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Consider the polynomial function [tex]f(x)=(x-1)^2(x+3)^3(x+1)[/tex].

Use the equation to complete each statement about this function.

The zero located at [tex]x=1[/tex] has a multiplicity of [tex]$\square$[/tex]. The zero located at [tex]x=-3[/tex] has a multiplicity of [tex]$\square$[/tex].

The graph of the function will touch, but not cross, the [tex]$x$[/tex]-axis at an [tex]$x$[/tex]-value of [tex]$\square$[/tex].

Sagot :

GinnyAnswer
To solve this problem, we analyze the given polynomial function [tex]\( f(x) = (x-1)^2(x+3)^3(x+1) \)[/tex].

1. Zero at [tex]\( x = 1 \)[/tex] and its Multiplicity:
- The term [tex]\((x-1)^2\)[/tex] indicates that [tex]\( x = 1 \)[/tex] is a root of the polynomial.
- The exponent 2 tells us the multiplicity of this root.
- Thus, the zero at [tex]\( x = 1 \)[/tex] has a multiplicity of 2.

2. Zero at [tex]\( x = -3 \)[/tex] and its Multiplicity:
- The term [tex]\((x+3)^3\)[/tex] shows that [tex]\( x = -3 \)[/tex] is a root of the polynomial.
- The exponent 3 gives the multiplicity of this root.
- Therefore, the zero at [tex]\( x = -3 \)[/tex] has a multiplicity of 3.

3. Behavior of the Graph at [tex]\( x = 1 \)[/tex]:
- A root with an even multiplicity (like 2) causes the graph to touch the x-axis at that point but not cross it.
- Hence, at [tex]\( x = 1 \)[/tex], the graph will touch, but not cross, the x-axis.

Now, let's complete the statements:

- The zero located at [tex]\( x = 1 \)[/tex] has a multiplicity of 2.
- The zero located at [tex]\( x = -3 \)[/tex] has a multiplicity of 3.
- The graph of the function will touch, but not cross, the x-axis at an [tex]\( x \)[/tex]-value of 1.
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