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) = x³ + 8x² + 20x + 16 -2, multiplicity 2, crosses the x - axis; -4, multiplicity 1, touches the x - axis and turns around 2, multiplicity 1, crosses the x - axis; -2, multiplicity 1, crosses the x - axis; -4, multiplicity 1, crosses the x - axis. -2, multiplicity 2, touches the x - axis and turns around; -4, multiplicity 1, crosses the x - axis. 2, multiplicity 1, crosses the x - axis; -2, multiplicity 2, touches the x - axis and turns around; -4, multiplicity 1, crosses the x - axis.

Explanation:

Step1: Test rational roots via Rational Root Theorem

Possible rational roots: $\pm1, \pm2, \pm4, \pm8, \pm16$. Test $x=-2$:
$f(-2) = (-2)^3 + 8(-2)^2 + 20(-2) + 16 = -8 + 32 - 40 + 16 = 0$

Step2: Factor out $(x+2)$ via polynomial division

Divide $x^3 + 8x^2 + 20x + 16$ by $(x+2)$:
$x^3 + 8x^2 + 20x + 16 = (x+2)(x^2 + 6x + 8)$

Step3: Factor the quadratic

Factor $x^2 + 6x + 8$:
$x^2 + 6x + 8 = (x+2)(x+4)$

Step4: Write fully factored form

$f(x) = (x+2)^2(x+4)$

Step5: Identify zeros, multiplicities, and graph behavior

• For $x=-2$: multiplicity 2 (even), so graph touches x-axis and turns around.
• For $x=-4$: multiplicity 1 (odd), so graph crosses x-axis.

Answer:

-2, multiplicity 2, touches the x-axis and turns around;
-4, multiplicity 1, crosses the x-axis.