7/9 would be irrational, I think.
The value of b to 2.
By doing that you are applying a vertical translation to the function
First, factor out a 3.
3(x² - 9)
In any quadratic ax² + bx + c, we can split the bx term up into two new terms which we want to equal the product of a and c.
In this case, we have x² + 0x - 9. (the 0x is a placeholder)
We want two numbers that add to 0 and multiply to get -9.
Obviously, these numbers are 3 and -3.
Now we have 3(x² + 3x - 3x - 9).
Let's factor.
3(x(x+3)-3(x+3))
<u>3(x-3)(x+3)</u>
There are multiple shortcuts which you could make here, FYI:
Instead of splitting the middle, if your a value is 1, you can go straight to that step (x+number)(x+other number).
Whenever you have a difference of squares, like a²-b², that factors to (a+b)(a-b).
<span>a = 1/2 x b x h
h = 3b
so 1/2 x b x 3b = (1/2)(3b^2) = 150
(150 x 2)/3 = b^2
100 = b^2
b = square root of 100 = 10
h = 3b = 3 x 10 = 30</span>
Answer:
Step-by-step explanation:
hello :
an Degree 3 polynomial with zeros 4, 6, and -2 is :
f(x) = (x-4)(x-6)(x+2)
all polynomial are : a (x-4)(x-6)(x+2) a ≠ 0
