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2 years ago

9

Find a degree 3 polynomial having zeros -5, 1 and 5 and the coefficient of x^3 equal 1

1 answer:

mote1985 [20]2 years ago

7 0

Answer:

$x^3 -x^2 -25x +25$

Step-by-step explanation:

$\text{The roots are,}~ \alpha = -5,~ \beta = 1 ~ \text{and}~ \gamma = 5\\\\\text{The polynomial is,}\\\\~~~x^3 - (\alpha + \beta + \gamma)x^2+(\alpha \beta + \beta \gamma+ \gamma \alpha)x - \alpha\beta \gamma\\\\=x^3 - (-5+1+5)x^2 +(-5 \cdot 1 + 1 \cdot 5 + 5 \cdot -5)x - (-5)(1)(5)\\\\=x^3 -(0+1)x^2 + (-5+5-25)x +25\\\\=x^3 -x^2 -25x +25$

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Rina8888 [55]

Answer:

y= -7

Step-by-step explanation:

So first distribute the 8

4=8y+32-4y

now combine like terms (the -4y+8y)

4=32+4y

Subtract 32 on both sides to isolate 4y

-28=4y

now divide 4 on both sides to isolate Y

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so y= -7

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3 years ago

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3 years ago

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3 years ago

Each child in the Martin family decided to save money for the family vacation.

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Answer:

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Step-by-step explanation:

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2 years ago

Generate the nest three terms of each arithmetic sequence shown below.

o-na [289]

Answer:

A)2,6,10

B)2,-6,-18

C)-1,-3,-5

Step-by-step explanation:

<u>A)a1=-2 and d=4</u>

We know that the arithmetic sequence formula is

$a_{n}=a_{1}+(n-1)d$

Now

$a_{2}=a_{1}+(2-1)d$

$a_{2}=a_{1}+(1)d$

Substituting the given value we get

$a_{2}= -2+(1)4$

$a_{2}= -2+4$

$a_{2}= 2$

------------------------------------------

Similarly

$a_{3}=a_{1}+(3-1)d$

$a_{3}=a_{1}+(2)d$

Substituting the given value we get

$a_{3}= -2+(2)4$

$a_{3}= -2+8$

$a_{3}= 6$

-------------------------------------------------

$a_{4}=a_{1}+(4-1)d$

$a_{4}=a_{1}+(3)d$

Substituting the given value we get

$a_{4}= -2+(3)4$

$a_{4}= -2+16$

$a_{4}= 10$

-------------------------------------------------------------------------------------------

<u>B</u><u> $a_n=a_{(n-1)}-8$ with $a_1=10$ </u>

$a_2=a_{(2-1)}-8$

$a_2=a_{1}-8$

Substituting the given value

$a_2= 10-8$

$a_2=2$

---------------------------------------------------------------------

$a_3=a_{(3-1)}-8$

$a_3=a_{2}-8$

Substituting the value

$a_3=2-8$

$a_3= -6$

---------------------------------------------------------------------

$a_4=a_{(4-1)}-8$

$a_4=a_{3}-8$

Substituting the value

$a_4= -6-8$

$a_4= -14$

-------------------------------------------------------------------------------------------

<u>C) $a_1=3, a_2=1$ </u>

Here the difference is -2

the arithmetic sequence formula is

$a_{n}=a_{1}+(n-1)d$

Now

$a_{3}=a_{1}+(3-1)d$

$a_{3}=a_{1}+(2)d$

Substituting the value we get

$a_{3}= 3+(2)-2$

$a_{3}= 3-4$

$a_{3}= -1$

------------------------------------------------------------------------------

$a_{4}=a_{1}+(4-1)d$

$a_{4}=a_{1}+(3)d$

Substituting the value we get

$a_{4}= 3+(3)-2$

$a_{4}= 3-6$

$a_{4}= -3$

----------------------------------------------------------------------------------

$a_{5}=a_{1}+(5-1)d$

$a_{5}=a_{1}+(4)d$

Substituting the value we get

$a_{5}= 3+(4)-2$

$a_{5}= 3-8$

$a_{5}= -5$

4 0

3 years ago