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Math  /  Algebra

Question5. The number of real solutions for x3+x2+x+1x^{3}+x^{2}+x+1 is

Studdy Solution

STEP 1

1. We are tasked with finding the number of real solutions for the polynomial equation x3+x2+x+1=0 x^3 + x^2 + x + 1 = 0 .
2. The polynomial is a cubic equation, which means it can have up to three real roots.
3. We will analyze the behavior of the polynomial to determine the number of real roots.

STEP 2

1. Analyze the polynomial for any obvious roots.
2. Use the derivative to find critical points.
3. Evaluate the polynomial at critical points to determine the behavior of the function.
4. Use the Intermediate Value Theorem to determine the number of real roots.

STEP 3

Check for any obvious roots by substituting small integer values into the polynomial x3+x2+x+1=0 x^3 + x^2 + x + 1 = 0 .
Try x=−1 x = -1 :
(−1)3+(−1)2+(−1)+1=−1+1−1+1=0 (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0
Thus, x=−1 x = -1 is a root.

STEP 4

Find the derivative of the polynomial to locate critical points:
f(x)=x3+x2+x+1 f(x) = x^3 + x^2 + x + 1 f′(x)=3x2+2x+1 f'(x) = 3x^2 + 2x + 1
Set the derivative equal to zero to find critical points:
3x2+2x+1=0 3x^2 + 2x + 1 = 0

STEP 5

Solve the quadratic equation 3x2+2x+1=0 3x^2 + 2x + 1 = 0 using the quadratic formula:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=3 a = 3 , b=2 b = 2 , and c=1 c = 1 .
x=−2±22−4⋅3⋅12⋅3 x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} x=−2±4−126 x = \frac{-2 \pm \sqrt{4 - 12}}{6} x=−2±−86 x = \frac{-2 \pm \sqrt{-8}}{6}
Since the discriminant is negative, there are no real critical points.

STEP 6

Since there are no real critical points, the polynomial x3+x2+x+1 x^3 + x^2 + x + 1 is either always increasing or always decreasing. We already found one real root at x=−1 x = -1 .

STEP 7

Use the Intermediate Value Theorem to check for other roots. Since the polynomial is continuous and we found one root at x=−1 x = -1 , we need to check the behavior of the polynomial as x→±∞ x \to \pm \infty .
As x→∞ x \to \infty , x3+x2+x+1→∞ x^3 + x^2 + x + 1 \to \infty .
As x→−∞ x \to -\infty , x3+x2+x+1→−∞ x^3 + x^2 + x + 1 \to -\infty .
Since the polynomial changes sign and is continuous, there must be exactly one real root.
The number of real solutions is 1 \boxed{1} .

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