Question

8-15. what are the roots (the solutions when y = 0) of each of the following polynomial functions?
a. $y = x^2 - 6x + 8$
hint (a):
factor the right side and use the zero product property
answer (a):
$x = 2, x = 4$
b. $f(x) = x^2 - 6x + 9$
hint (b):
the roots are the x - intercepts on the graph of the function.
answer (b):
$x = 3$
c. $y = x^2 - 4x$
hint (c):
$x$ is a common factor.
more help (c):
see part (a).

Explanation:

Step1: Set y=0, factor quadratic

Set $y=0$: $x^2 - 6x + 8 = 0$
Factor: $(x-2)(x-4)=0$

Step2: Apply Zero Product Property

Solve $x-2=0$ and $x-4=0$:
$x=2$; $x=4$

Step3: Set f(x)=0, factor quadratic

Set $f(x)=0$: $x^2 - 6x + 9 = 0$
Factor: $(x-3)^2=0$

Step4: Solve for x

Solve $(x-3)^2=0$:
$x=3$ (repeated root)

Step5: Set y=0, factor polynomial

Set $y=0$: $x^3 - 4x = 0$
Factor out $x$: $x(x^2-4)=0$
Factor difference of squares: $x(x-2)(x+2)=0$

Step6: Apply Zero Product Property

Solve $x=0$, $x-2=0$, $x+2=0$:
$x=0$; $x=2$; $x=-2$

Answer:

a. $x=2$, $x=4$
b. $x=3$
c. $x=-2$, $x=0$, $x=2$