The rational roots of the equation \({x^4} + 8{x^3} + 7{x^2} - 40x - 60 = 0\) is \(\boxed{{\mathbf{x = - 2, - 6 }}}\) .
Further explanation:
It is given that the equation is \({x^4} + 8{x^3} + 7{x^2} - 40x - 60 = 0\).
Substitute \(-2\) for \(x\) in the above equation to check whether it satisfy the equation or not.
\(\begin{aligned}{\left({ - 2}\right)^4} + 8{\left({ - 2} \right)^3} + 7{\left({ - 2}\right)^2} - 40\left({ - 2}\right) - 60\mathop&=\limits^?0\hfill\\16 - 64 + 28 + 80 - 60\mathop&=\limits^?0\hfill\\124 - 124\mathop&=\limits^? 0\hfill\\\end{aligned}\)
Therefore, \(x = - 2\) satisfies the equation so we simplify the given equation in the factor of 2.
Now, the simplification of the given equation is as follows:
\(\begin{aligned}{x^4} + 8{x^3} + 7{x^2} - 40x - 60&= 0\hfill\\{x^4} + \left({2 + 6}\right){x^3}+\left({12 - 5}\right){x^2} - \left({10 + 30}\right)x - 60&=0\hfill\\{x^4} + 2{x^3} + 6{x^3} + 12{x^2} - 5{x^2} - 10x - 30x - 60&= 0\hfill\\\end{aligned}\)
Now, make groups of the common term in the above equation as follows:
\(\begin{aligned}{x^4} + 2{x^3} + 6{x^3} + 12{x^2} - 5{x^2} - 10x - 30x - 60&= 0 \hfill \\{x^3}\left({x + 2}\right) + 6{x^2}\left({x + 2} \right) - 5x\left( {x + 2} \right)30\left({x + 2}\right)&=0\hfill\\\end{aligned}\)
Now, \(\left({x + 2}\right)\) is common term in the above equation as follows:
\(\left({x + 2}\right)\left({{x^3} + 6{x^2} - 5x - 30}\right)=0{\text{}}\) ......(1)
Simplify the term \(\left({{x^3} + 6{x^2} - 5x - 30}\right)\) as follows:
\(\begin{aligned}{x^3} + 6{x^2} - 5x - 30\hfill\\{x^2}\left({x + 6} \right) - 5\left({x + 6}\right)\hfill\\\left({x + 6}\right)\left({{x^2} - 5}\right)\hfill\\\end{aligned}\)
Therefore, the simplification of the term \({x^3} + 6{x^2} - 5x - 30\) is \(\left({x + 6}\right)\left({{x^2} - 5}\right)\).
Substitute \(\left({x + 6}\right)\left({{x^2} - 5} \right)\) for \({x^3} + 6{x^2} - 5x - 30\) in equation (1) as follows:
\(\left({x + 2}\right)\left({x + 6}\right)\left({{x^2} - 5}\right)=0\)
Now, simplify the above expression to obtain the value of \(x\) as follows:
\(\begin{aligned}\left({x + 2}\right)&=0\\x&= - 2\\\end{aligned}\)
Therefore, the value of \(x\) is \(-2\).
\(\begin{aligned}\left({x + 6}\right)&=0\\x&= - 6\\\end{aligned}\)
Therefore, the value of \(x\) is \(-2\).
\(\begin{aligned}\left({{x^2} - 5}\right)&=0\\{x^2}&=5\\x&=\pm\sqrt5\\\end{aligned}\)
Therefore, the value of \(x\) is \(\pm \sqrt5\).
Here, \(\pm \sqrt 5\) is not a rational number.
Since \(\pm\sqrt5\) is an irrational number, it will not be considered the obtained solution ofthegiven equation.
Therefore, the other two values of \(x\) that is \(-2\) and \(-6\) is considered the answer of the equation because they are rational numbers.
Thus, the rational roots of the equation \({x^4} + 8{x^3} + 7{x^2} - 40x - 60 = 0\) is \(\boxed{{\mathbf{x = - 2, - 6 }}}\) .
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Answer Details
Grade: Junior High School
Subject: Mathematics
Chapter: Coordinate Geometry
Keywords:Coordinate Geometry, linear equation, roots, variables, mathematics,equation of line, \({x^4} + 8{x^3} + 7{x^2} - 40x - 60 = 0\) , value of \(x\)
