How to solve higher degree polynomial equations?

Hi! I am currently taking an Algebra II classs and we just started learning about Polynomials and Polynomial Functions. I missed the lesson on "Solving a Higher-Degree Polynomial Equation" and my neither my teacher nor book are doing a very good job of explaining it to me. Can someone please explain, as best as you can, how do to this? Also when explaining can you use these problems from my homework? 1) 81x^3-192=0 2) x^4-64=0

Thank You!

4 Answers

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  • 8 years ago
    Favorite Answer

    You will probably get into other methods also, but the first thing you should try is factoring.

    Some formulas you need to know:

    (a^2- b^2)= (a+b)(a-b)

    (a^3- b^3)= (a-b)( a^2 + ab + b^2)

    (a^3 + b^3) = (a+ b)( a^2 -ab + b^2)

    1) 81x^3- 192= 0

    First factor out the gcf, 3

    3(27x^3 - 64)= 0.... Notice that 27 and 64 are both perfect cubes. So use the second formula above:

    27x^3= (3x)^3 and 64= 4^3

    3( 3x -4)( 9x^2 + 12x + 16)

    3x-4= 0 or 9x^2 + 12x + 16= 0

    X= 4/3; the second equation has no solution (by trying the quadratic formula)

    2) X^4 -64= 0: factor using the first formula (difference of squares)

    (x^2 + 8)(x^2 -8) = 0

    X^2 + 8 = 0 or x^2 -8 = 0

    X^2= -8 has no real solutions ;

    Or x^2 = 8

    X= +/- sqr(8)

    X= +/-2* sqr(2)

    Hoping this helps!

  • Anonymous
    4 years ago

    Higher Degree Polynomial Equations

  • 8 years ago

    for solving a polynomial of degree two we have a formula but for degree higher than two we don't have any specific formula. we have to use hit or trial method or there might be any particular pattern in question, . like in

    1)

    81x^3=192

    -> 27x^3=64

    -> (3x)^3=(4)^3

    ->3x=4

    in case we don't find any pattern ,we have to use hit and trial method .there is no other way but to guess one root of given three degree equation eg. x^3-x^2 +1=0 .we have to randomly guess one root of this equation. start from simple integers like 0,1,-1,2,-2...... here suppose we got 1 as a root ,so equation is (x-1)(x^2-x-1)=0 solve the remaining with the formula and hence obtain the three roots

    in case of four degree equation guess two roots and find the remaining two by formula . or sometimes the equation is pseudo quadratic(false two degree eqn) .like your question 2)

    put x^2=t

    -> t^2=64

    ->t=8 ( not -8 since x^2 can't be equal to -8)

    x^2=8

    other example of this like x^4+ x^2 -2=0 . put x^2=t and solve.

  • satszn
    Lv 6
    8 years ago

    (1) 81x^3 = 192 , x^3 = 192/81 = 64/27 , = 4/3 . (2) x^4 = 64 , x^2 = 8 or -8 , x = 2sqrt2 or - 2sqrt2 or 2isqrt2 or -2isqrt2

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