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marishachu [46]
3 years ago
7

How many complex zeros does the polynomial function have? F(x)=2x^4 +5x^3 - x^2 +6x-1

Mathematics
1 answer:
klio [65]3 years ago
7 0

Answer:

Two complex roots.

Step-by-step explanation:

F(x)=2x^4 +5x^3 - x^2 +6x-1

is a polynomial in x of degree 4.

Hence F(x) has 4 roots.  There can be 0 or 2 or 4 complex roots to this polynomial since complex roots occur in conjugate pairs.

Use remainder theorem to find the roots of the polynomial.

F(0) = -1 and F(1) = 2+5-1+6-1 = 11>0

There is a change of sign in F from 0 to 1

Thus there is a real root between 0 and 1.

Similarly by trial and error let us find other real root.

F(-3) = -1 and F(-4) = 94

SInce there is a change of sign, from -4 to -3 there exists a real root between -3 and -4.

Other two roots are complex roots since no other place F changes its sign

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How do you factor 2x^3+5y^3
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All you do is...
\mathrm{Apply\:sum\:of\:cubes\:rule:\:}x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)

2x^3+5y^3=\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(\left(\sqrt[3]{2}\right)^2x^2-\sqrt[3]{2}\sqrt[3]{5}xy+\left(\sqrt[3]{5}\right)^2y^2\right)
\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(\left(\sqrt[3]{2}\right)^2x^2-\sqrt[3]{2}\sqrt[3]{5}xy+\left(\sqrt[3]{5}\right)^2y^2\right) \ \textgreater \  Refine

\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(2^{\frac{2}{3}}x^2-\sqrt[3]{10}xy+5^{\frac{2}{3}}y^2\right)

Hope this helps!
4 0
3 years ago
Write 12.375 with number names
BaLLatris [955]
Sixty-two thousand, one hundred thirty-seven.
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3 years ago
45+ 1250 + x = 3049<br><br> What is x?
vovangra [49]

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x is equal to 1754, so it's 1754

4 0
3 years ago
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If tñ=2ñ+3, find S10​
Allisa [31]

Answer:

S_{10}=280

Step-by-step explanation:

Given that,

t_n=2n+3 ...(1)

We need to find the value of S_{10}.

Put n = 1 to find the first term.

t_1=2(1)+3=5

Put n = 2 to find the second term.

t_2=2(2)+3=7

Put n = 3 to find the third term.

t_3=2(3)+3=9

Put n = 10 to find the tenth term.

t_{10}=2(10)+3=23

It means we need to find the sum of 5,7,9,.....,23.

The formula for the sum of n terms is given by :

S_=\dfrac{n}{2}(a+a_n)

We have, n = 10, a = 5 and a_n=23

So,

S=\dfrac{10}{2}(5+23)\\\\S_{10}=10\times 28\\\\=280

So, the value of S_{10} is equal to 280.

3 0
3 years ago
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