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Find all roots of x^3 + 3x^2 - 10x + 6

  • Thread starter stat643
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  • #1
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find all roots of x^3 + 3x^2 - 10x + 6

the solution:
identify the easy root of x=1,
find the remaining roots from (x-1)(x^2+4x) using quadratic formula.

The only thing i dont understand here is how to factorize to (x-1)(x^2+4x)... namely the (x^2+4x) part.
 

Answers and Replies

  • #2
1,752
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factor out a common term in x^2+4x

and you're pretty much done
 
  • #3
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but how did i get to x^2+4x in the first place?... the original equation was x^3 + 3x^2 - 10x + 6.. i merely copied the solution... so i find the easy root of 1.. then what?
 
Last edited:
  • #4
1,752
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use synthetic division
 
  • #5
matt grime
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No, don't use synthetic division (just yet). Pause for a moment and think: is it plausible that x^2+4x is a factor? It isn't. Copying out the answer is never a good idea.
 
  • #6
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i just looked up synthetic devision on wikipedia and tried it but it didnt work
 
  • #7
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No, don't use synthetic division (just yet). Pause for a moment and think: is it plausible that x^2+4x is a factor? It isn't. Copying out the answer is never a good idea.
should i take out the common term x first?
 
  • #8
arildno
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i just looked up synthetic devision on wikipedia and tried it but it didnt work
Then you should practice synthetic division once more!

Further, ponder over matt grime's words:

WHY should you be suspicious of that particular factorization?

Hint:
How could you ascertain whether the factorization is correct or false?
 
  • #9
16
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oh sorry i copied it wrong, it should be (x-1)(x^2+4x-6).. now expanding that get: x^3 + 4x^2 -6x -x^2 -4x + 6 = x^3 + 3x^2 - 10x + 6.. so yeh its right now.. though i still cant get the synthetic devision right (its new to me)

i tried to learn it now from http://en.wikipedia.org/wiki/Synthetic_division

though i keep getting 1,2,-12,18

can someone help show how i would use synthetic devision for the original polynomial ?
 
Last edited:
  • #10
arildno
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Okay, we wish to find a second-order polynomial so that:
[tex](x-1)(ax^{2}+bx+c)=x^{3}+3x^{2}-10x+6[/tex] holds for all x.
I.e, we must determine a,b and c!!

Multiplying out the left-hand side, and organizing in powers of x, the lefthandside can be rewritten as:
[tex]ax^{3}+(b-a)x^{2}+(c-b)x-c= x^{3}+3x^{2}-10x+6[/tex]

NOw, the coefficients of each power must be equal on the right and left sides, yielding the system of equations:
a=1
b-a=3
c-b=-10
-c=6

This yields:
a=1
b=4
c=-6
 
  • #11
16
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thanks arildno, very helpful
 

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