Two consecutive odd integers such that 4 times the larger is 29 more than 3 times the smaller
1 answer:
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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
I'll go through to steps to factor that second expression below:
Solving both of the expressions when f(x) = 0 gets us our final two zeroes:
So, the remaining zeroes are 1/2 and 3.