = 2, x = - 0.382, x = -2.668

Gosh, have I gone wrong, somebody?
Question Author
Nah, thats what I got in the end I think.

I'll get my homework back next week, I'll post the correct answer on here. Although, I think we've got it correct anyway.

1 to 4 of 4rss feed

Do you know the answer?

Maths question - solve (x-2)(x^2+3x+1)=0

Answer Question >>