How Do You Simplify a Cubic Polynomial?

  • Thread starter Thread starter Patdon10
  • Start date Start date
  • Tags Tags
    Simplifying
AI Thread Summary
To simplify the cubic polynomial -x^3 - 3x^2 - 4x - 2, it is essential to factorize it rather than simplify in the traditional sense. The incorrect assumption that x-1 is a factor is clarified, as substituting x=1 does not yield zero. Instead, x+1 is identified as a factor since substituting x=-1 results in zero. Dividing the polynomial by x+1 will reveal the other factor. Accurate factorization is crucial for correctly simplifying cubic polynomials.
Patdon10
Messages
85
Reaction score
0

Homework Statement



simplify -x^3 - 3x^2 - 4x - 2
It is equal to -(x-1)(x^2 + 2x + 2)

Not sure how to get that answer, nor how to start it.
 
Physics news on Phys.org
Patdon10 said:

Homework Statement



simplify -x^3 - 3x^2 - 4x - 2
It is equal to -(x-1)(x^2 + 2x + 2)

Not sure how to get that answer, nor how to start it.

The word you're looking for is to factorize it, not simplify it. You can't simplify it in the sense that you can simply \frac{\sin(x)}{\cos(x)} to become \tan(x) for example.

Ok so first of all, in order to factorize that cubic you'll need to know one of its roots. Do you know how to check if a polynomial has rational roots?
 
Patdon10 said:

Homework Statement



simplify -x^3 - 3x^2 - 4x - 2
It is equal to -(x-1)(x^2 + 2x + 2)

Not sure how to get that answer, nor how to start it.
Good! Because you shouldn't get that "answer". It is wrong.

Setting x= 1 in that polynomial gives -(1)- 3(1)- 4(1)- 2= -(1+3+ 4+ 2)= -10, not 0. Since x= 1 does NOT make that polynomial 0, x- 1 is NOT a factor. -x^3- 3x^2- 4x- 2 is NOT equal to -(x- 1)(x^2+ 2x+ 2).

However, setting x= -1 gives -(-1)- 3(1)- 4(-1)- 2= 1- 3+ 4- 2= 0 so x-(-1)= x+ 1 is a factor. Divide -x^3- 3x^2- 4x- 2 by x+1 to get the other factor.
 
HallsofIvy said:
Good! Because you shouldn't get that "answer". It is wrong.

Setting x= 1 in that polynomial gives -(1)- 3(1)- 4(1)- 2= -(1+3+ 4+ 2)= -10, not 0. Since x= 1 does NOT make that polynomial 0, x- 1 is NOT a factor. -x^3- 3x^2- 4x- 2 is NOT equal to -(x- 1)(x^2+ 2x+ 2).

However, setting x= -1 gives -(-1)- 3(1)- 4(-1)- 2= 1- 3+ 4- 2= 0 so x-(-1)= x+ 1 is a factor. Divide -x^3- 3x^2- 4x- 2 by x+1 to get the other factor.

Ah yes I shouldn't noticed that at quick glance since -x^3 - 3x^2 - 4x - 2 = -(x^3+3x^2+4x+2) which should have all same sign coefficients in its factors
 
Patdon10 said:

Homework Statement



simplify -x^3 - 3x^2 - 4x - 2
It is equal to -(x-1)(x^2 + 2x + 2)

Not sure how to get that answer, nor how to start it.
-(x-1)(x^2 + 2x + 2)

is equivalent to:

-x^3 -x^2 + 2
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top