4n - 6m = -2
Let n = -2
4(-2) - 6m = -2
-8 - 6m = -2
-6m = -2 +8
-6m = 6
m = 6/-6
m = -1
So, when n is -2, m is -1.
Now, do the same when n is -1 and again when n is 0.
Answer:
y=-(6/4)x
so if x=4 the equation will be
y=-(6/4) *-4
y=6
These are both rational numbers, because they are (our could be written as) a fraction. 1.45 = 145/100
There is no smaller subset of real numbers that works, because the next smaller subset is integers, but these aren’t integers.
Answer:
1/2, 3
Step-by-step explanation:
This is a pretty involved problem, so I'm going to start by laying out two facts that our going to help us get there.
- The Fundamental Theorem of Algebra tells us that any polynomial has <em>as many zeroes as its degree</em>. Our function f(x) has a degree of 4, so we'll have 4 zeroes. Also,
- Complex zeroes come in pairs. Specifically, they come in <em>conjugate pairs</em>. If -2i is a zero, 2i must be a zero, too. The "why" is beyond the scope of this response, but this result is called the "complex conjugate root theorem".
In 2., I mentioned that both -2i and 2i must be zeroes of f(x). This means that both
and
are factors of f(x), and furthermore, their product,
, is <em>also</em> a factor. To see what's left after we factor out that product, we can use polynomial long division to find that
![2x^4+5x^3+5x^2+20x-12=(x^2+4)(2x^2+5x-3)](https://tex.z-dn.net/?f=2x%5E4%2B5x%5E3%2B5x%5E2%2B20x-12%3D%28x%5E2%2B4%29%282x%5E2%2B5x-3%29)
I'll go through to steps to factor that second expression below:
![2x^2+5x-3=2x^2+6x-x-3\\=2x(x+3)-(x+3)\\=(2x-1)(x+3)](https://tex.z-dn.net/?f=2x%5E2%2B5x-3%3D2x%5E2%2B6x-x-3%5C%5C%3D2x%28x%2B3%29-%28x%2B3%29%5C%5C%3D%282x-1%29%28x%2B3%29)
Solving both of the expressions when f(x) = 0 gets us our final two zeroes:
![2x-1=0\\2x=1\\x=1/2](https://tex.z-dn.net/?f=2x-1%3D0%5C%5C2x%3D1%5C%5Cx%3D1%2F2)
![x+3=0\\x=-3](https://tex.z-dn.net/?f=x%2B3%3D0%5C%5Cx%3D-3)
So, the remaining zeroes are 1/2 and 3.