Let's analyze the given function step by step to determine its x-intercepts and create a sign chart.

Step 1: Find the x-intercepts

The function is: $f(x) = (x + 1)(2x - 2)(x - 3)(x - 5)$

To find the x-intercepts, set $f(x) = 0$ and solve for $x$:

$(x + 1)(2x - 2)(x - 3)(x - 5) = 0$

Now, solve each factor separately:

1. $x + 1 = 0$ → $x = -1$
2. $2x - 2 = 0$ → $x = 1$ (divide by 2: $x - 1 = 0$)
3. $x - 3 = 0$ → $x = 3$
4. $x - 5 = 0$ → $x = 5$

x-intercepts: $x = -1, 1, 3, 5$

Step 2: Create a sign chart

The sign chart helps determine whether the function is positive or negative in the intervals defined by the x-intercepts.

The x-intercepts divide the x-axis into the following intervals:

• $(-\infty, -1)$
• $(-1, 1)$
• $(1, 3)$
• $(3, 5)$
• $(5, \infty)$

We'll check the sign of $f(x)$ in each interval by testing a point in each region.

1. Interval $(-\infty, -1)$ → Test with $x = -2$:

   • $f(-2) = (-2+1)(2(-2)-2)((-2)-3)((-2)-5) = (-1)(-6)(-5)(-7)$
   • The result is positive (four negatives).

2. Interval $(-1, 1)$ → Test with $x = 0$:

   • $f(0) = (0+1)(2(0)-2)(0-3)(0-5) = (1)(-2)(-3)(-5)$
   • The result is negative (one positive, three negatives).

3. Interval $(1, 3)$ → Test with $x = 2$:

   • $f(2) = (2+1)(2(2)-2)(2-3)(2-5) = (3)(2)(-1)(-3)$
   • The result is positive (two negatives).

4. Interval $(3, 5)$ → Test with $x = 4$:

   • $f(4) = (4+1)(2(4)-2)(4-3)(4-5) = (5)(6)(1)(-1)$
   • The result is negative (one negative).

5. Interval $(5, \infty)$ → Test with $x = 6$:

   • $f(6) = (6+1)(2(6)-2)(6-3)(6-5) = (7)(10)(3)(1)$
   • The result is positive (all positive).

Sign Chart Summary

• $(-\infty, -1)$: $f(x) > 0$
• $(-1, 1)$: $f(x) < 0$
• $(1, 3)$: $f(x) > 0$
• $(3, 5)$: $f(x) < 0$
• $(5, \infty)$: $f(x) > 0$

Step 3: Plot the x-intercepts and sign chart

The x-intercepts are at $x = -1, 1, 3, 5$, and the function changes sign at these points.

Would you like to see the graph of this function and a visual representation of the sign chart?

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5 related questions:

1. How does the behavior of the function near each x-intercept affect the shape of the graph?
2. Can we determine if the function crosses or touches the x-axis at each intercept?
3. How does the degree of the polynomial affect the end behavior of the graph?
4. What impact does the multiplicity of roots have on the sign chart?
5. How would the sign chart change if one of the factors were squared?

Tip: A function changes sign at a simple root (odd multiplicity) but does not change sign at a repeated root (even multiplicity).