Question

Solve for real x
${x^4} - 2{x^3} + x - 380 = 0$

Answer 1

Hint: In this particular type of question use the concept that in a fourth order polynomial there are at most 4 roots or zeros, as the complexity of the fourth order polynomial is high so first find out its real roots by hit and trial method so use this concept to reach the solution of the question.

Complete step-by-step answer:
Given polynomial:
${x^4} - 2{x^3} + x - 380 = 0$...................... (1)
As we see that the given polynomial is a fourth order polynomial so there are at most 4 roots or zeros of the given polynomial.
Now we have to find the solution for the real value of x.
Let us suppose that two the roots of the given polynomial are real, let these roots are, a and b.
So (x – a) and (x – b) are the factors of the given polynomial.
And the rest of the two factors together makes a quadratic equation, so when this quadratic equation multiplied by the assumed factors it will give us the original given polynomial.
Let the quadratic equation be $c{x^2} + dx + e$, so the multiplication of this quadratic equation with the assumed factors we will get the resultant polynomial.
Where, a, b , c , d and e are some constant values.
Therefore,
\[\left( {x - a} \right)\left( {x - b} \right)\left( {c{x^2} + dx + e} \right) = {x^4} - 2{x^3} + x - 380\]................ (2)
Now as it is a fourth order polynomial so it is very complex to solve like this, so we use a hit and trial method to solve this equation.
Let x = 5 be the first root of the given polynomial.
So substitute x = 5 in equation (1) we have,
$ \Rightarrow {5^4} - {2.5^3} + 5 - 380 = 0$
$ \Rightarrow 625 - 250 + 5 - 380 = 0$
$ \Rightarrow 630 - 630 = 0$
$ \Rightarrow 0 = 0$
So, x = 5 is the root of the equation.
Therefore, (x – 5) is the factor of the polynomial.
Now let x = -4 be the second root of the given polynomial.
So substitute x = -4 in equation (1) we have,
$ \Rightarrow {\left( { - 4} \right)^4} - 2.{\left( { - 4} \right)^3} + \left( { - 4} \right) - 380 = 0$
$ \Rightarrow 256 + 128 - 4 - 380 = 0$
$ \Rightarrow 384 - 384 = 0$
$ \Rightarrow 0 = 0$
So, x = -4 is the root of the equation.
Therefore, (x + 4) is the factor of the polynomial.
Now from equation (2) we have,
a = 5 and b = -4
\[\left( {x - 5} \right)\left( {x + 4} \right)\left( {c{x^2} + dx + e} \right) = {x^4} - 2{x^3} + x - 380\]
Now simplify the above equation we have,
\[ \Rightarrow \left( {{x^2} - x - 20} \right)\left( {c{x^2} + dx + e} \right) = {x^4} - 2{x^3} + x - 380\]
\[ \Rightarrow c{x^4} + d{x^3} + e{x^2} - c{x^3} - d{x^2} - ex - 20c{x^2} - 20dx - 20e = {x^4} - 2{x^3} + x - 380\]
\[ \Rightarrow c{x^4} + \left( {d - c} \right){x^3} + \left( {e - d - 20c} \right){x^2} + \left( { - e - 20d} \right)x - 20e = {x^4} - 2{x^3} + x - 380\]
Now compare its terms we have,
$ \Rightarrow c = 1$........................ (1)
$ \Rightarrow d - c = - 2$................. (2)
$ \Rightarrow e - d - 20c = 0$.................. (3)
$ \Rightarrow - e - 20d = 1$.................. (4)
$ \Rightarrow 20e = 380$............. (5)
Now from equation (5) we have,
$ \Rightarrow e = \dfrac{{380}}{{20}} = 19$
From equation (4) we have,
$ \Rightarrow - 19 - 20d = 1$
$ \Rightarrow - 20d = 20$
$ \Rightarrow d = - 1$
So the quadratic equation become,
$ \Rightarrow {x^2} - x + 19 = 0$
Now apply quadratic formula we have,
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where a = 1, b = -1 and c = 19
$ \Rightarrow x = \dfrac{{1 \pm \sqrt {1 - 4\left( {19} \right)} }}{2} = \dfrac{{1 \pm \sqrt { - 75} }}{2} = \dfrac{{1 \pm i5\sqrt 3 }}{2}$, $\left[ {\because i = \sqrt { - 1} ,\sqrt {75} = 5\sqrt 3 } \right]$
Where, i is a complex notation.
So the other two roots of the polynomial are complex.
So the real solution of the equation is, x = 5 and x = -4.
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the quadratic formula to solve the complex quadratic equation which is stated above, so first find the quadratic equation as above, then use this formula we will get the required other zeroes of the given polynomial.