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Question: Solve the following equation: \({x^4} - 3{x^2} - 42x - 40 = 0\)...

Solve the following equation:
x4−3x2−42x−40=0{x^4} - 3{x^2} - 42x - 40 = 0

Explanation

Solution

In this problem, we have to solve the given equation. This polynomial is the four degrees of the equation. First, find the factor of the given equation by inspection. And find the root. After that, we get, quadratic equation. solving that quadratic equation we get the solution for the given equation.
The formula used in this problem:
The formula for quadratic equation,
x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}};
a=a = coefficient of x2{x^2}
b=b = coefficient of xx
c=c = constant term

Complete step-by-step solution:
In this problem, we are going to solve the given equation.
The given equation is the fourth-degree equation.
The given equation is x4−3x2−42x−40=0{x^4} - 3{x^2} - 42x - 40 = 0
Let us take the given equation as
f(x)=x4−3x2−42x−40=0f(x) = {x^4} - 3{x^2} - 42x - 40 = 0 …………………………………. (1)\left( 1 \right)
By inspection, we have to find the factor of the given equation.
For assumption first, we take 1,−11, - 1 as the factors of the equation.
Now we have to check if the factors are true.
To apply for the numbers, we have
f(1)=14−3(1)2−42(1)−40=0f\left( 1 \right) = {1^4} - 3{\left( 1 \right)^2} - 42\left( 1 \right) - 40 = 0
Now simplify this we get,
=1−3−42−40= 1 - 3 - 42 - 40
Now we get,
=1−85= 1 - 85
−84≠0- 84 \ne 0
Therefore, (x−1)\left( {x - 1} \right) is not a factor.
Now apply for the number −1 - 1 as a factor.
f(−1)=−14−3(−1)2−42(−1)−40=0f\left( { - 1} \right) = - {1^4} - 3{\left( { - 1} \right)^2} - 42\left( { - 1} \right) - 40 = 0
=1−3+42−40= 1 - 3 + 42 - 40
Now simplify this, we get
=43−43= 43 - 43
f(−1)=0f\left( { - 1} \right) = 0.
Therefore, (x+1)\left( {x + 1} \right) is a factor of the given number.
Therefore f(x)=(x+1).g(x)f\left( x \right) = \left( {x + 1} \right).g\left( x \right)
g(x)=x3−x2−2x−40g\left( x \right) = {x^3} - {x^2} - 2x - 40
Now the equation g(x)=x3−x2−2x−40g\left( x \right) = {x^3} - {x^2} - 2x - 40 is the third-degree equation.
Now we find the root of g(x)=0g\left( x \right) = 0
Now we are going to rewrite the g(x)=x3−x2−2x−40g\left( x \right) = {x^3} - {x^2} - 2x - 40 as g(x)=x3+3x2+10x−4x2−12x−40=0g\left( x \right) = {x^3} + 3{x^2} + 10x - 4{x^2} - 12x - 40 = 0
Now take the common terms out, we get
=(x−4)(x2+3x+10)=0= \left( {x - 4} \right)\left( {{x^2} + 3x + 10} \right) = 0
Now we are going to solve the quadratic equation.
Already we know the formula for quadratic formulas.
Apply the values at the formula,
a=1a = 1
b=3 c=1  b = 3 \\\ c = 1 \\\
Therefore,
x=−3±32−4(1)(10)2(1)x = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4\left( 1 \right)\left( {10} \right)} }}{{2\left( 1 \right)}}
Now simplify this we get,
x=−3±9−402x = \dfrac{{ - 3 \pm \sqrt {9 - 40} }}{2}
Hence we get,
x=−3±31i2x = \dfrac{{ - 3 \pm \sqrt {31i} }}{2}
Therefore the solution roots of the given equation are
x=−1,4,−3±31i2x = - 1,4,\dfrac{{ - 3 \pm \sqrt {31i} }}{2}
These are the roots of the given question.
Hence the problem is solved.

Note: In this problem, first, we find the root for the fourth-degree equation by using some inspection. In the next step, we find the root for the third-degree equation by adding and subtracting some terms which do not make any changes in the equation. At last, we find the root for the quadratic equation by using the quadratic formula. This is a method to solve this problem.