Fundamental Theorem of Algebra...

Mathematics

1 answer:

Fittoniya [83]3 years ago

4 0

Answer:

A real root of fifth-grade multiplicity/No complex roots.

Step-by-step explanation:

The Fundamental Theorem of Algebra states that every polynomial with real coefficients and a grade greater than zero has at least a real root. Let be $f(x) = (x+7)^{5}$ , if such expression is equalized to zero and handled algebraically:

1) $(x+7)^{5} = 0$ Given.

2) $(x+7)\cdot (x+7)\cdot (x+7)\cdot (x+7)\cdot (x+7) = 0$ Definition of power.

3) $x+7=0$ Given.

4) $x = -7$ Compatibility with the addition/Existence of the additive inverse/Modulative property/Result.

This expression has a real root of fifth-grade multiplicity. No complex roots.

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50+51+52+...+101=...<br>Can you help me? I do not understand

Dmitriy789 [7]

Suppose

$S=50+51+52+\cdots+100+101$

At the same time, we can write

$S^*=101+100+99+\cdots+51+50$

Note that $S=S^*$ (just reverse the sum). Let's pair the first terms of $S$ and $S^*$ , and the second, and the third, and so on:

$S+S^*=(50+101)+(51+100)+(52+99)+\cdots+(100+51)+(101+50)$

Now, each grouped term in the sum on the right side adds to 151. There are 52 grouped terms on that same side (because there are 50 numbers in the range of integers 51-100, plus 50 and 101), which menas

$S+S^*=52\cdot151$