Such quadratic equation must have factors (x-8) and (x-2) in its left side
(x-8)*(x-2) = 0.
If you make FOIL, you will get this quadratic equation in the STANDARD form
x^2 - 10x + 16 = 0.
The term "zero" of a function is the same as a root.
For real numbers, the root is the x intercept.
The given roots are 8 and 2
x = 8 leads to x-8 = 0, so (x-8) is one factor
x = 2 leads to (x-2) being the other factor
We then need to expand out (x-8)(x-2)
We could use the FOIL rule to expand it out, but I'll use the box method
First place the terms along the left and top edge like so
x
-8
x
-2
Then fill each box with the product of the headers
Example: top left corner is x^2 because x*x = x^2
Another example: bottom right corner is 16 because -2*(-8) = 16
This is what the table would look like when everything is filled in
x
-8
x
x^2
-8x
-2
-2x
16
Then we add up the terms.
Combine like terms if possible
x^2 + (-8x) + (-2x) + 16
x^2 - 10x + 16
Therefore, (x-8)(x-2) = x^2 - 10x + 16
The equation
x^2 - 10x + 16 = 0
leads to the roots x = 8 and x = 2
Other equations lead to these roots because we can scale up or down the equation. For instance, triple both sides to go from x^2-10x+16 = 0 to 3x^2-30x+48 = 0. Both of those equations have the same roots.
Check:
Let's try x = 8
x^2 - 10x + 16 = 0
8^2 - 10*8 + 16 = 0
64 - 10*8 + 16 = 0
64 - 80 + 16 = 0
-16 + 16 = 0
0 = 0
This confirms x = 8 as a root
I'll let you try x = 2.
The goal is to get the left hand side to be zero.
Another way to verify is to use a graphing tool like Desmos as shown here https://www.desmos.com/calculator/gcibt9azif
The parabola has x intercepts 2 and 8 where the graph crosses the x axis.
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