Answer :
We want to solve the equation
[tex]$$
x^4 + 10x^3 + 24x^2 - 19x - 70 = 0.
$$[/tex]
A good strategy is to look for easy-to-test roots that let us factor the polynomial step-by-step.
1. Testing for a Factor of the Form [tex]\((x + 2)\)[/tex]:
Test [tex]\(x = -2\)[/tex]:
[tex]$$
(-2)^4 + 10(-2)^3 + 24(-2)^2 - 19(-2) - 70 = 16 - 80 + 96 + 38 - 70 = 0.
$$[/tex]
Since the polynomial evaluates to zero, [tex]\(x = -2\)[/tex] is a root, meaning [tex]\((x + 2)\)[/tex] is a factor.
2. Dividing by [tex]\((x + 2)\)[/tex]:
Dividing the original polynomial by [tex]\((x + 2)\)[/tex] gives a cubic polynomial:
[tex]$$
x^3 + 8x^2 + 8x - 35.
$$[/tex]
3. Testing for a Factor of the New Cubic:
Next, test [tex]\(x = -5\)[/tex] in the cubic:
[tex]$$
(-5)^3 + 8(-5)^2 + 8(-5) - 35 = -125 + 200 - 40 - 35 = 0.
$$[/tex]
With the result zero, [tex]\(x = -5\)[/tex] is a root, and [tex]\((x + 5)\)[/tex] is a factor of the cubic.
4. Dividing the Cubic by [tex]\((x + 5)\)[/tex]:
Dividing [tex]\(x^3 + 8x^2 + 8x - 35\)[/tex] by [tex]\((x + 5)\)[/tex] leaves a quadratic:
[tex]$$
x^2 + 3x - 7.
$$[/tex]
5. Solving the Quadratic Equation:
To find the remaining roots, solve the quadratic
[tex]$$
x^2 + 3x - 7 = 0.
$$[/tex]
Using the quadratic formula
[tex]$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$[/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = -7\)[/tex], we obtain
[tex]$$
x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-7)}}{2} = \frac{-3 \pm \sqrt{9 + 28}}{2} = \frac{-3 \pm \sqrt{37}}{2}.
$$[/tex]
6. Gathering All the Roots:
Combining our results, the four real solutions are:
[tex]$$
x = -2, \quad x = -5, \quad x = \frac{-3 + \sqrt{37}}{2}, \quad \text{and} \quad x = \frac{-3 - \sqrt{37}}{2}.
$$[/tex]
Thus, the real solutions to the equation are
[tex]$$
-2,\, -5,\, \frac{-3 + \sqrt{37}}{2},\, \frac{-3 - \sqrt{37}}{2}.
$$[/tex]
[tex]$$
x^4 + 10x^3 + 24x^2 - 19x - 70 = 0.
$$[/tex]
A good strategy is to look for easy-to-test roots that let us factor the polynomial step-by-step.
1. Testing for a Factor of the Form [tex]\((x + 2)\)[/tex]:
Test [tex]\(x = -2\)[/tex]:
[tex]$$
(-2)^4 + 10(-2)^3 + 24(-2)^2 - 19(-2) - 70 = 16 - 80 + 96 + 38 - 70 = 0.
$$[/tex]
Since the polynomial evaluates to zero, [tex]\(x = -2\)[/tex] is a root, meaning [tex]\((x + 2)\)[/tex] is a factor.
2. Dividing by [tex]\((x + 2)\)[/tex]:
Dividing the original polynomial by [tex]\((x + 2)\)[/tex] gives a cubic polynomial:
[tex]$$
x^3 + 8x^2 + 8x - 35.
$$[/tex]
3. Testing for a Factor of the New Cubic:
Next, test [tex]\(x = -5\)[/tex] in the cubic:
[tex]$$
(-5)^3 + 8(-5)^2 + 8(-5) - 35 = -125 + 200 - 40 - 35 = 0.
$$[/tex]
With the result zero, [tex]\(x = -5\)[/tex] is a root, and [tex]\((x + 5)\)[/tex] is a factor of the cubic.
4. Dividing the Cubic by [tex]\((x + 5)\)[/tex]:
Dividing [tex]\(x^3 + 8x^2 + 8x - 35\)[/tex] by [tex]\((x + 5)\)[/tex] leaves a quadratic:
[tex]$$
x^2 + 3x - 7.
$$[/tex]
5. Solving the Quadratic Equation:
To find the remaining roots, solve the quadratic
[tex]$$
x^2 + 3x - 7 = 0.
$$[/tex]
Using the quadratic formula
[tex]$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$[/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = -7\)[/tex], we obtain
[tex]$$
x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-7)}}{2} = \frac{-3 \pm \sqrt{9 + 28}}{2} = \frac{-3 \pm \sqrt{37}}{2}.
$$[/tex]
6. Gathering All the Roots:
Combining our results, the four real solutions are:
[tex]$$
x = -2, \quad x = -5, \quad x = \frac{-3 + \sqrt{37}}{2}, \quad \text{and} \quad x = \frac{-3 - \sqrt{37}}{2}.
$$[/tex]
Thus, the real solutions to the equation are
[tex]$$
-2,\, -5,\, \frac{-3 + \sqrt{37}}{2},\, \frac{-3 - \sqrt{37}}{2}.
$$[/tex]
