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Stuck in algebra factoring polynomial

  1. Jul 27, 2016 #1
    1. The problem statement, all variables and given/known data
    I did some refresher exercises in khan academy. I tried to wrap my head around the factoring problems, but it did not go well...

    I did not understand the khan academy tips and hints. (about factoring)

    one particular difficult polynomial was thus: factorize the polynomial.

    ##6x^2 +11x -35##

    according to khan academy this can yet be further factorized (in some manner)

    I looked at the hints section of the problem, but I could not understand the method by which the factorization was found. (it was not apparently the "guessing inside the parenthesis" method)
    2. Relevant equations
    3. The attempt at a solution


    looking at the x-term
    It can be seen that 11 is a prime number therefore factor cannot be found for 11, unless it is 1?

    greatest common factor does not apparently exist because of 11 is prime, and it's not factor of 6 and neither of -35

    I was taught many years ago in high school to use the guessing-method of trying to find the factor...

    But I was not very good at this technique. Is there a better way to help in factoring?
    I'm sorry I cannot be more productive at this problem... But I was quite stuck on this problem.
     
  2. jcsd
  3. Jul 27, 2016 #2

    fresh_42

    Staff: Mentor

    If you can write ##6x^2 + 11x - 35 = c(x-a)(x-b)## then ##6a^2 + 11a - 35 = 0## and ##6b^2 + 11b - 35 = 0##.
    You now try to find the ##x## where ##6x^2 + 11x - 35 = 0## by either using the formula for it, or you can try to write ##0 = 6x^2 + 11x - 35 = 6(x+d)^2 + e## for some ##d## and ##e## which have to be found and solve for ##x##.
    (But my back of the envelope calculation didn't produce nice numbers.)
     
  4. Jul 27, 2016 #3
    the entire problem was as follows

    factorize originally:

    $$12x^4 ~~+22x^3~~ -70x^2$$

    ##2x^2*(6x^2+11x-35)##

    I was reading that the webguide "for dummies" portion online about algebra.

    Apparently there is an algebraic method "for dummies" called British method (foil method?) for finding factorization

    But at this late hour I was not able to decipher much of the "idea behind the process" (i.e. I failed to understand the "for dummies" method oops:woot:)
     
  5. Jul 27, 2016 #4
    What are the factors of 35? What are the factors of 6? What is 3x7? What is 2x5? What is 3x7 - 2x5?
     
  6. Jul 27, 2016 #5

    fresh_42

    Staff: Mentor

    I know only one other way than the ones I described above, Vieta's formulas. With the definitions for ##a## and ##b## I already used, they say
    ##\frac{11}{6} = -(a+b)## and ##-\frac{35}{6} = a \cdot b##.
    Beside that there are probably only numerical or geometric approximations left. Guessing won't help a lot, I guess, but likely refers to Vieta's formulas. Every small prime shows up here: ##6=2\cdot 3 \; , \; 35 = 5 \cdot 7 \; , \; 11 \cdot 1##. That makes the solution a bit unpleasant.

    Not all quadratic polynomials can be solved over the reals. Even less by integers or fractions. E.g. ##x^2 + 1## is irreducible over ##\mathbb{R}##. If complex numbers are allowed, then you can always find the (at most) two zeros ##a## and ##b##.
     
  7. Jul 27, 2016 #6

    Factors for 6 are. 2x3 (both primes)
    3×7=21
    2×5=10

    21-10=11
    Factors for 35 are 5×7 (both primes)
    But Im not sure where the 3x7 comes from

    Where 2x5 comes from

    Looks like I will need to study more after some sleep
     
  8. Jul 27, 2016 #7
    You just have to be methodical. The factors of 6 are either 1 and 6 or 2 and 3. Those for 35 are either 5 and 7 or 1 and 35. The third term is negative, and is the product of the two second terms of the original binomials, which means that one of the second terms is positive and one is negative. The middle term is positive, so the larger factor is also. Using 1 and 35 won't get us a difference of 11, so that leaves us with four possibilities:
    (6x- )(x+ )
    (6x+ )(x- )
    (3x- )(2x+ )
    (3x+ )(2x- )
    To get a difference of 11 is easier with the 3 and 2 combinations. 3x7=21 and 2x5=10. 21-10=11, so there's the right combination. Using the FOIL method:
    (3x-5)(2x+7) = 6x^2+21x-10x-35 = 6x^2+11x-35
     
  9. Jul 27, 2016 #8
    In order to get 6x2, you can either multiply 6x times 1x, or you can multiply 3x times 2x. There are no other possibilities. So that means that the first terms inside the parentheses looks like either:
    (6x )(x ) or (3x )(2x )

    Now looking at the -35 number, it's kind of conspicuous that 7 x 5 = 35. Nevertheless, the only ways to get 35 is to multiply 35 times 1, or multiply 7 times 5. And since it is -35, that means one of those integers (but not both) has to be negative. So that means that the second terms inside the parentheses look like either:
    ( -35)( +1) or ( +35)( -1) or ( +5)( -7) or ( -5)( +7)

    So putting all that together, there are only a total of 8 possibilities, which are:
    (6x-35)(x+1)
    (6x+35)(x-1)
    (6x+5)(x-7)
    (6x-5)(x+7)
    (3x-35)(2x+1)
    (3x+35)(2x-1)
    (3x+5)(2x-7)
    (3x-5)(2x+7)

    One of those HAS TO BE the solution.

