F(-3) = 2(-3)^2 - (-3) = 2(9)+3 = 18 + 3 = 21
Since both α and β are in the first quadrant, we know each of cos(α), sin(α), cos(β), and sin(β) are positive. So when we invoke the Pythagorean identity,
sin²(x) + cos²(x) = 1
we always take the positive square root when solving for either sin(x) or cos(x).
Given that cos(α) = √11/7 and sin(β) = √11/4, we find
sin(α) = √(1 - cos²(α)) = √38/7
cos(β) = √(1 - sin²(β)) = √5/4
Now, recall the sum identity for cosine,
cos(x + y) = cos(x) cos(y) - sin(x) sin(y)
It follows that
cos(α + β) = √11/7 × √5/4 - √38/7 × √11/4 = (√55 - √418)/28
To factor a trinomial, first we should factor out the GCF, or greatest common factor, shared by all of the coefficients and constants. The greatest common factor of -3, -21, and -30 is -3, therefore, this is the number we must factor out first.
When we factor out the GCF of -3, we get:
-3(q^2 + 7q + 10)
Next, we must look for two numbers that will serve as coefficients for the variables in factored form. These two numbers must add to 7 (the middle term) and multiply to 10 (the constant or last term). The two numbers that satisfy these parameters are 5 and 2.
Therefore, we must use these numbers to create factors of the innermost factor that we have left, to get:
-3(q+5)(q+2)
The above expression is your final answer, because you cannot factor any more. And remember: you can always check to make sure that your factoring is correct by multiplying out the factors and making sure you get your original trinomial.
I hope this helps!
Sadly, the cube you promised us is not shown.
IF the cube is resting on the table in the normal way ... with
two of its faces horizontal and the other four faces vertical,
then any horizontal or vertical slice is a square.
You know what's fun ? Try doing this problem with cubes of
chocolate cake, and after you make slices, look at them and
then eat them.
