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3 years ago

A polynomial has root that equals 4+17i. Name other of this polynmal

Mathematics

2 answers:

Ilia_Sergeevich [38]3 years ago

8 0

If we know the polynomial has real coefficients, the complex roots come in conjugate pairs so another root is

4 - 17 i

joja [24]3 years ago

4 0

If there is a root with i, there is also a root with -i.

If one root is 4+17i, then another root is 4-17i.

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D. 88

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Anyone knows how to do questions 7 and 8? 15 pts!!

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7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: $2x^2-4x+5$ . With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

$E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3$

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

$2x^2+5=4x\\\\ 2x^2-4x+5=0$

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

$E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)$

Multiplying by (2-1) in the both sides:

$(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ...$ Repeating the process, we obtain: $...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}$

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4.2 divided by 8.19 =

Damm [24]

Well, 4.2 divided by 8.19 = 0.51282051282 But if you switch it to 8.19 divided by 4.2 that would equal 1.95

Hope I helped!

- Debbie <span />

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