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ss7ja [257]
3 years ago
8

Find a degree 4 polynomial having zeros -8,-1,4 and 5 and the coefficient of x^4 equal 1.

Mathematics
1 answer:
Keith_Richards [23]3 years ago
4 0

Answer:

  x^4 -53x^2 +108x +160

Step-by-step explanation:

If <em>a</em> is a zero, then (<em>x-a</em>) is a factor. For the given zeros, the factors are ...

  p(x) = (x +8)(x +1)(x -4)(x -5)

Multiplying these out gives the polynomial in standard form.

  = (x^2 +9x +8)(x^2 -9x +20)

We note that these factors have a sum and difference with the same pair of values, x^2 and 9x. We can use the special form for the product of these to simplify our working out.

  = (x^2 +9x)(x^2 -9x) +20(x^2 +9x) +8(x^2 -9x) +8(20)

  = x^4 -81x^2 +20x^2 +180x +8x^2 -72x +160

  p(x) = x^4 -53x^2 +108x +160

_____

The graph shows this polynomial has the required zeros.

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A park is mapped on a coordinate plane, where C1, C2, C3, and C4 represent chairs and SW1, SW2, and SW3 represent swings. How fa
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Answer:

Option (B)

Step-by-step explanation:

To calculate the distance between C2 and SW1 we will use the formula of distance between two points (x_1,y_1) and (x_2,y_2).

d = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^2 }

Coordinates representing positions of C2 and SW1 are (2, 2) and (-6, -7) respectively.

By substituting these coordinates in the formula,

Distance between these points = \sqrt{(-6-2)^2+(-7-2)^2}

                                                     = \sqrt{(64)+(81)}

                                                     = \sqrt{145} units

Therefore, Option (B) will be the correct option.

5 0
3 years ago
Read 2 more answers
What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove
wolverine [178]

Answer:

The solutions are:

b=0,\:b=4

Step-by-step explanation:

Considering the expression

  • \frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

Solving the expression

\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c

5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2

5b^3-10=6b^3-4b^2-10

\mathrm{Switch\:sides}

6b^3-4b^2-10=5b^3-10

6b^3-4b^2-10+10=5b^3-10+10

6b^3-4b^2=5b^3

\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}

6b^3-4b^2-5b^3=5b^3-5b^3

b^3-4b^2=0

Using\:the\:Zero\:Factor\:Principle: if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

So,

b=0,b-4=0

b=0,b=4

Therefore, the solutions are:

b=0,\:b=4

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3 years ago
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189.271 in expanded form
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Answer:

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Step-by-step explanation: hope this helps <3

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Answer:

Step-by-step explanatio

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