Question

Find the rational roots of x4 + 8x3 + 7x2 – 40x – 60 = 0 (1 point)2, 6. –6, –2. –2, 6. –6, 2.

Answer 1

If you would like to find the rational roots of x^4 + 8 * x^3 + 7 * x^2 - 40 * x - 60 = 0, you can do this using the following steps:

x^4 + 8 * x^3 + 7 * x^2 - 40 * x - 60 = 0
(x^2 - 5) * (x^2 + 8 * x + 12) = 0
(x^2 - 5) * (x + 6) * (x + 2) = 0
1. x = - sqrt(5)
2. x = sqrt(5)
3. x = - 6
4. x = - 2

The correct result would be -6 and -2.

Answer 2

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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3. What are the values of x?brainly.com/question/2093003

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 $x$