The rational roots of the equation [tex]{x^4} + 8{x^3} + 7{x^2} - 40x - 60 = 0[/tex] is [tex]\boxed{{\mathbf{x = - 2, - 6 }}}[/tex] .
Further explanation:
It is given that the equation is [tex]{x^4} + 8{x^3} + 7{x^2} - 40x - 60 = 0[/tex].
Substitute [tex]-2[/tex] for [tex]x[/tex] in the above equation to check whether it satisfy the equation or not.
[tex]\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}[/tex]
Therefore, [tex]x = - 2[/tex] satisfies the equation so we simplify the given equation in the factor of 2.
Now, the simplification of the given equation is as follows:
[tex]\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}[/tex]
Now, make groups of the common term in the above equation as follows:
[tex]\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}[/tex]
Now, [tex]\left({x + 2}\right)[/tex] is common term in the above equation as follows:
[tex]\left({x + 2}\right)\left({{x^3} + 6{x^2} - 5x - 30}\right)=0{\text{}}[/tex] ......(1)
Simplify the term [tex]\left({{x^3} + 6{x^2} - 5x - 30}\right)[/tex] as follows:
[tex]\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}[/tex]
Therefore, the simplification of the term [tex]{x^3} + 6{x^2} - 5x - 30[/tex] is [tex]\left({x + 6}\right)\left({{x^2} - 5}\right)[/tex].
Substitute [tex]\left({x + 6}\right)\left({{x^2} - 5} \right)[/tex] for [tex]{x^3} + 6{x^2} - 5x - 30[/tex] in equation (1) as follows:
[tex]\left({x + 2}\right)\left({x + 6}\right)\left({{x^2} - 5}\right)=0[/tex]
Now, simplify the above expression to obtain the value of [tex]x[/tex] as follows:
[tex]\begin{aligned}\left({x + 2}\right)&=0\\x&= - 2\\\end{aligned}[/tex]
Therefore, the value of [tex]x[/tex] is [tex]-2[/tex].
[tex]\begin{aligned}\left({x + 6}\right)&=0\\x&= - 6\\\end{aligned}[/tex]
Therefore, the value of [tex]x[/tex] is [tex]-2[/tex].
[tex]\begin{aligned}\left({{x^2} - 5}\right)&=0\\{x^2}&=5\\x&=\pm\sqrt5\\\end{aligned}[/tex]
Therefore, the value of [tex]x[/tex] is [tex]\pm \sqrt5[/tex].
Here, [tex]\pm \sqrt 5[/tex] is not a rational number.
Since [tex]\pm\sqrt5[/tex] is an irrational number, it will not be considered the obtained solution ofthegiven equation.
Therefore, the other two values of [tex]x[/tex] that is [tex]-2[/tex] and [tex]-6[/tex] is considered the answer of the equation because they are rational numbers.
Thus, the rational roots of the equation [tex]{x^4} + 8{x^3} + 7{x^2} - 40x - 60 = 0[/tex] is [tex]\boxed{{\mathbf{x = - 2, - 6 }}}[/tex] .
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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, [tex]{x^4} + 8{x^3} + 7{x^2} - 40x - 60 = 0[/tex] , value of [tex]x[/tex]
