Question

(4x^2-4x+1)(x^2+6x+5)《0

Answer 1

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Answer 2

To determine the intervals where the expression $(4x^2-4x+1)(x^2+6x+5)$ is less than zero ($< 0$), we need to find the values of $x$ that make the expression negative.

Step 1: Factorize the quadratic expressions. $(4x^2-4x+1) = (2x-1)^2$ (using the perfect square trinomial formula) $(x^2+6x+5) = (x+1)(x+5)$

Step 2: Set each factor equal to zero and solve for $x$:

From $2x-1 = 0$, we get $x = \frac{1}{2}$. From $x+1 = 0$, we get $x = -1$. From $x+5 = 0$, we get $x = -5$.

Step 3: Create a sign chart to analyze the expression's sign in different intervals.

Interval 1: $x < -5$ In this interval, both factors $(2x-1)^2$ and $(x+1)(x+5)$ are negative since $x+1$ and $x+5$ are both negative when $x < -5$. The product of two negatives is positive.

Interval 2: $-5 < x < \frac{1}{2}$ In this interval, $(2x-1)^2$ is positive (a square of a real number is always positive), and $(x+1)(x+5)$ is still negative since $x+1$ is negative while $x+5$ is positive. The product of a positive number and a negative number is negative.

Interval 3: $x > \frac{1}{2}$ In this interval, both factors $(2x-1)^2$ and $(x+1)(x+5)$ are positive since both $2x-1$ and $x+1$ and $x+5$ are positive when $x > \frac{1}{2}$. The product of two positives is positive.

Step 4: Determine the solution based on the sign chart. The expression is less than zero ($<0$) in the interval $-5 < x < \frac{1}{2}$.

So, the solution to the inequality $(4x^2-4x+1)(x^2+6x+5) < 0$ is $x \in (-5, \frac{1}{2})$.