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Nikolay [14]
3 years ago
8

Write a polynomial of degree 5 with zero x=0,i square root 7, -2i

Mathematics
1 answer:
professor190 [17]3 years ago
7 0

Answer:

P(x)=x^5+11x^3+28x

Step-by-step explanation:

<u>Roots of a polynomial</u>

If we know the roots of a polynomial, say x1,x2,x3,...,xn, we can construct the polynomial using the formula

P(x)=a(x-x_1)(x-x_2)(x-x_3)...(x-x_n)

Where a is an arbitrary constant.

We know three of the roots of the degree-5 polynomial are:

x_1=0;\ x_2=\sqrt{7}\boldsymbol{i}:\ x_3=-2\boldsymbol{i}

We can complete the two remaining roots by knowing the complex roots in a polynomial with real coefficients, always come paired with their conjugates. This means that the fourth and fifth roots are:

x_4=-\sqrt{7}\boldsymbol{i}:\ x_3=+2\boldsymbol{i}

Let's build up the polynomial, assuming a=1:

P(x)=(x-0)(x-\sqrt{7}\boldsymbol{i})(x+\sqrt{7}\boldsymbol{i})(x-2\boldsymbol{i})(x+2\boldsymbol{i})

Since:

(a+b\boldsymbol{i})\cdot (a-b\boldsymbol{i})=a^2+b^2

P(x)=(x)(x^2+7)(x^2+4)

Operating the last two factors:

P(x)=(x)(x^4+11x^2+28)

Operating, we have the required polynomial:

\boxed{P(x)=x^5+11x^3+28x}

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Step-by-step explanation:

6 0
3 years ago
The volume of a cylindrical pipe 6' long with an inside diameter of 4' is
malfutka [58]
----------------------------------------
Formula:
----------------------------------------
Volume of a cylindrical pipe = πr²h

----------------------------------------
Find Radius:
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Radius = Diameter ÷ 2 
Radius = 4 ÷ 2 
Radius = 2 in

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Find Volume of the pipe:
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Volume of a cylindrical pipe = π x 2² x 6

Volume of a cylindrical pipe = 24π
 
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3 0
3 years ago
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What is the perimeter of triangle LMN
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4 0
3 years ago
Given: Quadrilateral DEFG is a parallelogram.
levacccp [35]

Answer:

The correct option is A.

Step-by-step explanation:

It is given that  DEFG is a parallelogram.

Draw the diagonals DF and EG. Place point H where DF and EG intersect.

In triangle HGD and HEF

,

∠HGD ≅ ∠HEF                            (Alternate Interior angle)

∠HDG ≅ ∠HFE                      (Alternate Interior angle)

By the definition of a parallelogram, the opposite sides of a parallelogram are congruent.

DG ≅ EF                                      (Opposite sides of parallelogram)

According to ASA postulate, two triangles are congruent if any two angles and their included side are equal in both triangles.

So, by using ASA criterion for congruence we get,

ΔDGH ≅ ΔFEH

Since corresponding sides of congruent triangles are congruent, therefore

GH ≅ EH                      (CPCTC)

DH ≅ FH                     (CPCTC)

Option A is correct.

8 0
4 years ago
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