    What Chestermiller was trying to get you to ask yourself is what will produce 6x2, what will produce -35, and what combination will result in +11x. That's probably why he's a mentor and I'm not. He is trying to get people to think. Maybe I'm just a pushover.
     
  10. Jul 28, 2016 #9

    pasmith

    User Avatar
    Homework Helper

    No guesswork needed:
    [tex]
    6x^2 + 11x - 35 = 6 \left(x - \frac{-11 + \sqrt{11^2 + 4\times6\times35}}{2\times6}\right)\left(x - \frac{-11 - \sqrt{11^2 + 4\times6\times35}}{2\times6}\right)
    = 6 \left(x + \frac{11 - 31}{12}\right)\left(x + \frac{11 + 31}{12}\right) \\
    = 6\left(x - \frac53\right)\left(x + \frac72\right) \\
    = (3x - 5)(2x + 7)[/tex]
     
  11. Jul 28, 2016 #10
    Curiously enough, I understood TomHart's method and message but I didnt understand mathdroid's answer.

    I did not understand adequitely the reason why this term (-35) should be the product of the two second terms of original binomial (something to that effect was written)

    Similar stuff was said in khan academy video.

    Im not sure I understood the part of the method which involves the product (-35)
     
  12. Jul 28, 2016 #11

    fresh_42

    Staff: Mentor

    If you have ##a_2 x^2 + a_1 x + a_0 = a_2 (x-A) (x-B)## then you can multiply the right hand side and get ##a_2 x^2 + a_2(-A-B) x +(a_2AB).##

    Now you can compare the terms at ##x^2\, , \,x\, , \,1=x^0## and get
    ##a_2 = a_2\, , \,-a_2(A+B) = a_1\, , \,a_2AB = a_0## or for símplicity if ##a_2 = 1## it reads
    ##A+B=-a_1 \, (=-\frac{11}{6}## in the example) and ##AB = a_0 \, \, (=-\frac{35}{6})##.
     
  13. Jul 28, 2016 #12

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The factorization has to be one of the above only if the roots are rational. Of course, one can employ the rational root theorem to check if that is the case. The method in post #9 does not need to make any such assumptions,or perform any such checks.
     
  14. Jul 29, 2016 #13
    I had to look up my old high school math textbook (which I had stowed away in the garage). I think the textbook gave a good explanation and reasoning for the idea about factoring quadratics.
    . The textbook did not contain elaborate and rigorous proof of the factorization, but it began with some assumptions.
    These things were at the end of the final chapter of the book basically, so I doubt that I ever really studied these back in the day very thoroughly...The "proof" was just an algebraic reasoning which explained the right hand side of the equation, which equals to the original trinomial (ax^2+bx+c)
    The algebraic calculations were quite laborious looking, but they seemed to add-up to make equal value on both sides of the equation. Thereby proving sort of the entire rationale behind the process ( at least algebraically.) .

    it looks like it was assumed inititially that there is the trinomial

    ##ax^2 +bx +c= a (x-x_1)(x-x_2)##

    x_1 and x_2 look like they are the roots of the equation of ##ax^2 +bx +c = 0##


    working from the book definitions.
    My textbook said that for the trinomials which are of the form ## Ax^2 +Bx +C##

    Essentially finding the quadratic equation roots will aid us. Also the size of the discriminant, from the equation, will aid us.
    If the discriminant is non-negative, then we still have hope of finding factorization, for the trinomial itself.
    However, if the discrimiannt is negative, then we will not be able to further factorize the trinomial with any first degree polynomial factors.
    And, if the discriminant is zero, then the factorization would take such form as this.
    ##Ax^2+Bx+C= A(x-x_1)^2##
    the root of the equation will be the singular x_1 (essentially x_1 and x_2 are equal in this case)

    If the discriminant would be positive, then there are two separate roots x_1 and x_2
    Then, it will be that the factors for ## Ax^2 +Bx +C##
    would be something like ##A(x-x_1)(x-x_2)##

    we find the roots of the equation first. We have two roots x_1 = 5/3
    and x_2 = -7/2

    ##6x^2 + 11x -35 = 6*(x-\frac{5}{3})(x+\frac{7}{2})##

    ##6*(\frac{x^2}{1}+\frac{7x}{2}-\frac{5x}{3}-\frac{35}{6})##

    expand all the fractions until they all have 6 in the denominator. The coefficient 6 cancels out the denominator after that.

    ##6x^2 +21x -10x-35## clearly this equals the original expression of the trinomial

    But now comes the part about the grouping of the terms.

    It looks like we will use +21x and -10x in order to form one term, by which we multiply the other (x+ something) term
    There are only a couple options here but one option which seemed to work at the second attempt was as follows

    2x(3x-5) + (3x-5)7

    which itself can be transformed into
    (3x-5)*(2x+7)

    and foiling out the brackets gives us indeed 6x^2 +[21x-10x] -35= 6x + 11x -35


    On the other hand I was thinking about pasmith's earlier answer also especially the final step
    He managed to get neatly packaged terms inside each bracket also quite easily

    I wonder if he re-arrenged the multiplication as follows

    6* (x- 5/3) ( x+ 7/2)

    2*3*(x- 5/3) (x+ 7/2)
    3*(x- 5/3)*2*(x+ 7/2)

    =(3x-5)(2x+7)
     
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