Question

find the zeros for the polynomial function and give the multiplicity for each zero. state whether the graph crosses the x - axis, or touches the x - axis and turns around, at each zero.
f(x)=4(x + 8)(x - 5)^2
the zeros are. (use a comma to separate answers as needed.)

Explanation:

Step1: Set the function equal to zero

$4(x + 8)(x - 5)^2=0$

Step2: Use the zero - product property

If $ab = 0$, then $a = 0$ or $b = 0$. So $x+8 = 0$ or $(x - 5)^2=0$.

Step3: Solve for $x$ in $x+8 = 0$

$x=-8$

Step4: Solve for $x$ in $(x - 5)^2=0$

Taking the square - root of both sides, we get $x - 5=0$, so $x = 5$.

The multiplicity of a zero is the exponent of the corresponding factor. For the factor $(x + 8)$, the exponent is 1, so the zero $x=-8$ has multiplicity 1. For the factor $(x - 5)^2$, the zero $x = 5$ has multiplicity 2.

If the multiplicity of a zero is odd, the graph of the function crosses the $x$-axis at that zero. If the multiplicity is even, the graph of the function touches the $x$-axis and turns around at that zero.

Answer:

The zeros are $-8,5$. The zero $x=-8$ has multiplicity 1 and the graph crosses the $x$-axis at $x=-8$. The zero $x = 5$ has multiplicity 2 and the graph touches the $x$-axis and turns around at $x = 5$.