Question

How do you find the roots of \[{x^3} - {x^2} - 34x - 56 = 0\]?

Answer 1

Hint: First, we need to know about the concept of finding the roots of the polynomial, and then we can simply find the roots for given polynomial equations.
We will use the concept of the factoring theorem to solve the given problem further to get the simplification equation.
The factorize theorem is useful for the simplification of the cubic equation to a quadratic equation and then quadratic equation to linear degree equation.

Complete step by step answer:
From the given that we have, \[{x^3} - {x^2} - 34x - 56 = 0\]and we have to find the roots of the given equation.
Using the factors theorem, first rewrite the given equation into a generalized form.
Since \[{x^3} - {x^2} - 34x - 56 = 0\]now changing the value $ - {x^2} $ as $ - {x^2} = 6{x^2} - 7{x^2} $ and change the third value $ - 34x $ as $ - 34x = - 42x + 8x $ (we are changing the function to get the common term and then solve further)
Hence the equation can be rewritten as \[{x^3} - {x^2} - 34x - 56 = 0 \Rightarrow {x^3} - 7{x^2} + 6{x^2} - 42x + 8x - 56 = 0\]
Now taking common value $ x - 7 $ in all the two consecutive terms, thus we get \[{x^3} - 7{x^2} + 6{x^2} - 42x + 8x - 56 = 0 \Rightarrow {x^2}(x - 7) + 6x(x - 7) + 8(x - 7) = 0\]
Further taking common we get, \[{x^2}(x - 7) + 6x(x - 7) + 8(x - 7) = 0 \Rightarrow (x - 7)({x^2} + 6x + 8) = 0\]
Hence $ x - 7 = 0 $ is the first factor of the given term and thus $ x = 7 $ is the first root for the given equation.
Further approaching the same method for the second value, \[({x^2} + 6x + 8) = 0\]
Now $ 6x $ can be rewritten as $ 6x = 2x + 4x $ and thus apply it in the above equation we get, \[({x^2} + 6x + 8) = 0 \Rightarrow ({x^2} + 4x + 2x + 8) = 0\]
Now taking common in the general values we get; \[({x^2} + 4x + 2x + 8) = 0 \Rightarrow x(x + 4) + 2(x + 4)\]in the first two values $ x $ is common and in last two values $ 2 $ is common.
Further taking common terms again we get, \[x(x + 4) + 2(x + 4) \Rightarrow (x + 4)(x + 2) = 0\]
Hence, we get, $ x + 4 = 0 $ is a factor and $ x + 2 = 0 $ is a factor.
Thus, we get $ x = - 4, - 2 $ is also the roots of the given polynomial.
Hence finally we have the roots of the equation as $ x = 7, - 2, - 4 $

Note: We can also able to solve the given cubic equation as fix a value of $ x $ any number and then the remainder, that needs to be zero while factorizing the terms and that particular value is one root, and we need to solve the quadratic equation
In quadratic form, we can also apply the formula $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ and then we get the same values in this method also.